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Functions
In this functions activity, 11th graders solve 10 different problems that include various functions of domain. First, they determine the domain of a given function. Then, students determine the range of another function. They also determine the description that best describes each function. | 677.169 | 1 |
College Algebra with Trigonometry – Barnett, Ziegler – 9th Edition
Barnett, Ziegler, Byleen, and Sobecki's College Algebra with Trigonometry text is designed to be user friendly and to maximize student comprehension by emphasizing computational skills, ideas, and problem solving as opposed to mathematical theory.
The large number of pedagogical devices employed in this text will guide a student through the course. Integrated throughout the text, students and instructors will find Explore-Discuss boxes which encourage students to think critically about mathematical concepts.
In each section, the worked examples are followed by matched problems that reinforce the concept being taught. In addition, the text contains an abundance of exercises and applications that will convince students that math is useful. A MathZone site featuring algorithmic exercises, videos, and other resources accompanies the text | 677.169 | 1 |
This unique textbook focuses on the structure of fields and is intended for a second course in abstract algebra. Besides providing proofs of the transcendence of pi and e, the book includes material on differential Galois groups and a proof of Hilbert's irreducibility theorem. The reader will hear about equations, both polynomial and differential, and about the algebraic structure of their solutions. In explaining these concepts, the author also provides comments on their historical development and leads the reader along many interesting paths. In addition, there are theorems from analysis: as stated before, the transcendence of the numbers pi and e, the fact that the complex numbers form an algebraically closed field, and also Puiseux's theorem that shows how one can parametrize the roots of polynomial equations, the coefficients of which are allowed to vary. There are exercises at the end of each chapter, varying in degree from easy to difficult. To make the book more lively, the author has incorporated pictures from the history of mathematics, including scans of mathematical stamps and pictures of mathematicians. Antoine Chambert-Loir taught this book when he was Professor at École Polytechnique, Palaiseau, France. He is now Professor at Université de Rennes 1. | 677.169 | 1 |
Summary and Info
''This book illustrates theories and associated mathematical expressions with numerical examples using various methods, leading to exact solutions, more accurate results, and more computationally efficient techniques. It presents the derivations of the equations of motion for all structure foundations using either the continuous model or the discrete model. It discusses applications for students taking courses including vibration mechanics, dynamics of structures, and finite element analyses of structures, the transfer matrix method, and Jacobi method'' | 677.169 | 1 |
This book presents activities designed to promote students' understanding of important mathematical ideas as well as to provide learning experiences intended to let students practice and refine their skills. Chapter topics include: (1) understanding and using numbers; (2) computation; (3) problem solving; (4) calculators; (5) fractions and decimals; (6) graphs and statistics; (7) geometry; and (8) measurement skills. Teaching materials, many including line drawings, are provided for each chapter. Each of the activities contains objectives, a list of materials, preparation, time frame, procedures, evaluation and an extension activity. (YP) | 677.169 | 1 |
10.1 Polar Coordinates The origin for rectangular coordinates becomes the pole in the for polar coordinates, the positive x-axis becomes the polar axis, and (x, y) becomes (r, ). In this system, r > 0is the distance from the pole to a point and is the ang
MAC1147
12.1 Sequences
A sequence is a function whose domain is the set of positive integers (natural numbers); that is, the terms are found by calculating f(1), f(2), f(3) . f(n). Subscripted letters are usually used to represent the terms; e.g. a1 = the
uu ur A vector is a quantity that has both magnitude and direction. This is related to a directed line segment PQ from P (the initial point) to Q (the terminal point), in which the distance between the two points is the magnitude of the vector. Two vector
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Professor Remesar is a great Precalc and Trig professor! He teaches the concepts, and problems extremely thoroughly and clearly. You learn a lot in his class as it is a course compiled with two, but you learn it well. He is great! Highly recommended.
Course highlights:
Professor Remesar teaches math in a different light. You can learn it much more implicitly and understand it clearly. He also offers a lot of help.
Hours per week:
3-5 hours
Advice for students:
Pay attention in his class, ask questions, and do the reviews and math homework he hands before the test! They are beneficial!
Course Term:Fall 2016
Professor:Remesar Jose
Course Required?Yes
Course Tags:Math-heavy
Apr 12, 2016
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
The teacher is very helpful when questions are asked and will help you understand the material | 677.169 | 1 |
Algebraic Expressions: Application to Physical Objects
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2.97 MB | 9 pages
PRODUCT DESCRIPTION
Algebraic Expressions such as 2x + 3, x-y, 2m - 4n and others are like alien or foreign language to many students. They do not fully comprehend the meaning behind them.
This lesson and worksheet combined together shows the representation of Algebraic Expressions using physical objects, thus, making it more integated to the real world. It aims to create awareness on the part of the students about the existence of algebraic expression in the real world.Moreover, it has a lot of drawing tasks which students will enjoy as they do the worksheet.
The lesson and worksheet is divided into several pages:
Page 1:Motivation Part of the Lesson and 1st Example on how to represent Algebraic Expressions through objects.
Page 2: It shows a 2nd Example on how to represent Algebraic Expressions through objects.An exercise problem follows.
Pages 3, 4 and 5: Worksheet
Pages 6,7,8 and 9: Answer Key
For questions or comments; please e-mail me: thematrixtrilogy123@gmail.com
Note: When you print it as a pdf, click the "FIT" which will appear when you print. Algebraic Expressions: Application to Physical Objects by Nelson D. Ocampo is licensed under a Creative Commons Attribution | 677.169 | 1 |
Why is there a giant gap in the completion schedule?
I'm expecting a baby in October and need to plan accordingly :)
What grade level(s) is this curriculum designed for?
This curriculum is intended for 8th grade students or advanced 7th grade students. It will work well for anyone taking Algebra 1 the following year. This curriculum is purposely written to be challenging and better prepare students for Algebra 1!
How do I find out about and access new materials as they are posted?
Continue to follow me to get the notification emails of new products. Then, simply re-download the bundle and copy over the new items.
This purchase is for one teacher only. This resource is not to be shared with colleagues or used by an entire grade level, school, or district without purchasing the proper number of licenses. Licenses are non-transferable. If you are a coach, principal, or district interested in transferable licenses to accommodate yearly staff changes, please contact me for a quote at allthingsalgebra@gmail.com. This resource may not be uploaded to the internet in any form, including classroom/personal websites or network drives, unless the site is password protected and can only be accessed by students.95.00. | 677.169 | 1 |
Mathercise™ Book C Classroom Warm-Up Exercises
112 Downloads
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files.
0.97 MB | 2 pages
PRODUCT DESCRIPTION
Class Starters Keep Your Students' Skills in Shape
Each Mathercise™ worksheet is a blackline-master "class starter" for high school and middle school students. Use the activities with an overhead projector, or make copies for your students, to "Mathercise" them once or twice a week at the beginning of class. You'll keep your students in shape, on track, and on time!
Discovering Geometry author Michael Serra wrote the original Mathercise™: Book C as warm-up activities for his geometry students. He found the first few minutes of class to be the perfect time to sharpen skills and provide additional algebra practice. The original Mathercise™ was so successful that he created four additional Mathercise™ books for different levels.
Each Mathercise™ takes ten minutes and includes one reasoning exercise, one solving exercise, one sketching or graphing exercise, and space for a review exercise of your own design.
Mathercise™ Book C: For students taking Algebra and Geometry (Grades 9-10) or second year high school | 677.169 | 1 |
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Maths for Everyday Life English | 2009 | 32 pages | PDF This short Table of Contents Lesson 1: Course Outline 3
Lesson 2: Practical Maths
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Lesson 3: Converting Things
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Lesson 4: What Are the Odds?
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Lesson 5: Probability Rules
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Lesson 6: Understanding Statistics
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Lesson 7: Understanding Surveys
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Lesson 8: Is the Price Right?
26
Lesson 9: Personal Finances
29
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Course Name: Maths for Everyday Life
Lesson 1: Course Outline This It's important to stress that regardless of your current level of math skills or knowledge, your beliefs about your ability to do math, or whatever level of frustration you feel when doing maths, you absolutely can learn the maths you need. All that's needed is a bit of motivation. Now, granted, you won't become a maths expert overnight—it's an incremental process and it takes some time. But if you complete the course, your confidence for practical maths will continue to grow. What are the benefits? There are three benefits to becoming a "maths person": practical, psychological, and intellectual. Financial rewards. When you know how to reconcile your bank accounts, understand the true cost of using credit cards, start to invest, and learn to buy high-deductible insurance, the money you save or earn can be considerable. Another practical benefit is an increased know-how when it comes to things like figuring how much carpeting or paint to buy, understanding how to convert to foreign currency when travelling, and figuring discounts, and interest rates. Learning maths is empowering. Psychological benefits. In our culture, maths skills are often equated with general intelligence. So, as you become more comfortable using maths, others are likely to be impressed—and you may actually feel smarter. Then there is the feeling of satisfaction and pride you feel when you successfully solve a maths problem. You may also enjoy being able to help your children or your friends with maths, and take pride in being the one who can instantly announce the discount amount when shopping, or the percent increase in this year's sales over last year's. Intellectual benefits. Business, government, economics, medicine, finance, science, and technology all use the language of maths. The more maths you know, the better able you are to understand these subjects. Once you know the necessary maths, you'll have a better grasp of news stories about government spending, the scale of our solar system. It's difficult to get a handle on scientific things, such as the human genome project or what relative humidity or barometric pressure mean, without a knowledge of basic mathematics. When your maths confidence grows, you'll also have a clearer understanding of political polls, medical studies, and the countless surveys you see in the media. You'll also feel less intimidated when reading about and working with technology, such as digital cameras and hand-held computers. While a mathematical view is certainly not the only way to look at the world, your perspective is incomplete if you don't know at least the basics of mathematics. Use a Calculator Throughout this course, we'll give you calculator tips and shortcuts for solving certain problems. It's important to know how to use a calculator effectively. (Some problems are impossibly difficult or time-consuming without one.) But it's also a good idea not to become overly reliant on them.
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this will help to get your brain in shape. Now. and many will insist that they can't do it in their heads. we don't mean to discourage you from using a calculator. if each of the 15 people received £2. you'd need £30. But it's a snap if you cut the problem down 4 x 25 = 100 and 4 goes into 21 5 times therefore you go 5 x £100 plus £25 (remainder) = £525
Estimate 4750 ÷ 15 in your head. or the total value of all the merchandise in a store. but it gets easier the more you do it.25). this requires that you develop your capacity for doing math on paper or in your head without relying on a calculator. A facility for estimating lies at the heart of the mathematical way of thinking. such as 40 x 25. your estimate is 325 (instead of 3.50). Practice makes perfect. you'd need $45. It is the uncertain nature of estimating that makes it so valuable: Estimating forces you to develop your mathematical judgment. you have to learn to trust your judgment without cut-and-dried formulas that provide automatic answers. Approximately how much will each person receive? Well. or 1/3 of 60. so each person gets a little more than £3—call it £3.25. In part.50. Here are a few examples:
What is 21 x 25? Most people will reach for their calculators for this problem. since our problem actually involved the pence amount (4750).
Another key to becoming a math person is to get in the habit of estimating mathematical things. it helps you to build your maths muscles. Now imagine you have £47. think of the 4750 as 4750 pence.50 to divide among 15 people. First. It's important to know how to use a calculator effectively. especially if you're not used to it. You've got a little more than $45.
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. it may be hard at first. it helps to go through the extra effort of doing math on paper or in your head. force yourself to do some maths in your head. If your math is rusty.000.One of the main objectives of this course is to give you greater confidence in your maths skills. the number of words on the page you're reading. which equals £47. Here's a great tip for doing basic arithmetic in your head: Think of money when you need to do mental maths. Estimating may not be easy at first. Look how good this estimate is: The exact answer of 4750 ÷ 15 = 316 2/3. and. if each received £3.
By the way. Estimate anything: the number of square feet of area or cubic feet of volume of the room in which you're sitting. Do some simple problems on paper rather than with your calculator. And just like working out. or a 15% tip for a dinner bill of £65. Doing math this way is similar to giving your mind a workout. but if you want to become a maths person. not the pounds amount (47. at least occasionally.
000. Begin by using the aforementioned 150 million taxpayer number. Assume that half of them (or 150 million people) pay an average income tax bill of $10. therefore.000 that would come to $150 billion.Say you'd like to know the cost of petrol for a journey in your 25 miles-per-gallon car. and never know it.
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. residents. 1. A billion is much bigger than a million.
Not only does this method reduce errors. When you do a problem the exact way—with a formula and perhaps using a calculator—you might use the wrong formula or punch a wrong button on your calculator. take a moment to ask yourself: "What's a sensible answer here? What's a ballpark estimate?" If your calculation is not near this benchmark. "I have absolutely no idea of how to solve that. If you don't estimate—if you blindly follow formulas or punch buttons on your calculator—you're at the mercy of mysterious rules and methods. You now know that you will need 120 gallons. you'll have taken a huge step toward becoming a maths person. So.000. Say you read in the newspaper that the estimated cost of a US government initiative is $100 billion. or about $650. whenever you do a math problem. When you estimate. a thousand billions is a trillion. When you do. and you want to determine the amount per US taxpayer. We will assume that petrol costs £2 per gallon.000 miles. if each taxpayer paid $1. So. or about $2. the cost of the US government program would be 3 x $650. One gallon of petrol takes you 25 miles. This is for the average taxpayer who pays about $10. it will cost you approximately £240. you're in control.000). 4 gallons takes you 100 miles." it's time to push yourself to learn how to estimate such things. it puts you in the driver's seat. you've likely made a mistake somewhere. Now. get the wrong answer. we were able to easily estimate the cost. Formulas and your calculator become tools that you use to confirm your independent mathematical judgment. Such problems are really quite easy when you get used to them.
The numbers in the millions and billions that we see in the media on a daily basis are generally so far beyond our everyday experience that our eyes tend to glaze over. The estimated project cost was 2/3 of that—$100 billion—so the cost per taxpayer is 2/3 of $1000. you can use these very rough numbers: Assume that there are 300 million U. 40 gallons takes you 1. You might not realize your answer is wrong because you have nothing to compare it to. how can you make sense of these astronomical numbers? One trick is to convert a big number into a much smaller amount per person—this brings the astronomical number down to earth. If you pay three times that much ($30.000 x 150 million = 150 billion. Estimating also has a very practical benefit: It's a great way to catch errors. Estimating becomes much easier once you make it a habit. since a thousand millions is a billion. the journey is approximately 3. and a trillion is much bigger than a billion in fact a thousand thousands is a million. a thousand millions is a billion.000 in income taxes.000 miles. If you're thinking.
By thinking logically and breaking the problem down into manageable figures.
First. and so forth.S. So. For these conversions.
000 tax bill pays $2.Here are a couple of handy benchmarks that you can use. If you need more basic maths help your local library has information about national schemes through local classes or on the web.000 for the $100 billion program. Hopefully that's got you thinking about maths and given you some tools to develop – we will cover more practical tools in the next lesson. From now on. for this taxpayer. The above hypothetical American taxpayer with a $30. If you do the math. That's $20 per billion. you will know how to relate those numbers to your life. So.000 average. each billion in federal spending costs him or her approximately 20 dollars. Now you can work out your own personal benchmarks using the same formula for the UK – 55m population – 25m taxpayers at £10. whenever you see astronomical figures reported in the news. it follows that each $50 million in federal spending costs that taxpayer $1.
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you have one more slice than half a pizza. Decimals. To demonstrate this technique of understanding fractions by visualizing pizza slices. However. This is also a little more than a half. if you have 5/8. the extra slice in 5/8 is more than the extra slice in 7/12. and percentages are three sides of the same coin—three ways of expressing a portion. since slices from an 8-slice pizza are bigger than slices from a 12-slice pizza. Which is greater. Because slices from a 12-slice pizza are bigger than slices from a 13-slice pizza. Which is bigger. 12/11 is more than 13/12.) 7/12 is like 7 slices of a 12-slice pizza. and 7/12 is one more slice than half of the 12-slice pizza. 5/8 or 7/12? This is easy to answer if you visualize the pizza slices. The good news is that. But. you've got one slice less than a whole pizza. and percentages. And for practical maths. since 6 is half of 12. Therefore. and Percentages Fractions. so 1/12 is a bit more than 1/15. Fractions An easy way to understand fractions is by visualizing pizza slices. decimals. that means that you have 5 slices from an 8-slice pizza. For example. The top number in a fraction (the numerator) indicates the number of slices that you get. let's consider the following question: Which fraction is larger.75 pounds. 5/8 is one more slice than half of the 8-slice pizza. So in both cases.Course Name: Maths for Everyday Life
Lesson 2: Practical Maths Welcome back There is some good news and some bad news: The bad news is that you can't possibly do practical math without fractions. 1/12 or 1/15? Slices from a 12-slice pizza are a bit bigger than slices from a 15-slice pizza. the bottom number (the denominator) indicates how many slices the pizza is cut into. 3/4 of a pound is the same as 75% of a pound. approximations are often good enough. If you focus on the pizza analogy. 11/12 or 12/13? In both cases. Assume that two pizzas are the same size. the missing slice from 11/12 is bigger than the missing slice from 12/13 but you're missing more with 11/12.)
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. it's a bit less than 12/13. decimals. you should have no difficulty developing a solid grasp of the approximate size of different fractions. Numbers Matter: Fractions. (And since half of an 8-slice pizza is 4 slices. which is the same as 0. 5/8 is a bit larger than 7/12. regardless of any difficulty you might have had with these topics before there's really nothing to them. 5/8 is a little more than a half. and that one has been cut into 8 slices and the other has been cut into 12 slices. (If you are having difficulty visualizing this concept. try sketching it out on a piece of paper.
01 as 1p. the denominator of the fraction is always 10 or a power of 10:100 (10 x 10). 0.09. We used the multiplication approach above with the pizza idea.
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.01 because £0. You should use whichever approach seems easier to you depending on the nature of the particular problem.09.) . you can often just round decimals to two places and then add or subtract just like you'd add or subtract money.05 is the same as 5/100. just being aware of this dual nature of fractions may help clear things up a bit. and comparing decimals to fractions. 2. On the other hand. Since two places of decimal accuracy is often good enough for practical math. of course.09. for example. So 23/6 is one slice shy of 4 pizzas. One thing that makes the subject of fractions confusing is that a fraction is simultaneously a multiplication problem and a division problem.00917 is less than 1 pence. and so on. 1000 (10 x 10 x 10). Since multiplication is a simpler idea than division. 5/2 or 21/10? Two 10-slice pizzas give you 20 slices.3 is more than . With decimals. at the same time. We saw in Lesson 1 that thinking of money is an easy method for doing simple arithmetic in your head. 5. Since 2 1/2 is substantially more than 2.022 as about £5. you can use the same technique to work with decimals.7 is the same as 2 7/10.30 is more than approximately £0. . Which is greater. For people who have had difficulty with fractions in the past.72 is the same as 1 72/100 or 172/100. 5/2 is the larger fraction. so 21/10 is just one more small slice than 2 pizzas (2 1/10 to be exact). 24 slices would be 4 whole pizzas. 7/8. and then the money analogy works like a charm. it's usually easier to think about a fraction like 7/8 as 7 x 1/8 or 7 eighths than to try to picture dividing 7 into 8 pieces. 0. while it's a good idea to review how to work out these problems on paper.087 because £0. and so on.6 is the same as 6/10. Using the money method. you should have no difficulty estimating the size of a decimal number. 7/8 is 7 slices from an 8-slice pizza—that's 7 times a 1/8 slice.10 is more than £3. comparing different decimals. with a fraction such as 24/6. it's probably easier to see that 24 ÷ 6 = 4 than to do 24 x 1/6 = 4. means 7 x 1/8 and. because £6.7 as 70p.How much is 23/6? 23/6 means 23 slices of a 6-slice pizza.02.00917 is less than . For multiplication and division. For addition and subtraction. 7 ÷ 8. Here are some examples: 6. (You can always add one or more zeros to the end of a decimal number without changing its value. 0. 30p v 9p. as our money is based on the decimal system – 100p + £1. you'll usually want to do these operations with your calculator. Think of 0. you can often round off decimals to two places. Decimals Decimals are simply fractions written in a different form.1 is more than 3. and thus it's a little less than 4. 1.
62. you can reduce the even tenths to fifths: 2/10 is 1/5. So 18% of something is the amount of it that compares to the whole thing in the same way that 18p compares to one pound. and thus 65% equals about 2/3. Seeing the location of 65%. you'll develop a quick. After a while. You might like to simply round percents off to the nearest 10% and then use tenths as your benchmarks since they're so easy to picture. 3. So. or about a twelfth (they might then confirm this intuition by noting that 8 x 12 is about 100). either write it out on paper or just imagine it in your head—to locate any percentage from 1% to 100%.5%. for example. note the portions that equal 25%. which is 4/10 or 2/5.5%. and 7 slices (1/8. You may be able to easily estimate that you've got a little less than 1/5 of a pound as it is 2p less than a 20p piece.
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. of course.Percentages Percentages. Imagine a pizza cut into 8 slices: We know that 25% or 1/4 of the pizza is two slices. So. and 75% as shown below. 50%.5%. 25% means 25/100. First. 5/10 is 1/2. or £48—the sale price. And. you may find it easier if you convert the percent into a fraction with 100 in the denominator. and some people have an instant sense that 65/100 is about 2/3. 15% is halfway between 1/10 and 2/10. Imagine 18p in the palm of your hand. So. are fractions written in a different form. Percentages are hundredths. you can use a scale. and uses eighths instead of tenths. 37. And. 5. A different method involves the pizza idea. 18p is a little less than 1/5—call it about 1/6 of a pound—and thus 18% of something equals about 1/6 of it. so 5 x 15 is 75. For example. 50% is four slices. Remember if you need more basic maths help your local library has information about national schemes through local classes or on the web that can help you improve your skills. since one slice is half of 25%. 60 x 80p = 4800p. A second thing you can try is to convert percentages into pence. may help you see that it's about two thirds of the way to the top. great! If not. 15% of 500 means 15/100 of 500. leaving 80p of each pound. 4/10 is 2/5. 5% is 5/100. 83% becomes about 80%. for example. and some can intuit that. here are a couple of methods you can try. £60 marked down 20 percent means that 20p of each pound is subtracted. or 12. Focusing on the fact that a percentage is the same thing as a hundredth is all some people need to do to get a quick feel for the size of a percentage. 8% is a little less than that. you can easily note four more benchmarks for 1. and 7/8): 12. and 87. Don't forget that if you find yourself puzzled by some percentage problem. If this works for you. 6/10 equals 3/5. and 8/10 equals 4/5. which is 8/10. Here's a different sort of problem. You may come up with other shortcuts for picturing percentages on your own. since 10% is a tenth.5%. 8% is 8 hundredths. If you like. intuitive feel for percentages. You should be able to do this one in your head: One hundredth of 500 is 5. like decimals. For example. which equals 2/10. or again use the money tip system.5%. 38% of something is about 40%. 3/8. 5/8. and pence are hundredths of a pound. and 75% is six slices. A percent is simply a hundredth. Whatever methods you use. 65% is 65 hundredths. 19% is rounded off to 20%. the important thing is to practice estimating what quantity a particular percentage stands for.
Let's consider some additional examples: 4/5 = 4 ÷ 5. which is commonly written out as £. or 0.875 In this way. and 0. Move the decimal point two places to the left. which equals 0.35 = 35% /. you can see that it is easy to convert from a fraction to a decimal. Converting a percentage to a decimal is very simple. or 0.000 We can move further in our understanding by converting from a percentage to a fraction. we can convert any fraction to a decimal. (Note that with percents such as 400% and 213%.) In all of these examples.80 = . This is a simple two-step method using the information we've already learned.037. we have simply moved the decimal point two places to the left to convert from a percentage to a decimal.5 represents five tenths. This makes sense to us because we know from our experience using money that one half of a dollar is expressed as $. there's sort of an imaginary decimal point to the right of the last digit.065 = 6. Likewise.70. 0. This can be expressed as 5/10.8/ 95% = . and 213% = 2. into a percent.13. any percentage can be converted easily to a decimal.25. and percentages are different ways of expressing the same mathematical idea. and we know how to convert from a decimal to a fraction. 400% = 4.25. Let's consider some additional examples: . For example.15 = 115% To convert from a fraction to a decimal. Let's continue by converting decimals to fractions. the fraction 7/8 converts to the decimal
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. As we can convert from a percentage to a fraction.2438 = 2.49 = 349/100 / 0.76 = 76% / 1.80 / 5/6 = 5 ÷ 6. Simply move the decimal point two places to the right. Imagine that we need to convert the decimal. To convert from a fraction to a percent. Mastering this skill will be helpful to you as you continue in your pursuit of mathematical self-reliance.40 = .833 / 7/8 = 7 ÷ 8. many mathematical ideas are more immediately clear to us if we think in terms of money. 1/2 = 1 ÷ 2. or 25p which is ¼ of a pound or 25%. For example. we can also convert from a fraction to a percentage.95 / Likewise. we can convert 70% to 0. Let's consider some additional examples.4 / 65% = . As we've seen. We will convert it to a fraction by expressing it as 375 thousandths. and then convert the decimal to a percentage. fractions. 25% is equal to .50.037 converts to the fraction 37/1000. All we have done to compute this conversion is to express the five tenths in fraction form.438/10. Now let's convert from a decimal to a percentage.0. There is no need to use a calculator.50. you will convert the fraction to a decimal.Course Name: Maths for Everyday Life
Lesson 3: Converting Things As we've seen. with which we are already familiar.175 = 175/1000 / 3.375 is written as a decimal and it is written to the thousandths place. or 0.065.65/ 80% = . Using this method of converting.70 can easily be converted to 7/10. 3.5%. As we've already learned.7% converts to 0. 0. You can use this simple conversion method any time you need to convert from a percentage to a fraction. 0. The decimal 0. and 0. or 375/1000. For example.25 = 25% /. Keep this in mind as we do several more conversions: 40% = . decimals. Move the decimal point two places to the right: 0. you can easily do this conversion in your head or on paper. you divide the numerator by the denominator.
and that is expressed as 66. Let's consider some additional examples: 2/3 converts to 0. 7/12 converts to 0. quarts to cups. you can get the multiplier in one step by just putting the 7 over the 5.
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. The class is 1/5 boys. What if you want 7 dozen cookies? Well. we have also strengthened our understanding of what these different mathematical expressions mean. By the way. the multiplier is 7/5. Since 7 dozen is 2 dozen more than 5 dozen. That's what happens when you keep the ratio of cups of flour to dozens of cookies at 2:5. and percentages. Now we have learned simple methods of converting between fractions. liters to gallons. For example. Take note of the common error discussed there: a boy-girl ratio of 1:4 does not mean that the class is 1/4 boys. It's difficult to overemphasize how important it is to use your common sense when doing math. 13/8 converts to 1. It can really come in handy to be able to at least estimate—conversions from. and that is expressed as 162. because there are 1 boy and 4 girls for every 5 students. that tells you that there are 2 boys for every 3 girls. The same applies to any fraction. decimals. and thus that there are 2 boys for every 5 students and 3 girls for every 5 students. but it should be obvious that regardless of how many cookies you make. cups to ounces. which equals. Proportions are used when you want one ratio to equal another. if the ratio of boys to girls in a classroom is 2 to 3 (or 2:3).3%.5%. 1. and it is easy to convert that to a percentage by moving the decimal point: 87. you'll need twice as much flour (4 cups). 1 dozen would be 1/5 of a batch. it's 2/5 of a batch more than 5 dozen. So 7 dozen is 1 2/5 batches.5%. For example. miles to kilometers. let's say your chocolate chip cookie recipe makes five dozen cookies and calls for two cups of flour. And that tells you that 2/5 of the class is boys and 3/5 is girls. Ratios and Proportions Ratios and proportions are both very closely related to fractions.4. this percentage (25%) or fraction (1/4) of flour needs to remain unchanged.7%. the flour makes up about 25% of the cookie mix. and that is expressed as 58. It's common sense that if you want to make twice as many cookies (10 dozen). and percentages. This simply guarantees that the percentage or fraction of flour in the cookie mix remains the same. You don't need to know this when doing a proportion problem. which equals 2 4/5 cups of flour.4) to increase the amount of any other ingredient when you're making 7 dozen cookies instead of 5 dozen. Conversions These are of great practical value.875. In learning how to accomplish these conversions.625. say. or square feet to square yards. as well as how they relate to each other.0. Imagine that in this recipe. Whenever possible. and will help you in any number of ways in your everyday life. Ratios are basically just fractions written a different way. again. decimals. You therefore multiply the 2 cups of flour by 1 2/5.667. You would use this same multiplier (1 2/5 or 1.583.
you get fewer of them. Converting units always involves multiplication or division by the appropriate conversion number from a conversion table. map books.7 kilograms into grams. or 0.05 kilograms—move the imaginary decimal in 50 (50. Conversions Within the Metric System Converting units within the metric system is even easier because you're usually multiplying or dividing by numbers such as 10." If you convert from a number of miles (a big thing) to the equivalent number of feet (a smaller thing). you can't go wrong as long as you know whether your answer should be bigger or smaller than the number with which you start. chart. When converting from a small thing to a big thing. you'll obviously get a greater number of feet than the number of miles with which you started. Since there are obviously more than 12 feet in 12 miles. you must multiply 12 miles by 5280 to obtain the correct answer of 63. Remember: Don't just memorize this rule. which you can easily do without a calculator.7 three places to the right. One half-pound is how many grams? Solution: 454 grams = 1 pound. If you focus on these simple rules. and you'd divide by 4 to convert 8 quarts into the equivalent number of gallons. you'll never miss a conversion problem. 100. but you might see a different type of conversion table that lists conversions like 1 quart = 1/4 gallon. reference books and on the web. pay attention to whether your answer is sensible or not. Of course. and 1000. If you want to convert 20 quarts into gallons using the 1/4 conversion number. you'd multiply by 4 to convert 5 gallons into the equivalent number of quarts. you'd have gotten the ridiculous answer of 12 ÷ 5280. you'll always multiply by the given conversion number to convert from a big unit to a small one. Tables list the equivalences in the standard way. When using such a conversion table (usually in a diary. you get them in diaries. rather than simply following formulas and methods on autopilot. For example. you divide 50 by 1000 for an answer of 0. To convert 50 grams to kilograms. multiplication by 1/4 gives you the sensible answer of 5 gallons.
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.0023 feet in 12 miles. For example. half that number (227) must be half a pound. and then drop the meaningless zero at the far right and add a zero to the left of the decimal point. let's say you want to convert 12 miles into feet and 5280 feet = 1 mile. Always ask yourself whether the answer to a problem should be a larger or a smaller number than the number with which you start. Since you're converting from a big thing (pounds) to a smaller thing (grams). Again. cook books and so on). Your response to this should be: "Of course. you multiply by 1000 to obtain 3700 grams—move the decimal in 3. you get more of them. One of these—multiplication or division—will give you a sensible answer that follows the above rules. you could also have simply reasoned that since 454 grams is 1 pound. so to convert 3.360 feet. if you choose to use a conversion table. Here's the common sense for doing conversions: When converting from a big thing to a small thing.) three places to the left and fill in the zero (. If you had divided instead of multiplied. you multiply by the conversion number: 1/2 x 454 = 227 grams.050). and divide to convert from a small unit to a big one. 1000 grams = 1 kilogram.bolster your knowledge of maths with your common sense. If you instead divide 20 by 1/4 you'll get the unrealistic answer of 80 gallons. The other will give you an unrealistic answer.
Note that 1 kilometers = 1000 meters. Since a meter is roughly 3 feet.6 kilometers = 1 mile.
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. 5280.6 = 12. 20 kilometers is how many miles? Solution: The table gives you the conversion number: 1. (Remember that kilometers are smaller than miles. which is less than the 5280 feet in a mile. so you divide: 20 ÷ 1.79 liters = 1 gallon.10 liters is how many gallons? Solution: 3.5. You're converting from a small thing (kilometers) to a bigger thing (miles).64 gallons. it turns out that 1 kilometers equals 3280 feet—an easy number to remember because of its coincidental similarity to the number of feet in a mile.) By the way. This time you're converting from a small thing (liters) to a bigger thing (gallons). so you divide by the table number: 10 ÷ 3.79 = about 2. 1 kilometers or 1000 meters is roughly 3000 feet.
those are the only two possibilities. We will discuss them here. that the chances of it happening are 1-in-2. and the highest possible probability.Course Name: Maths for Everyday Life
Lesson 4: What Are the Odds? Probability is also known as "odds" or "chance. A probability of 100% means the thing will happen with absolute certainty. Being a maths person requires familiarity with the six basic probability rules and principles. Halfway up from 0 to 100 is 50%. you've got to give 110% effort. you can convert probability percentages into fractions. you can also convert probability percentages into decimals. The probabilities going up from 0% to 100% indicate greater and greater likelihood of some event occurring. or that the chance of rain is 50%. once out of every 2 chances. And that means that. on average. which means that the chances of the event in question occurring are 1-in-10. For instance. equals the number 1. you'd expect the event to occur about once in every 10 chances. equals the number 0.) Likewise. roughly 4 out of every 5 times. Whenever you hear a news report warning that a percentage of people are at risk for some illness. so a probability of 15% is about a 1-in-7 chance. on average. If you buy a lottery ticket. and a probability of 85% is the same as a 0. 50% converts to 50/100 or 1/2. When something has a 50% probability of occurring. it is as likely to occur as to not occur. in the long run. you'd expect the event in question to occur roughly 1/7 of the time or. you are hearing about probability. (Don't make the common error of concluding that just because something must either occur or not occur. The lowest a probability can be is 0%—which means that the thing in question will definitely not happen. so a 50% probability means that something would occur. or 50%. 80% converts to 80/100 or 8/10 or 4/5. For example. Probability is helpful because it provides a way for us to measure and discuss the uncertainties associated with future events. the only two things that can happen. a probability of 10% is the same as 10/100 or 1/10. It's the same with probability.85 probability.
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. In other words. a probability of 25% can be expressed as 0. you will necessarily either win or lose. 15% equals about 1/7. 1 time in every 7 chances. The probability of something occurring can't be more than 100%. far less than 50%. in the long run. or about half the time. As with any percent problem. you could say the chances are 50-50. The scenarios involving probability are all around us. but it's literally impossible to give more than 100% effort." That may work for motivating athletes. so a probability of 80% means you'd expect the event in question to occur about 4/5 of the time or. "If you want to win. or about 1/10 of the time. and it dictates our insurance premiums. 100%. The lowest possible probability.25. The Six Rules of Probability Rule 1: A Probability Is Always a Percent from 0% to 100% or a Number from 0 to 1 Coaches love to say things such as. But your chances of winning are far. Finally. So. 0%. probability tells us the comparative dangers of driving versus flying." It would be difficult to go through a single day without encountering a reference to probability.
whether or not you know the formula. so the answer is 3/8 or 37. five pence. their probabilities will always add up to 100%. these are the only two possibilities. If you hear a prediction that Birmingham have a 15% probability of winning against Chelsea in their next game. In other words. if the weather forecast predicts a 70% chance of rain today. so the answer is 3/10 or 30%. (Note that the answer is not 3/7. the answers to all probability problems are based on this simple formula—though few problems are as straightforward as the above example. A closely related idea is that if you consider all the different things that could occur in a given situation.) In a sense. Therefore. let's say that London has seven football teams in the premier league in any one season. 5p and 10p. then it follows that there's a 100%-15% (or 85%) probability that Birmingham will lose.3 of the 8 outcomes are winners (bold). what's the probability of getting one heads and two tails? Solution: There are three ways to "win"—the penny. For example. or ten pence could land heads up—so the numerator in the formula is 3. Total 100%
Rule 3: The Basic Probability Formula You are probably familiar with the way this basic idea works. Penny Five pence Ten pence Heads Heads Heads Heads Tails Tails Tails Tails Heads Heads Tails Tails Heads Heads Tails Tails Heads Tails Heads Tails Heads Tails Heads Tails
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. Here's a tougher one: If you toss a penny.Rule 2: Probabilities Always Add Up to 100% An event must either happen or not happen. then there must also be a 30% chance that it will not rain. the probability of an event happening plus the probability of it not happening must add up to 100%. It will either rain today or it will not rain today—there is not a third possibility. Here's a simple example: What's the probability of drawing a black marble from a jar containing 3 black and 7 white marbles? 3 out of a total of 10 marbles are black. The denominator is 8 because there are the following 8 possible outcomes .5%. The probabilities of each team winning the division (and remember that the probabilities in this context are only somebody's best guesses) must add up to precisely 100%. One possibility could be: Chelsea 26% Arsenal 22% Spurs 18% West Ham 15% Charlton 8% Fulham 6% Crystal Palace 5%.
There's no guarantee. nonetheless. there's the same uncertainty about rain as there would be about tossing a heads with a coin. that exact answer doesn't guarantee anything. But. So. There's no way to compute a numerator or denominator for such a weather prediction. You're dealing with probability. The prediction of 25% equals 1/4. If there are 3 black marbles and 7 white ones in a jar. or 30%. Its accuracy is limited by the weathercaster's skills. All probability involves uncertainty. the probability of drawing a black one is exactly 3/10. assuming the rain prediction of 25% is correct. The prediction of 50% is only an estimate. Unlike the 50% chance of tossing a heads—an exact probability—the 50% probability of rain is only the weathercaster's best educated guess. If the weatherman says there's a 50% chance of rain tomorrow. But predictions like this involve a second type of uncertainty. for example. the likelihood of rain tomorrow would be the same as the likelihood of drawing a black marble from a jar containing 1 black marble and 3 white ones—which is 1 winning marble out of a total of 4 marbles. you might get lucky and draw a black 5 times in a row. Most predictions about the future involve this second type of uncertainty as well as the uncertainty inherent in every probability problem. which is also a 50% chance. But note the following connection between such a prediction and the marbles-in-the-jar idea. that if you draw a marble 10 times (replacing the marble each time) you'll draw a black 3 times out of 10. the 30% probability is an exact answer that can be precisely computed from the formula. or 8 times out of 10—or you could get unlucky and strike out 10 times in a row. As anyone who's been to a casino knows. the true probability might be 30% or 70% or anything else. as well as by the current level of sophistication of the science of meteorology. so nothing's guaranteed. Predictions are different.Probability statements such as "There's a 25% chance of rain tomorrow" aren't as cut-and-dried as the above two problems. But you can compute exact answers for the marble and coin problems. Now. granted. The 50% number could be wrong. There's an important distinction between problems such as the ones involving marbles or coins and problems involving predictions such as the probability of rain or the probability of Manchester United winning the League.
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and that only one person will be offered the job. Call it 50% since you're dealing with estimates. so there remains a 30% probability that the job will be offered to a candidate other than one of these three people. Assuming these estimates are correct. the probabilities must add up to 100%. what's the probability that both girls will be accepted? Just multiply the probabilities: 70% = 0. let's consider why simple addition won't work in this case. For example.7 x 0. and you want to determine the probability of at least one of them happening.. Common sense should also tell you that the answer will be less than 100%. In order to use this rule.7 = 0. if one happens. note that since it is possible that he could be accepted by both. and Tom and Bill each have a 15% probability of being selected. which means that whether or not the first event happens will have no bearing on whether or not the second event happens. We will now learn the method to solve such a problem. or 12. so the probability of tossing 3 heads in a row is.. Tom. If you estimate that his chances of being accepted at each of the two safety colleges is 90%—that's 90% for each one individually—what is the probability of his being accepted by at least one of the safety colleges? First. because there is a very slim chance that both will reject your son. these are not mutually exclusive events.e. and their tutor estimates that each girl has about a 70% chance of being accepted. if Safety College #1 accepts your son. Rule 5: The Addition Rule for Mutually Exclusive Events When you have two or more mutually exclusive events (i. For example. the probability of tossing a heads is 50% or 1/2. or 49%. Rule 6: What to Do when Events Are Not Mutually Exclusive When you have two or more events that are not mutually exclusive (i. then there is a 70% probability that one of the three will be offered this job.5%. Here's another one: Let's say your daughter and her best friend have applied to the same college. you simply add up their individual probabilities. For instance. If you did use the addition rule here. you can't just add up their individual probabilities like you can with mutually exclusive events. none of the others can happen). you'd obtain an answer that is obviously impossible: 90% + 90% = 180%. the events must be independent. but it's a bit oversimplified. Let's say your son is applying to eight colleges. that has no
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.e. the addition rule does not apply here. you multiply their probabilities. This conveys the basic idea of the multiplication rule. Before showing you how to do these problems. As we have seen.7. and two of them he considers his "safety" colleges. If you estimate that Jane has a 40% probability of being selected. any number of them can occur at the same time). This result should seem reasonable because it's quite a bit harder to toss 3 heads in a row than to get heads on a single coin toss. and you want to figure the probability that any of them happens. there's no such thing as a probability greater than 100%. and Bill are all applying for the same job.49. imagine that Jane.Course Name: Maths for Everyday Life
Lesson 5: Probability Rules Rule 4: The Multiplication Rule When you know the individual probabilities of two or more things and you want to determine the probability that all of them will occur. and 0. therefore.
Last.) have no memory. things will tend to balance out in the long run. This is a fundamental law of probability. the law of averages does not say that future events (such as coin tosses) will tend to balance out past events. It tells you. but it is because in the long run (1000 tosses in this example). then use the multiplication rule to compute the probability of both things not happening (his not being accepted at either college). The probability of heads (or tails) on any toss is always 50%. Calculate the probability of his not being accepted at each college. heads and tails will tend to balance out. then 0. to determine how many different ways there are to order or rank several items. etc." and thus the probability of tails on the next toss would be something greater than the usual 50%. for instance. Law of Averages There's a common misconception about probability: Many people believe that the law of averages says that future events will balance out past events so that everything will come out even. It says only that if you start keeping track of results now. The two events are thus independent of each other. The law of averages never looks back in time—period. in fact) end up with between 450 and 550 heads out of 1000 tosses (that's between 45% and 55%). In other words. that while getting 7 heads out of 10 tosses (70% heads) would not be very unusual. if a coin has landed on heads six or eight times in a row. Multiply: 10% = 0. For example. subtract this probability from 100% for your final answer: Probability of not being accepted at Safety College #1: 100% – 90% = 10%. 700 heads out of 1000 tosses (which is also 70% heads) would be extremely unusual.1 = 0. We're ready to move on to our discussion of the four counting principles.1 x 0. On the other hand. Figuring It Out: The Four Counting Principles You can use these rules to achieve several objectives: to determine the total number of different combinations available given a set of variables. Subtract from 100%: 100% – 1% = 99%.bearing on whether Safety College #2 also accepts him. the past is irrelevant in computing the probability of future events. It's important to understand what the law of averages says and does not say. which equals 1%. to determine how many different groups of a certain size can be selected
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. You'd very likely (over 99% of the time. The last 10 or 50 or 1000 coin tosses have no effect whatever on the likelihood of tossing heads or tails on the next toss. Probability of not being accepted at Safety College #2: 100% – 90% = 10%.01. There's a 99% probability that your son will be accepted by at least one of his two safety schools.1. This may seem surprising to you. this mistaken belief holds that tails would be "overdue. That's your answer. This tells you that there is a 1% chance of your son being rejected from both schools. Mathematicians often express this idea as dice (or coins. But this is simply not true.
or 12. such as "Mum. the number of different combinations is 3 x 4.from a larger group. and you've narrowed your final decision down to 3 entrées and 4 wines. Bill. Amy always gets to go first. 3 entrées. For example. The Fundamental Counting Principle The fundamental counting principle is helpful when you need to determine the total number of different combinations available given a set of variables. If your calculator has a factorial button. Assuming that any wine can be paired with any entrée. or 24. and Diane. you can do this computation in one quick step by
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. Charlie. The Factorial Rule The factorial rule refers to a simple method of determining how many different ways there are to order or rank several items. How many different ways can you order them from 1 to 4? The answer is 4 factorial—written mathematically as 4!—which equals 4 x 3 x 2 x 1. 4 wines. and so on. Let's begin with the first: the fundamental counting principle. the grand total of all possible outcomes is given by multiplication. imagine that you have four children: Amy. the total number of different dinners you could serve would be 2 x 3 x 4 x 3. and to further determine the ranking of those groups. When you have a number of options or choices followed by a second set of options or choices followed by a third. or 72 dinners. You start with the total number (4 in this case) and multiply it by all of the numbers going down to 1. For example. imagine that you're planning a dinner party. The 12 options might look like this: Fish with Cabernet Merlot Zinfandel Pinot noir Steak with Cabernet Merlot Zinfandel Pinot noir Pasta with Cabernet Merlot Zinfandel Pinot noir If you had to decide among 2 salads." so you want to use all possible orders of your four children equally. and 3 desserts. You don't want to show any favoritism or hear complaints.
and Diane /Amy. here are the six orders that begin with Amy: Amy. Diane. Charlie. Charlie. n = 7 (the total number) and r = 3 (the subcommittee size). Bill. (By the way. you find this answer by simply punching: This gives you an answer of 35. Bill. Diane. Order means any ranking or designation for the members of the selected group. and Bill There are also six orders beginning with Bill.
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. If your calculator doesn't have the combinations button. Diane. Bill. For example. Charlie. and Diane/ Amy. Bill. In case you're curious. and Charlie / Amy. and Charlie / Amy. For combination problems. six beginning with Charlie. bringing the total to 24. Charlie. you can use the following formula: For this problem. the factorial symbol '!' has nothing to do with the grammatical mark—whoever came up with this symbol must have had a sense of humor?) The Combinations Rule The combinations rule is helpful when you want to determine how many different groups of a certain size can be selected from a larger group. Plugging 7 into n and 3 into r gives you: Note that this combinations problem involved selecting people to serve on a subcommittee without any concern about their positions or titles on the subcommittee. mathematicians say order doesn't matter.just punching the given number followed by the factorial button. let's say you're the chairperson of the special events committee at your child's school. and Bill/ Amy. and six beginning with Diane. How many different subcommittees of three can be selected from a group of seven people? If your calculator has a combinations button. and you have to appoint three members from the seven other committee members to serve on a fund-raising subcommittee. Diane.
80. The median height is the one in the middle (the 6th one counting from the left or from the right). One strength of the mean is that it correlates perfectly with the total. One of the best ways to get a handle on the data about some group—say UK household income. but technically. median and mode will help you to make sense of the different ways there are to express what is typical. for instance—don't pull the median upward the way they do with the mean. median. or average. in the above example about test scores. the weight of UK men or women.000 in 2003) than what's typical (somewhere between $35. median.000 in 2003) is a better measure of a "typical" household. 4'2" is the mode because that height occurs more often (three times) than any other height in the list. The mean is another word for the mathematical average that you learned about in school. and it's important to understand that none is a perfect indicator of what's typical. (Note that when people use the word "average. A drawback of the mean is that when the data contains some extremely high numbers that aren't balanced by low numbers (this often happens because no numbers can go below zero). 86. Median U. imagine that a student wants to calculate his classroom grade: He would need to calculate the mean of all the grades he has received. Say a student scores 80. The median of a set of numbers is simply the middle value. the mean gives a pretty good idea of what's typical. median. And when a set of numbers is relatively balanced around the middle values. For example." they are usually referring to the mean. The mode of a set of data is the value that occurs most frequently. there are three types of average: mean. But what makes the median a good measure in the above income example—not giving full weight to extremely high incomes—becomes a drawback in other cases. is 87. household income (about $41. or the heights of the girls on your daughter's volleyball team—is to consider the group's mean. 82.8.000 and $45. and 95. The mean of this student's test scores. all of the millionaires and billionaires in the US cause the mean US household income to be higher (about $55. the mean is determined by adding up the test scores and then dividing by the number of scores (5). The median score of 82 is not a fair indicator of the student's performance because it does not give full weight to the perfect score (the median would be the same if the 100 were replaced with an 83). 90. just add the numbers up and divide by the total number of items. For the same list of heights. If his test scores are 80. of a group of numbers.S. and 100 on five tests.Course Name: Maths for Everyday Life
Lesson 6: Understanding Statistics How do we know what's typical for a large group of people? An understanding of mean. then. Imagine that the illustration shown below charts the heights of the 11 players on a girls' volleyball team. and mode. the greater the total test points earned by a student. the higher the student's mean test score. 82. the mean is often higher than what's typical.000 in 2003). For instance. For example. An advantage of using the median is that extremely high values—billionaires. The mean. Each has its advantages and disadvantages. and mode. 88. and mode of a group or collection are three different indications of what's typical or middling for the group.
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.) To calculate the mean.
99. Note the distinction here between median and mean: The median home price in this scenario is £240. annual rainfall. Three standard deviations (3 times 15 or 45) to the left of 100 is 55.000. and so on. £180.
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. Understanding the Bell Curve Many sets of data have a shape that resembles a bell like the one shown below. In general.7% of all people.The median is also often used to discuss typical home prices for a given area. That span of IQs constitutes 99.000.000.7% of all values are within three standard deviations of the middle value. As you might guess from the symmetry of this so-called "bell curve. IQ scores.000. £290.000. The standard deviation for IQs is 15. the mode is often meaningless. the larger the set of numbers. and three to the right is 145. the median gives a much better idea of what's typical. you get about 68 percent of the total area under the bell. This is instructive because it tells us that a one-child household is the most common. Bell curves are another way of interpreting data. for example. and £4. £220. Much of the ordinary data that we want to understand can be expressed in this way. and that tells you that about 68 percent of all people have IQs between 85 and 115.000. For any "normal" distribution ("normal" in this context means the same thing as bell-shaped).000. If you go 15 to the left from 100 (to 85) and 15 to the right from 100 (to 115). Understanding the bell curve and its standard deviation will enhance your comprehension of data such as. the weights of newborn babies. £210. 95% of all values are within two standard deviations of the middle value. There is no question that in this case. In fact. Two standard deviations for IQs would be 2 times 15. All sets of data that have this bell shape follow this standard deviation scheme: 68% of all values are within one standard deviation of the middle value. This number (one child) gives a better idea of what the typical household is like than the mean number of children would." the mean. the heights of women or men. it tells you what is most common. Understanding the strengths and weaknesses of the mean.000. and the mode will give you a better grasp of the averages and medians of various things that are reported so often in the news.000.000. For very small sets of numbers. you get about 95 percent of all people. the mode for the number of children is one child. If you go 30 to the left from 100 (down to 70) and 30 to the right (up to 130). the standard deviation indicates how tightly the values are bunched near the middle value. the mode is therefore exactly what many people would think of as most typical. A strength of the mode is that. by definition. Imagine that 11 houses have sold in one town in a given year. median. For some sets of data. £350. £375.000. £240. the median. or the date of the first frost in a given city. and that the sale prices were £50. or 30. £150.000. Among families with children living at home. but the mean home price is £620.000. £250. the more telling the mode will be.000.500. there is nothing typical at all about the mean price of £620. and mode of such data sets are all the same and are all equal to the value at the centre of the bell.
Let's say four different pollsters in a town wanted to measure the level of public support for building a new soccer stadium. In the example above. in fact. Can you trust the results? And what does the margin of error tell you? While poll and survey results are often informative — they are often imperfect and sometimes just plain wrong. #3 and #4 because of the way the questions are worded. the 50% and the 45%. it is possible to feel supportive of the idea of a soccer stadium. and yet not believe that other
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. Even if the poll had shown Candidate A ahead of Candidate B. It's not guaranteed. it's often difficult to determine whether a survey or poll was done properly. the margin of error would be ± 3%. Another limitation of polls concerns the way poll or survey questions are worded. Someone who answers "Yes" in Poll #1 might answer "No" in Polls #2. 52% to 44%—thus satisfying the margin of error because the difference (8%) is larger than twice the margin of error (2 x 3% or 6%)—what newscasters never tell you is that satisfying the margin of error doesn't mean the poll result is 100% guaranteed. there's still a very small chance that the poll results are wrong. Say a political poll shows Candidate A leading Candidate B. The maths is beyond the scope of this course. But the margin of error must be applied to both numbers.Course Name: Maths for Everyday Life
Lesson 7: Understanding Surveys
Polls. Surveys. 52% of those polled favored Candidate A with a 3% margin of error. Below are four possible polls and their results:
Poll #1: Are you in favor of building a new soccer stadium in our town? Yes: 65% No: 35% Poll #2: Are you in favor of building a new soccer stadium in our town if it means decreased funding for other public projects? Yes: 40% No: 60% Poll #3: Do you agree that building a new soccer stadium is the top priority for our town? Yes: 35% No: 65% Poll #4: Do you agree that your taxes should be significantly raised to support a new soccer stadium? Yes: 25% No: 75% Note that the way a poll question is worded can have a big effect on the outcome. Since 50% is 5% more than 45% and the margin of error is only ± 3%. 50% to 45% with 5% undecided. The poll is. unfortunately. That means that the support for Candidate A in the population as a whole is probably somewhere between 49% (52% – 3%) and 55% (52% + 3%)—but only 95% of the time. If 1000 people were polled. you might think that Candidate A is definitely ahead of Candidate B. and Their Limitations It would be difficult to go through a week without hearing or reading about the results of some poll or survey. And. Let's first look at the margin of error. but just be aware that even when the margin of error is satisfied. inconclusive—the race would be called a statistical dead heat—because 50% – 3% (or 47%) is less than 45% + 3% (or 48%).
34. Consider the first and third polls. Say you read an article bemoaning the huge salaries earned by finance executives. This tells you that the value of the pound shrunk by a factor of 2.340.000.4—that's about 2. "Are you willing to pay £10 more in taxes per year to support a new stadium?" the results would certainly have been different.000.000. and to call the article sloppy would be kind.
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.projects should suffer for it or that taxes should be raised significantly. Another way a survey can be flawed comes up a lot in the informal surveys you see in magazines and hear on news shows that report data such as. Perhaps no shy people responded to the survey because the questions were too personal. and the results could be used to bolster an argument either in support of or in opposition to it. Since polls are often conducted or paid for by groups that want a certain result.000 today—up from £1. The four polls show four different levels of support for this project. the CPI is about 193. the polls are often designed to achieve the desired result. check to see whether the author of the report corrected for inflation. Perhaps people who responded didn't answer the questions truthfully.000 would give you the average finance executive salary for 1980 in 2004 pounds. These questions are very different. the report is seriously flawed. this comparison is meaningless. Errors and misconceptions abound regard statistics figures we'll consider two of the most important errors that you should watch out for. Look up the CPI for 1980—that's 82.34 = £2.000. If he or she did not. For 2004. Had Poll #4 asked instead." There are so many possible flaws in such a survey that it's difficult to know whether the 30% result is even close to the true percentage. by this factor: £1.000 x 2. With many informal surveys. people may have complicated opinions.000 to £2.000. Imagine how people would respond to the question. and polls—which reduce subjects to simple Yes-or-No questions—may be designed to manipulate a response.000 amount has not been corrected for inflation. that's not a valid cross-section of the population. The author states that the average finance executive earns £2. did the magazine survey only its readership? If so.000 in 1980 (these numbers are invented for the purposes of this lecture). A person might feel strongly supportive of a new soccer stadium.340.000. Here's how: Look up or estimate the Consumer Price Index (CPI) for the current year. If the £1. The Effect of Inflation Whenever you see a report that compares an amount of money today to an amount of money from a number of years ago. there isn't even an attempt to reduce potential biases. Multiply the 1980 amount.000.500. Let's assume that the £1. You can do the correcting yourself.34 between 1980 and 2004. So take such results with a grain of salt. Divide 193 by 82. It can be difficult even for professional statisticians to eliminate all of the things that might skew the results of a survey.000. This answer of £2. £1. "30% of those who responded to the survey do such and such. "Are you ecstatic about paying a bigger tax bill?" Who would respond "Yes" to such a question? The obvious response of "No" to this question wouldn't necessarily mean that the person polled would object to a small tax increase for some good purpose.4. and it would show that the average finance executive salary increase from 1980 to 2004 was actually very little: from £2.000 amount from 1980 has not been corrected for inflation. For example. and yet at the same time might not believe that it should be the town's top priority.340. As we have seen from this example.500.
of course.What's important to remember is that to make valid comparisons between pound amounts from different years. Since it seems possible that something about television-watching could cause aggressive behavior. by itself. If this were true. the more aggressive they are. in fact. the more violent crime there is. is that cities with more people have both more hospitals and more crime. because it suggests that a greater number of hospitals causes crime. the less they tend to exercise. then it would be possible that television-watching. it's easy to jump to the conclusion that there is a causal connection even when none has been established. despite the correlation. comparisons are virtually meaningless. and that it's the lack of exercise that causes the aggressive behavior. it's very easy to jump to the conclusion that causality has been shown. causes no increase in aggressive behavior. The trouble is that when you hear about some study that shows a correlation where a causal connection seems plausible or likely. imagine that some new study showed that the more television children watch." This is. though. the past amounts must be corrected for inflation. The explanation. Causation Just because two things go hand in hand doesn't mean one of them causes the other. But the explanation might instead be that the more television children watch. This is obvious when you consider the absurd statement. "The more hospitals a city has. It's absurd. Correlation vs. a true correlation. For example. If they are not.
25
.
26%.) £51 is approx 30% of £167. Now that you know that 10% of the original price is approximately £17.95 is rounded up to £170. Note that if you had wanted to compute a 30% mark up on a product that costs £167. Is this one of the reasons that you are working to improve your math skills? You will learn two easy methods for estimating discounted and marked-up prices. You can also use Method One to compute discounts (or mark ups) of 5%. you could estimate this as £170 + £50 = £220.00.50.95. First. 15%. you will learn to calculate the math involved in a discount of 15% or 40%. To determine the actual sale price. There is a simple way to do this. In this way.95. For instance. Round
26
. Add £8. you will determine the price when £121. For example. Both methods are easy to use. If you prefer to use the first method. you would add instead of subtract: £168 + £50 = £218.95. Many people find the first method faster. We already know how to determine 30% of £167. 40%. the second method is more accurate. and so on.95 is marked up by 18%. The approximate sale price is £118. Compute 10% of the purchase price: 10% of £170. if you prefer the second method.00. and you may choose to use one or the other depending on the percentage with which you're dealing. and the item is discounted by 30%. if it is easier.00 is £17.Course Name: Maths for Everyday Life
Lesson 8: Is the Price Right? A knowledge of mathematics will help you to manage your finances—at home and abroad—more effectively.95 is approximately £17. Round that to £60 and subtract that from £168 for your final answer: £108. we wanted to figure out 30% of £170. and we need to add an additional 5% to that figure. The first method is designed for you to use when you need to figure out discounts and mark ups (increases) that are multiples of 5. round off the purchase price to the nearest £5 or £10.50. subtract £51 from £167. Multiply $17 x 3 and the total is £51. and so on. for a total of £51. It might be easier for you to now round £167.95 product is 35%. You now know that the discount is approximately £51 on this purchase. 30%. plus 3 x £7. you will learn to calculate the math involved in a price that has a 22% mark up. (You could also calculate it as 3 x £10. 35%. In this example. In the example given above. you can use it to determine a percentage that is not in a multiple of 5 (such as a 22% mark up). We determined earlier that 10% of £167. half of that (or 5%) is £8.Imagine that you are considering a purchase of a product with a sales tag of £167. by multiplying.50 (or 5%) to the previously determined £51 (or 30%) and you know that 35% of £167.95. The second method is intended for you to use when you need to figure out all other percents that are not multiples of 5. 25%.) Now you know how to use Method One to compute discounts (and mark ups) in multiples of 10. Method One . We know that 10% is £17. as long as rounding to 20% and figuring the 20% amount is a good enough estimate for your purposes. you can use it for all types of percent problems. or a markup of 20% or 35%. Imagine that the discount of the £167.95 is approximately £59. £167. Method One gives you an easy way to compute that discount. or £30. Method Two will teach you to determine discounts and mark ups when the percentage is not a multiple of 5. and so on. you can easily compute 20%. 22%.95. or £21. We will now compute 5% by figuring that it is half of 10%. and then subtract £50. (Or. On the other hand. such as 17%.95 up to £168. but for percents that are not in multiples of 5. you will be able to easily compute percentages of any amount.
you always want to use the larger of the two numbers—the one that's more than 1. Add that to £18. say $120. Just multiply 18 x £1 (or £18) + 18 x 20p.199? Use the steps as shown above: £2. and the total is £21. just enter the new exchange rate into your calculator and forget about it.70 yen number. But you don't have to worry about that. One of these numbers will always be more than 1 and the other will always be less than 1. instead.199. note that exchange rates can be written in two ways: the number of the foreign currency in one pound.£480 = £1720. Using Method Two. you want to convert from pounds to yen. And £122 + £21. so you can see that Method Two produced an estimate that was very close to the actual marked-up price).95 to the nearest pound. But this should go without saying: yen. Likewise. For example. This is probably easier to do in your head than you think. Now imagine that you're in Switzerland.60.) This will give you the equivalent pound amount. Now determine 1% of the amount from Step 1: 1% of £120. the Times gave the following yen-to-pound exchange rates: Yen in 1 Dollar Dollars in 1 Yen 108. The beauty of this method is that it works the same regardless of what country you're in. In this instance.50 = £143. we are trying to determine 18%. Enter this in your calculator and press the memory or store button. you take out your calculator and press: (You may have to press slightly different buttons depending on your calculator model. Add this to the original price.£121. And when the exchange rate is extreme—say you see a coat in Japan with a 37. in the above example. you'd use the 108. When you travel to a second or third or fourth country. First. then.60.00. you can easily figure out sale prices and mark ups.50. is £484.90. You'll know which one to do because no matter where you are. For instance.70 0. £22 x 22 = £484. subtract that discount from the original price. That will give you the other exchange rate (the one that's more than 1).199 rounds up to £2. What is 22% of £2.20 by 18. Imagine that this morning. To figure out the sale price.300 yen price tag. in the problem outlined above.20. where one franc is worth more than one pound. Now when you see an item with a 34. imagine that you need to determine a 22% discount on a product that costs £2. euros. Let's call it £21.300 yen into the fewer number of pounds and multiplication to convert £120 into the greater number of yen.000 yen price tag—there can be a tendency to spend as if you're using play money. which is £122. For example.200 is £22. Use multiplication to make an amount bigger and division to make an amount smaller. press: Voilà! You've got the equivalent yen amount. Exchange rates fluctuate daily. This stores the exchange rate in your calculator's memory. If you happen to see only the exchange rate that's less than 1. Multiply the 1% amount by the percent that you want to compute. and the number of pounds in one unit of the foreign currency. we could use Method Two to determine a markup price by simply using addition instead of subtraction.00920 Let's use these yen exchange rates to illustrate how you can know what things cost in dollars when you're shopping abroad. If. (The exact answer is £143. 1% of £2. dollars. It may be easier to round that purchase price of £121. The only thing you have to pay attention to is whether you want to make an amount bigger or smaller. just take out your calculator and divide 1 by this number. which is £3. The discount. £484 rounds down to £480.200. For the method discussed below. £2200 . and the like are absolutely not play money.00 is £1.95 down to £120. your common sense will tell you whether the
27
. Simply multiply £1. and regardless of the exchange rate. Foreign currency can sometimes look and feel like play money. If you're travelling in Japan. we used division to convert 34. Use an easy way to figure out 18 x 20p. the conversions work the opposite way: division converts pounds to francs and multiplication converts francs to pounds.50.
28
.answer to the conversion you're doing should be larger or smaller than the amount with which you started.
Step Two suggests that you don't incur fees by withdrawing more than you have in your account. such as debit and ATM cards and online banking. there are seven steps only if you're starting the entire balancing process from scratch. Moreover. Now. Not so many years ago. And if you are able to avoid overdraft fees. you may have arranged to pay your utility bills by direct debit. and the chore of balancing the books was more difficult than it is today. we will cover three steps you can take to get your finances in order and save money: balancing your accounts. or record. this will be the new tool that you use from now on to keep your checking account in order. a good way of evening out bills. ATM withdrawals and fees. you'll really come out ahead. many people use computer programs to help them balance their accounts. Once you make a habit of reconciling your balance. for example. entering the balance from your monthly statement. This topic is of great value to everyone who needs to learn how to work with accounts effectively. All three are easy to do—all you have to do is to make up your mind to do them. The financial rewards can be substantial. Step Five asks you to read your monthly statement and compare it to your own record. but there's really nothing to it. if you need to do no more than reconcile your records with the bank statement. Most people today choose helpful tools. This sounds obvious. Regardless of any poor habits you may have had in the past. Reconciling means bringing the two documents into agreement. Turn over a new leaf and make this a habit. and buying less insurance. Neglecting this task can lead to paying unnecessary overdraft fees. Step One invites you to begin fresh by starting a new register. it's even easier. Here we will discuss a few of the important points in the seven-step approach. all checks were paper. you'll enjoy the peace of mind that comes from knowing that your finances are in order. It is important to track those expenditures so that you don't make costly overdraft errors. Step Four recommends using a calculator to figure your balance after every five or 10 transactions. If you've found it difficult to keep track of your account in the past. Unless your bank account is in order and you are already in the habit of reconciling the balance regularly. As long as you know you've got enough money in your account. keeping your books balanced will likely save you money in the long run simply because of your new organized and disciplined attitude about money. as well as the bank's automated phone service) may make the difference for you. adding to your record items that appear on your statement but not on your record. and the checks you write. and that may be an idea for you to consider. When the monthly statement
29
. This is the efficient and accurate way to compute your balance. there's no reason to do it more often. thanks to new innovations in banking. Don't forget to track any automatic withdrawals that are made on your account. making use of new banking tools (such as online bill paying and account tracking. you need to review this section. (By the way. paying off your credit cards.) This process may seem a bit tedious at first.Course Name: Maths for Everyday Life
Lesson 9: Personal Finances In this lesson. You'll be glad you did. to make checking easier and reconcile your records with the bank statements. Step Three asks you to make a habit of entering into your record all of your checking account transactions. including debit card purchases. and it is easier now than it used to be.
purchased on a 20% APR credit card who then carries the card's maximum balance of £5000 for 30 years without paying down the balance will find her outfit costs her £36. The example describes a realistic scenario. and add or subtract them from your statement balance.000. a woman with a £625 outfit.comes in the mail. Paying off the balance may seem overwhelming in the beginning. All you have to do is tighten your belt a little bit in the short term and make a commitment to yourself to forego just a little spending. If it does. Notice any fees that are being charged by your bank. You will make note of any debits and credits from your own record that don't appear on your statement. you may decide not to worry about it further. the true cost of the outfit would be "only" about £15. although many people do not adequately understand them. This exercise may also help you to track your spending and develop an awareness of where your discretionary income is going. if the discrepancy is large or troubling to you.750. Place a star next to any charges on your statement that were non-essential or luxury items. This exercise can be a helpful tool as you move forward and anticipate future purchases. Take a moment now to think about your credit card spending. However. there's hardly anything you could do—short of winning the lottery—that will so significantly improve your financial future for so little effort as making up your mind to gradually pay down your credit card balances.750 amount is computed in future pounds. If it doesn't. If you have never taken the time to carefully review your statement. but it will be very helpful for you to get into the habit of reading it. For example. it may be your habit to file it away. not today's pounds. and pay off the credit card debt. Granted. But no matter how you look at it. understanding this concept and what it implies is the most important thing for you to learn in this course. proceed to Step Seven. or clothes that you really didn't need. impulse purchases. this example ignores inflation. carrying credit card debt month after month can have devastating financial consequences. You may choose to just accept it and make the necessary "fudge" correction.
30
. the £36. and if the discrepancy is a small one. You will save yourself a great deal in the long term. but many people have carried a credit card balance for the past 10 years. sit down in a quiet place and review your credit card statements for the past three months. such as treats. If you carry a credit card balance month after month without paying down the balance. The Cost of Buying on Credit If you carry a balance on a credit card month after month without paying down the balance. This illustration drives home the financial consequences of credit card spending are very serious. but it is a manageable task if you make a commitment and stick to it. For dramatic effect. if you get the same discrepancy a second time. Nothing else even comes close in terms of the long-term consequences for your life. you're done. and make sure that you understand why you are being charged. If you take inflation into account.000 or £20. and that is a third of the way to 30 years. If you are carrying credit card debt. Once you become more adept at reconciling your books. This corrected statement balance should agree with the balance in your own record. 30 years is a long time to carry a credit card balance. there will probably not be many surprises of this sort. Step Six is to reconcile your own record with your monthly bank statement. Step Seven asks you to check your work on Step Six. you may choose to call your bank and ask for help in explaining the discrepancy. this may be a revelation to you.
(The best way to do this is to keep separate ledgers: Keep one for items such as consumer electronics. You might never recover financially from the loss. you'd be out of luck. People who create.) Subtract from this running total the costs that you incur because of the insurance you decided not to buy. and as long as you feel comfortable with the amount). Keep a running total of the amount that you saved by not buying such insurance. glasses. (the higher the better. It doesn't. The bottom line is that you should not buy insurance to protect against losses you could afford to pay for yourself. Imagine that there was no such thing as homeowner's insurance. They understand how insurance works for things like their houses. To neglect to insure your home would be foolish and irresponsible. but one that's misapplied when purchasing a VCR. a good idea. and compute how much you will end up paying for that £100 item. Think about £100 spent today on a luxury item purchased on credit. This applies to items such as consumer electronics. contact lenses. it will be helpful to curtail that practice as you begin to pay off the balance. Every time you're considering buying insurance for minor purchases. Don't buy it. Millions are wasted insuring small-value items. But with the great invention of insurance. This is the mistake that many. Record the amount that you saved on insurance. don't buy that insurance. items being mailed. This is what insurance is for: to protect you from devastating financial losses that you could not afford to suffer. thousands of homeowners are able. such as when you are buying a new DVD player. Ideally. and do the maths: Compute how long it will take you to pay off that £100 purchase. You may choose to try the following exercise. Insurance is a great thing. We couldn't get along without it. just because it makes sense to insure big items. and so on. One reason they can afford to offer such good bargains is that they know that many customers will insure their purchased items (often at the checkout).If you are compounding your long-term credit card debt with unnecessary spending. as long as you know you could afford to repair your car if you get into an accident. one for car insurance." The amount you save can be substantial. to pool their resources. and predictable expenses. Record the amount that you save each time you pay your premiums. say. again. when calamity strikes the unlucky few. Then call your car insurance agent. market. To insure high value items like your home is. those impulse purchases and treats become much less appealing. perhaps you might never be able to afford to buy another house. and increase your excess rate from. it doesn't follow that the same logic applies to small items. For many people. the vast majority of people who try this experiment will enjoy watching their balance move further and further into the "black. But. The VCR or computer that was such a great deal becomes a lousy deal as soon as you add on the insurance. cell phones. and what a huge profit the company makes from them." which is a perfectly natural and logical desire when purchasing a home. and they think the same logic works for insuring things like their phones and VCRs. instead of just
31
. Think of those huge consumer electronics stores that sell electronics very cheaply. many people make. £200 to $500. once they fully comprehend the damage they are doing to their financial future by spending on credit. contact lenses. this financial pool can be drawn upon to rebuild their houses. in a sense. If your house were destroyed by fire. and keep track of the policies that you didn't buy. you should put away the money you've saved. etc. glasses. and sell insurance for such small items are part of the problem: They take advantage of the desire most people have to feel "safe. While your ledger may dip into the "red" now and then. It must be a challenge for some of the salespeople to keep a straight face when selling these policies because they know what a bad deal the policies are for customers.
for each of the next 25 years.58. that's £216. say. Now that you've completed this course. This is the estimated growth rate of your investments after inflation and taxes are taken into account. This product—73. or something in between. If your VCR does break. the following step is required. like the deposits. you first have to pick an interest rate between 3% and 6%. that's 73. each year. the more money will go into your account instead of into your insurance company's pocket. look up the interest rates Table for 4% and 20 years (the number of years of withdrawals). And that's the important point to remember. subsequent deposits should grow slightly to keep pace with inflation. Next. that's 43.930 of today's pounds for 20 years of retirement. 216. the value of your retirement account in today's pounds when you retire in 25 years. and you can perform mathematical tasks with assurance. In a nutshell. That equivalent amount should allow you to purchase. Let's use 4% for the following problem. look up an interest rate table for 4% and 25 years (the number of years of deposits). Lastly.930 in today's money.5. There is one slight complication with this method. which takes both inflation and taxes into account. the following inexact method. Let's say you think you can afford to save £5000 per year until you retire in 25 years. the equivalent of saving £5000 this year. The higher the excess you can comfortably risk having to pay. may be as good as any more precise method for giving you an estimate of how your investments will grow and what you'll be able to withdraw during your retirement. remember that the answer of £15. To use this shortcut method.930 is in today's dollars. This shouldn't be hard to do. So. this account will likely grow. 6% for an optimistic estimate. You will be able to move forward with a new ability to compute figures.930—is the value in today's pounds of your annual withdrawals during retirement. roughly what you can buy with £15.550. and that you will then draw down these savings for the following 20 years. and navigate through the mathematical world. you will be able to buy yourself a new VCR using the money in your self-insurance savings. What will your withdrawals be worth in today's pounds? Here's what you do. Inflation/ Taxes/ Investment Since no one can predict what the inflation rate will be in the future or what the tax laws will be. This way you won't spend the money on something else. you'll be able to withdraw the equivalent of £15. First. Since you think you can afford to save £5000 per year. Use 3% for a conservative estimate. Your actual withdrawals will be quite a bit larger. your car insurance excess even higher.930 today. manage your finances. you have a strong understanding of the fundamentals that you need to be mathematically aware in your everyday life.31. because each deposit will feel about the same as putting away £5000 today.58 x 216. will grow slightly each year to keep pace with inflation. Once it grows large enough. Now multiply that by the number of thousands you begin with at retirement. you might feel comfortable raising.5. Congratulations!
32
. Results that aren't in today's pounds like this aren't realistic or practical.tracking it. consider opening a special "self-insurance" savings account. Because the method takes inflation into account. Multiply this by your annual deposit of £5000. Over time. But each withdrawal will be worth approximately £15. this shortcut method tells you that if you can afford to save. and the withdrawals. after your first deposit of £5000. you should be able to save £5000 in today's pounds per year. or £15. | 677.169 | 1 |
Covering material usually found in a linear algebra class, this text is written from the point of view that knowing why a mathematical fact is true is just as important as how it is true. Abstract concepts are given as they are needed. Featured are areas on development of vector spaces, applications, conceptual exercises. An emphasis on geometry and a selection of computer exercises are provided throughout the text. A student supplement is available that contains extra hints, worked out examples and computer exercise translations | 677.169 | 1 |
ENGINEERIN Documents
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Chapter 1
Limits and Continuity
1.1
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The limits is the fundamental notion of calculus. This underlying concept is the thread
that binds together virtually all of the calculus you are about to study.
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Chapter 5
Applications of Definite Integral
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Area Between Two Curves
In this section we use integrals to find areas of regions that lie between the graphs of
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Consider the region that lies between two curves y = f (x) and y = g(x) and bet
Chapter 6
Techniques of Integration
6.1
Integration by Parts
Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. The rule
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Chapter 2
Differentiation
2.1
Tangent Lines and Rates of Change
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Tangent Lines
If a curveC has equation y = f (x) and we want ti find the
tangent to C at the point
P a, f (a) , then we consider the nearby point Q x, f (x) , where x 6= a, and compute
Chapter 3
Applications of Differentiation
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In this section we shall study related rates problems. In such problems one tries to
find the rate at which some quantity is changing by relating it to other quantities whose
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Chapter 7
Infinite Sequence and Series
7.1
Sequences
A sequence can be thought of as a list of numbers written in a definite order:
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MA111: Prepared by Asst.Prof.Dr. Archara Pacheenburawana
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Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of
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Keywords
Match the word to its meaning using
arrows
Atom
Two or more different atoms that are
chemically joined
Element
Two or more atoms chemically joined
Compoun
d
Mixture
Molecule
The smallest particle
Made of only one type of atom
Two or more differen
Proteins
Enzymes as Biological
Catalysts
Increase reaction rates by
over 1,000,000-fold
Two fundamental
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Increase the reaction rate
with no alteration of the
enzyme
Increase the reaction rate
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Reduce the activatio
Separating mixtures
Cut out and stick the correct picture with its
name and what it separates into your book
Chromatograph
y
Separates liquids with
different boiling points
Distillation
Separates a liquid and solid
by boiling off the liquid
Filtration
Sep
Chapter 1
The Science of Chemistry
Chemistry
study of the composition,
structure, and properties of matter and
the changes it undergoes
Chemical any substance that has a
definite composition
Made of the same stuff no matter where it
comes from
Chemical
C1 - The Particulate Nature of
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C2 Experimental Techniques.
On your whiteboards:
1. Forge the signature below
2. How could you identify a
forged signature?
3. Make a note of the potential
differences of the ink and paper
of an original and
Separation techniques &
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CHOOSE JUST ONE ANSWER
A, B, C, OR D
Question 1
Which of these solids dissolves in water?
Sand
chalk
Salt
Sulphur
Wrong Answer!
salt
Question 2
When can filtration be used?
To separate a
solid from a
liquid
To separate
two
On your whiteboards:
1. Forge the signature below
2. How could you identify a
forged signature?
3. Make a note of the potential
differences of the ink and paper
of an original and a forged
signature
A forged signature might be identified by:
1. The signa
Chemistry GCSE practicals guide for England
From September 2016 there are new practical requirements for GCSE science in England and Wales.
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A solution of hydrogen peroxide is mixed with one containing potassium iodide, starch
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Enthalpy change of reaction
Shape of orbitals
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Electrons can only occupy so-called atomic orbitals with well defined
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Electronegativity of C: 2.5
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Why is fluorine more electronegative than carbon?
A simple dots-and-crosses diagram of a C-F bond is perfectly adequate to explain it.
The bonding | 677.169 | 1 |
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GRE Mathematics Subject Test
The Mathematics Subject Test of GRE is developed with the idea of determining the capabilities of a student with respect to Mathematics as a subject. The Mathematics
Subject test is now called the Mathematics Test (Rescaled). Though the
general test of GRE does have a quantitative section which tests the
mathematical skills of a student, it does not suffice. The purpose of
the Mathematics subject test is that
the knowledge and caliber of a student for high school level
mathematics can be checked. In the quantitative section of the general test the basics of mathematics are tested whereas in the Mathematics Test an advanced level of mathematics is tested. The scope of the Mathematics test is so as to gauge the knowledge and skills that a student must have in order to pursue mathematics as a subject in graduation.
This subject test has 66
questions. All the questions are multiple choice questions and they are
to be answered in two hours and fifty minutes. These multiple choice
questions of the Mathematics Test
typically have five answer choices out of which you have to choose the
correct answer. There will rarely be more than one question clubbed
together. There will be extra space provided for rough work.
The questions in the test are based on algebra, calculus and some other topics of mathematics. 50% of the questions in the test
are based on calculus. Calculus includes differential and integral
calculus (in one and more than one variables). In addition there will
be questions based on the relation of calculus with trigonometry,
coordinate geometry, differential equations and other branches of
mathematics. Some questions will also be application based.
25% of the questions in the Mathematics test
are based on algebra. Apart from the elementary algebraic techniques
which are widely used there will be questions based on linear algebra
and abstract algebra and number theory. Linear algebra includes matrix
algebra, systems of linear equations, vector spaces, linear
transformations, characteristic polynomials and Eigen values and Eigen
vectors.
Abstract algebra and number theory includes group theory, the theory of rings and modules, field theory, and number theory.
25% of the questions of the Mathematics test are
based on additional topics. These include topics from introductory real
analysis which are: sequences and series of numbers and functions,
continuity, differentiability and integrability and elementary topology
of real numbers. Also there are topics from discrete mathematics which
include logic, set theory, combinatorics, graph theory, and algorithms.
Further the questions are based on general topology, geometry, complex
variables, probability and statistics, and numerical analysis | 677.169 | 1 |
Financial Products provides a step-by-step guide to some of the most important ideas in financial mathematics. It describes and explains interest rates, discounting, arbitrage, risk neutral probabilities, forward contracts, futures, bonds, FRA and swaps. It shows how to construct both elementary and complex (Libor) zero curves. Options are described, illustrated and then priced using the Black Scholes formula and binomial trees. Finally, there is a chapter describing default probabilities, credit ratings and credit derivatives (CDS, TRS, CSO and CDO).
This text includes the following chapters and appendices: Introduction to MATLAB, Root approximations, Sinusoids and complex numbers, Matrices and determinants, Review of differential equations, Fourier, Taylor, and Maclaurin series,…
Scientific computing is the study of how to use computers effectively to solve problems that arise from the mathematical modeling of phenomena in science and engineering. It is based on mathematics, numerical and symbolic/algebraic computations and visualization. | 677.169 | 1 |
Math Prerequisite Flowchart
Mathematics is a cumulative subject and builds on previously learned concepts. The arrows indicate that the previous class is a prerequisite for the next class. For example, Calculus I (Math 1600) is a prerequisite for both: Calculus II (Math 1610) and Logic (Math 2000). | 677.169 | 1 |
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CAS in ME: Theory and practise PowerPoint PPT Presentation
CAS in ME: Theory and practise. Paul Drijvers Freudenthal Institute Utrecht University P.Drijvers@fi.uu.nl. Theories concerning CAS use. 1. Specific local theories, originating from CAS research 2. Originating from ME research in general, and applied to CAS use
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CAS in ME: Theory and practise General theories on ME
Practice
How can theories help us in interpreting and understanding student behaviour?
How can we link theory and practice?
Let us try in the case of a recent observation.
Came2001 1907-2001
Assignment
Given are functions y with Here a stands for a number that can also be negative, or a fraction.A. Sketch a 'comic' that indicates how the graph of the function changes as a gets bigger.B. What values of a are 'special'? Why?
Came2001 1907-2001
The case of Maurit (1)
P:For a = 0 you have a straight line. Can you see this in the formula, too?
M:Eh, no.
P:That's a pity.
M:Yeah, but with the calculator, I think it is much more clumsy, because normally I understand it very well, but such a formula, I don't see much in it if I just enter it into the calculator and it draws the graph.
Came2001 1907-2001
The case of Maurit (2)
P:And if you just look at it, without calculator, you take x, add a times the square root of x^2+1, what happens then if a = 0?
M:Well then it gets straight but I really don't know why, no idea.
P:What happens with a times that square root if a equals zero?
M:Ehm, well then the square root will be zero as well?
P:Yeah, so what will be left of the formula in fact?
M:x + a times x^2 +1, isn't it?
P:But a was zero, remember?
Came2001 1907-2001
The case of Maurit (3)
M:Yes.
P:And in this case
M:Let's look, well then, … well the square root is then zero en the square, yes zero squared is also zero, so in fact, then I think this complete part is skipped, or not?
P:And what will remain?
M:Eh, x + a times … +1 or something?
P:No x isn't zero but a equals zero, isn't it?
M:… O yeah … well then, then I think the square root is dropped.
P:Yes.
Came2001 1907-2001
The case of Maurit (4)
M:And the rest remains.
P:Yes, and what is the rest then?
M:Well x + a times x^2 +1, .. , or not?
P:But a was zero?
M:O then it is eh x + x^2 +1
P:No, because eh it says, for this a you should read a zero in this case,
M:mmm.
P:If a = 0, then you get x + 0 times, a whole part.
M:Yes.
P:But how much is zero times a whole part?
Came2001 1907-2001
The case of Maurit (5)
M:Zero.
P:Yes. So what will be dropped?
M:In fact the complete last part?
P:Yes
M:O.
P:So what will remain?
M:x + a?
P:No, because a = 0, yes, so
M:x.
P:Yes. Are you guessing now or eh?
M:No, I really think so.
Came2001 1907-2001
The case of Maurit (6)
P:OK, I also really think so.
M:Then it is only x.
(…)
M:O I understand it, that's why it is so!
P:Yes.
M:Yeah but I think it is a bit strange because normally you have a graph and you draw from point to point but here you suddenly have for each a a different graph. | 677.169 | 1 |
Product Description:
This is a widely accessible introductory treatment of infinite series of real numbers, bringing the reader from basic definitions and tests to advanced results. An up-to-date presentation is given, making infinite series accessible, interesting, and useful to a wide audience, including students, teachers, and researchers. Included are elementary and advanced tests for convergence or divergence, the harmonic series, the alternating harmonic series, and closely related results. One chapter offers 107 concise, crisp, surprising results about infinite series. Another gives problems on infinite series, and solutions, which have appeared on the annual William Lowell Putnam Mathematical Competition. The lighter side of infinite series is treated in the concluding chapter where three puzzles, eighteen visuals, and several fallacious proofs are made available. Three appendices provide a listing of true or false statements, answers to why the harmonic series is so named, and an extensive list of published works on infinite series.
REVIEWS for Real Infinite | 677.169 | 1 |
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Summary
The BPB team has created a book where the use of the graphing calculator is optional but visualizing the mathematics is not. By creating algebraic visual side-by-sides to solve various problems in the examples, the authors show students the relationship of the algebraic solution with the visual, often graphical, solution. In addition to helping students visualize the math with side-by-sides, the authors focus on helping students make the connection between x-intercepts, zeros, and solutions, both visually and algebraically. | 677.169 | 1 |
Statistics
This episode focuses on Statistics as the study of the collection, organization, analysis, interpretation, and presentation of data. It also includes the planning of data collection in terms of the design of surveys and experiments.
This episode focuses on Geometry (Ancient Greek: geo- "earth", -metron "measurement") as a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space.
This episode focuses on solving systems of linear equations by solving the value of one variable in terms of the other. Situations and examples in the episode will give students ideas on substituting less expensive goods for costly ones.
K High Math 3 - Exploring Set Concepts, Set Operations, Venn Diagrams and Sets Part 1
This episode focuses on the sources of energy that mother nature provided for us. It explains the flow of energy from its sources down to different equipment and devices. It further explains how these energy sources affect the growth of population and urbanization of a particular community.
This episode talks about angles as the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. In determining the type of angles, relationship of facts and points of measure are carefully considered.
This episode tackles Central tendency and relates to the way in which quantitative data tend to cluster around some value. A measure of central tendency is any of a number of ways of specifying this "central value." | 677.169 | 1 |
Cambridge IGCSE Mathematics: Core Practice Book
Cambridge IGCSE Mathematics: Core Practice Book offers a wealth of questions, with hints and tips along the way to reinforce skills and learning. It provides comprehensive and targeted exercises ensuring plenty of practice both for the classroom and for independent learning.
With concise reminders at the start of each topic, and hints in the margin, it is designed to work as a stand alone practice tool that will successfully accompany any IGCSE mathematics textbook. Those using the Cambridge Coursebook will find this a complementary asset with extra questions. For additional exam practice, access to full exam-style papers is also available.
Emphasis is on the main two Assessment Objectives: Maths techniques (mastering maths skills through practice) and Application of maths techniques to solve problems (problem solving in real-world and mathematical contexts).
Like-for-like chapter and section sequence to the Coursebook, offering targeted practice within the classroom or for home study.
Packed full or interesting and progressive exercises.
Summary of key points listing what students need to know to do the exercises.
IGCSE Mathematics Online is our brand new interactive online learning and teaching tool with assessment resources designed for both students and teachers. When used in conjunction with the Cambridge IGCSE Mathematics series written specifically for Cambridge IGCSE Mathematics 0580 this resource offers a fully blended learning and teaching solution, or alternatively can be used to supplement teaching at IGCSE Level. IGCSE Mathematics Online is a high quality flexible tool with lessons, tasks, questions, quizzes, widgets and games intended for both individuals or whole classes, for school and at home. | 677.169 | 1 |
How
to Prepare For Math Exam:
Students, Are you want to know
that How to Prepare for Examination of theMathematics if yes then you are on
correct
stead & right article. Mathematics is one of those subjects which
consume our time that is very required time for all to solving its problems or
doubts. We front various problems specially teenagers and younger's who have a
fear about mathematics but it is true & true that mathematics is most essential
to all like in market, domestic budget our full working and digest of work.
Therefore we may tell that mathematics
is joined portion of daily life. In Mathematics all work complete in practice
when we do well practice than our computation should have better than better therefore
we may tell math subject is compulsory for all student or persons.
How to Prepare
For Mathematics Exam – Details:
Elements of mathematics:
The whole syllabus divided into three parts-
Arithmetic
Algebra
Geometry
All mathematics is depend on these
three parts
Arithmetic:
Arithmetic called daily life mathematics whose contain our regular
domestic budget and scale the quantum of our work. in this math having simple
addition, subtraction, multiplication and division, and BODMAS sums, disadvantage
and advantage, simple interest, compound interest, percentage, work and days,
time and distance, boat and stream, power and exponents, surds, square
roots and cube roots etc.
Algebra:
Algebra is the very imperative of portion of mathematics in this
we have to red or study monomials, binomials trinomials, quadratic equation, Bi
quadratic equation, linear equation, factors etc.
Geometry:
In geometry we read all definitions related to-
Triangles
On the bases of sides and angles
Circle
Square
Rectangle
Rhombus
Perimeter and area
Sphere
Semi sphere
Cone
Cylinder their volume and areas etc
We attempt for good injunction on mathematics attend the
coaching class of math and try self study or read and take assist your parents,
friends, school or college teacher and books. Lots of books are also
available in market for short tricks & good preparation. In practice of
mathematics subject all square and cube should have learn or study very well
approximate 30 and our computation is very firm to solve all problem or doubts
than we do our work very easy. Learn all formulas like as momentum and time, circumference,
area, definition of angels, profit and loss then our work is sole very easily
& simply.
Some Important Formulas:
Perimeter of rectangle =2 x [length +breadth]
Area of square = side x side or (Side)2
Profit %= profit x100/cost price
Volume
of cuboids= length x breadth x height or (lxbxh)
These formulas are very helpful for candidate's calculation. We can
listen music to attempt calculation easily and to create a relaxing environment
which incites the flow of details. Having suitable stead of study or
learning, candidates may obtain maximum concentration. Of course, you must
steer clear of Pit-bull and Eminem, instrumental music is the best thing
in these times.
Note:
Dear aspirants if you have gain required
news about given topic then visit our website to get further details of all
kinds of newest jobs available in India nation. | 677.169 | 1 |
STEM-Prep Pathway
The STEM-prep pathway reenvisions the algebra-intensive course sequence through the lens of what will best serve students who are going on to calculus or who need the strong algebraic skills required in technical fields.
Why a new pathway?
As mathematics educators, we continually examine our own practice for ways to better meet the needs of our students in a changing world.
Recent research on the conceptual understanding and skills necessary for calculus and on how students learn inform the collaborative design of this more efficient and effective efficient pathway.
In January 2014, we embarked on the design of an improved STEM-prep pathway by collaborating with faculty and other experts from around the country.
This pathway will be supported by two new courses, Reasoning with Functions I & II, which prepare developmental-math-level students seeking STEM careers to enter calculus and succeed in future STEM coursework or other technical courses that require strong algebraic skills and a mastery of functions. | 677.169 | 1 |
Free problem-solving mathematics software allows you to work through more than 500 math problems with guided solutions, and encourages to learn through in-depth understanding of each solution step and repetition rather than through rote memorization. Tutoring material is available in three languages. A translation option offers a way to learn math lexicon in a foreign language. Test preparation options facilitate development of printable math tests and homeworks, and automate preparation of test variants around each constructed test. Covered subject areas are arithmetic, pre-algebra, algebra, trigonometry and hyperbolic trigonometry. This includes basic math and advanced topics, such as solutions of linear, quadratic, biquadratic, reciprocal, cubic and fractional algebraic identities, equations and inequalities, solutions of trigonometric and hyperbolic equations and more. The software supports a number of interface styles. This program is a trial version of the EMTeachline mathematics software7)
Free Number Base Converter 2.0
The Number Base converter is a tool which converts numbers from one base to another such as binary,decimal,hex,octal. The program is quite simple so even if you are a beginner you can use it without any trouble.
910)
MultyGraphiX 1.0
MultyGraphix designed to plot any mathematical 2D and 3D, explicit and parametric functions. It's allow to plot function in the Cartesian, polar, cylindrical and spherical systems of coordinates. It's useful for educators, and students.
New Mathematics / Home & Education programs
1)
Robot4 0.73
Robot2)Golden Rectangular Solid 1.03)Automaton Lab 3D Quantum 1.0
The automata which are modeled in this application are composed of a set of spheres whose size and axis are relative to one another, and where each sphere is rolling upon the surface of one other sphere in a fully deterministic pattern.
4)
MITCalc - Profiles Calculation 1.18
The calculation solves area characteristics of profiles and mass characteristics of solids created by extrusion or
rotation of the profile. Application is developed in MS Excel, is multi-language and supports Imperial and Metric
units.
5)
Lissajous 3D 3.0
Male6)
BREAKTRU QUICK CONVERSION 9.8.0
Lite version converts several units of length. Plus version converts length, weight and capacity measures. By typing a number into box provided will instantly display the results without the user having to search through a confusing menu of choices.
7)
BREAKTRU FRACTIONS N DECIMALS 9.6.0
Add8)
BREAKTRU PERCENT 6.6.0
Calculate the percentage of any number out of any number. Add percentage to an amount or calculate a discount percentage.
For tip calculation. Figure the selling price after a discount was applied. | 677.169 | 1 |
UEB Under the Hood: An Honest Look
Executive Summary TheOne reason teachers favor a unified code is because they hope it will be more consistent, and thus easier for students, than separate codes.
However, the UEB has separate modes which sometimes require the same information to be represented in different ways.
UEB mathematics is entirely different from Nemeth and is untried
in the classroom. It is hoped that the information in this article
will allow experienced braille teachers to anticipate some of the problems
that might be encountered were the UEB to be adopted as the basis for teaching mathematics. Feedback from teachers of early braille learners is most crucial
since many experts have already agreed that the UEB
may be inferior to the Nemeth code for
advanced users.
Areas of special concern for early braille learners
mode-switching may be required for worksheets
alignment in spatial arithmetic requires more planning ahead
expanded basic symbols are not "alphabetic", i.e. not one-for-one with print
numeric mode (use of letters as numbers) less print-like than
separate cells for letters and numbers
harder for sighted readers to dictate, no associated MathSpeak,
reader has to anticipate more
upper numbers are inconsistent with current approaches to the
teaching and use of electronic calculators
upper numbers are inconsistent with direct entry:
direct entry of braille from a standard keyboard (ASCII Braille) or
direct entry of print from a braille keyboard (computer braille)
cannot be based on upper numbers (same characters used for digits and letters)
mode-switching in word problems
contracted email and Web addresses cannot be cut and pasted;
some will be ambiguous
Background
TheThe UEB also includes representations for most of the symbols defined in the Computer Braille Code. However, since the UEB
uses the same cells for letters
as for digits, it is not possible to design an ASCII Braille or computer braille table like the ones built into notetakers that
would be consistent with the UEB. (Appendix E An ASCII-Braille Machine Code or Font Based upon UEB of the latest UEB Project Report suggests the
use of dot-6 numbers to address this issue.)
Introduction
One reason teachers favor a unified code is because they believe it will be more consistent, and thus easier for students, than two separate codes with two different ways of representing the same information.
An item that users may find surprising is that although the UEB is a single code, it has different modes. In fact, there are many cases where the UEB has different ways of representing the same information depending on the mode. These cases arise when technical material is embedded in ordinary text
and also in other situations.
One code but different modes = different representations
A common example of embedded technical material occurs when chemical formulas such as H2O, which is the well-known formula for water, are embedded in an ordinary contracted braille sentence. This formula is typeset in print with the two as a subscript. Since the UEB subscript indicator, (dots 26), is the same cell as the contraction for (en), the UEB requires a Grade 1 symbol indicator (dots 56) to be placed just before the subscript indicator when a subscripted item occurs in an Grade 2 sentence. However, the Grade 1 symbol indicator is not used for chemical formulas and other subscripted items that appear in displayed expressions where the Grade 1 passage indicator is already in force.
Same in print, different in UEB
There are also a number of cases
where UEB has different ways of representing information that
has the same print represention.
Numeric fractions vs. symbolic fractions
Simple fractions where the numerator and denominator are both numbers don't use fraction indicators and use a special numeric fraction line, (dots 34). However, simple fractions involving letters or symbols do use fraction indicators and use the general fraction line, (dots 456 34).
Phantom enclosures
Subscripted and superscripted items that don't qualify as single items require the insertion of braille grouping indicators.
Quotation marks
The UEB even has cases where it uses two different braille symbols for the same print character. The UEB specifies that two braille symbols, (dots 236) and (dots 356), be used to represent the one (neutral) double quotation mark character on a standard keyboard. The UEB has new symbols, (dots 6, 236) and (dots 6, 356), for the so-called smart double quotation marks. (A similar situation applies for the various single quotation marks as well.)
Modes
The UEB unifies contracted braille and a mathematics code through the use of modes. At any given point in a UEB text, the outer mode is either Grade 1 mode or Grade 2 mode. (There are also numerous inner modes. For example, the numeric mode can occur within either Grade 1 mode or Grade 2 mode and the Grade 1 symbol mode can occur in the middle of the Grade 2 passage mode. )
The name Grade 1 is a bit misleading; Grade 1 mode is really a "higher math mode." The Grade 1 mode indicators are used to signal when cells that are normally used for contractions have special, mathematical meanings such as when dots 12356 is used as the fraction open indicator instead of as the contraction for "of".
UEB Grade 1 mode doesn't mean quite the same thing as uncontracted. Braille text is permitted to be uncontracted in either Grade 1 or Grade 2 mode but it is required to be uncontracted in Grade 1 mode. In other words, braille is only permitted to be contracted in Grade 2 mode.
Early Mathematics in Grade 2 Mode
The UEB design attempts to minimize the need for early braille learners to switch out of Grade 2 mode for beginning mathematics by the use of two related strategies.
Numeric mode
First is the use of numeric mode, which is activated by the number sign in either Grade 2 or Grade 1 mode. In numeric mode, the letters a-j stand for the digits. Numeric mode also converts the meaning of (dots 256) from a period to a decimal point and the meaning of (dots 34) from the (st) contraction to a division slash so that simple numerical fractions, but not other fractions, can be handled in Grade 2 mode.
Expanded mathematical symbols
The second way that UEB avoids the need to switch out of Grade 2 mode for simple mathematics is by using expanded (two-cell) symbols for the key characters—plus, minus, times(dot), general fraction line, and the left and right parentheses—used in all levels of mathematics.
A New Look at Some Old Criticisms
These two strategies—the use of the same cells for letters and numbers and the use of two-cell rather than one-cell symbols for the key special characters—are perhaps the most controversial aspects of the UEB.
This controversy is usually framed in terms of indicator clutter and longer expressions such as the quadratic formula's requiring 29 cells in the UEB rather than the 22 required in Nemeth mathematics. However, there are other issues, which have not been as well researched, that may turn out to be more serious.
Expanded symbols may be more difficult for early readers who do
better with alphabetic braille that uses one braille cell for one
print symbol
The use of letters as numbers may create conceptual problems
for some learners
The less print-like symbols may impede interactions with teachers, other students, and family members
Miscellaneous Issues
UEB mathematics is completely different from Nemeth mathematics and there
are many
additional items which need to be carefully reviewed. This section includes short descriptions of some of these items.
Spatial Mathematics
Spatial mathematics requires the number sign to be used with all numbers,
including those appearing in results. Additional samples are needed in order
to understand how these changes will affect teaching. This is an area where the less print-like approach is expected to impede interactions with others.
Spoken Form
The Nemeth code has a natural spoken form which sighted persons, with no familiarity with either math or braille, can usually
learn in no more than 15 minutes. This spoken form allows sighted readers, parents, and teachers to
dictate print math to Nemeth users who then directly enter the corresponding
braille cells. According to the researchers currently developing a
MathSpeak™,
a fully-automated version of spoken Nemeth,
"[The Nemeth] code ... allows a student superior access to mathematics by conveying the information unambiguously and concisely using a special grammar and lexicon unique to mathematics."
There is currently no spoken form specified for the UEB math. It is
difficult to anticipate how this could be achieved given the need for
the reader to anticipate the UEB mode indicators.
Calculators and computers
The use of electronic calculators and of computers—even
just using internet-based search engines—often requires
the braille user to directly enter ASCII characters, including the
ASCII digits. Braille users
can now enter ASCII characters directly from braille notetakers by
using computer braille in "translate off" mode. Computer braille is
quite similar to Nemeth. However, the use of upper numbers precludes
the possibility of specifying a computer braille that is consistent
with the UEB.
The UEB represents email and web addresses in contracted braille.
This leads to ambiguity in some cases. Contracted braille addresses
cannot be copied and pasted directly into other applications.
Summary
Adopting the UEB in Canada would be a long and expensive process.
Teachers experienced in teaching Nemeth mathematics might find it
useful to consider
the following questions with respect to early braille learners in trying to decide whether the perceived
advantages of UEB mathematics would outweigh the disadvantages pointed out in the
present article. (These questions are modeled after those being
researched in the
ABC Study.)
What differences are anticipated in comprehension,
quantitative thinking, and mathematics achievement levels?
What differences are anticipated in setting up spatial arithmetic?
What differences are anticipated in attitudes towards mathematics?
What differences are anticipated in the quantity and quality of mathematical literacy and interactive experiences in general education classrooms, the home environment, and in the community? | 677.169 | 1 |
Mathematical Thinking and Writing: A Transition to Higher Mathematics ability to construct proofs is one of the most challenging aspects of the world of mathematics. It is, essentially, the defining moment for those testing the waters in a mathematical career. Instead of being submerged to the point of drowning, readers of Mathematical Thinking and Writing are given guidance and support while learning the language of proof construction and critical analysis.
Randall Maddox guides the reader with a warm, conversational style, through the task of gaining a thorough understanding of the proof process, and encourages inexperienced mathematicians to step up and learn how to think like a mathematician. A student's skills in critical analysis will develop and become more polished than previously conceived. Most significantly, Dr. Maddox has the unique approach of using analogy within his book to clarify abstract ideas and clearly demonstrate methods of mathematical precision.
Recommendations:
Save
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About this product
Description
Description
This fascinating work makes the link between the rarified world of maths and the down-to-earth one inhabited by engineers. It introduces and explains classical and modern mathematical procedures as applied to the real problems confronting engineers and geoscientists. Written in a manner that is understandable for students across the breadth of their studies, it lays out the foundations for mastering difficult and sometimes confusing mathematical methods. Arithmetic examples and figures fully support this approach, while all important mathematical techniques are detailed. Derived from the author's long experience teaching courses in applied mathematics, it is based on the lectures, exercises and lessons she has used in her classes.
Author Biography
In 1970 Olga Waelder was born in Moscow. She began her studies at the Technical University for Electronics and Mathematics in Moscow in 1987 from whence she received her degree in applied mathematics in 1993. From 1996 until 1999 she furthered her studies at the Technical University Mining Academy in Freiberg (Germany) from whence she received a degree in mathematics. Between 2001 and 2004 she was a research assistant at the Institute for Cartography, Dresden Technical University. In 2003 she began lecturing mathematical cartography. She continued her post doctorate studies there in 2004 and since 2005 has been a lecturer for adjustment theory. | 677.169 | 1 |
Mathematics
What Is Mathematics?
Mathematics is the science and study of numbers, shapes and spaces and how they apply to the universe in which we live. Mathematics examines operations, interrelations, combinations, generalizations, abstractions, configurations, structure, measurement and transformations.
Why Study Mathematics?
Mathematicians have the opportunity to make a lasting contribution to society by helping to solve problems that affect our daily lives. Weather forecasters, software programmers and ecologists apply mathematical principles every day in very different ways.
Perimeter College prepares students to transfer to a four-year college and offers math clubs and initiatives such as STEM or MESA to help increase academic performance. Students can take advantage of opportunities for tutoring, scholarships and grants, There also are opportunities to compete in national tests sponsored by the American Mathematical Association.
Students need basic math skills in any job, but a degree in mathematics opens the door to a variety of career opportunities. Many areas require a bachelor's degree or higher. Among the professions for which a math background is helpful are: | 677.169 | 1 |
Thursday, January 28, 2010
Periodically the Homeschool Buyer's Co-op provides access to Thinkwell courses on sale. I'm a fan of Thinkwell for high school and college prep homeschooling.
Maybe it's Professor Edward Burger that makes me like Thinkwell - he's a kinda cute, geeky, math guy and he makes me laugh. Or the fact that their course is so thorough - he obviously just loves his subject and knows it inside out. Either way, if you're looking for high school online software it's worth checking out Thinkwell.
Courses on offer this time around:
middle school, high school, and some college level math (except for Geometry, which I really wish they had), including college algebra, trigonometry and calculus
high school biology, chemistry and physical science
American government
macroecomomics and microeconomics
public speaking
Prices range from $49.95 to $84.95 with bigger discounts when there are more subscribers.
For some reason it's hard to access the deals if you just go to the Homeschool Buyers Co-op home page - go straight to Thinkwell courses on sale if you want to take a look. The current offer expires February 10th at noon (eastern). | 677.169 | 1 |
Cabri 3D
Modelling of selected math theorems and knowledge in ICT environments In this article we present some topics and ideas on how to use ICT in math education. We are concerned with the projection of solids, graphic approach to percent calculation and its application in chemistry, modeling the volume of a pyramid, geometric meaning of algebraic expression and quadratic function. We follow a simple goal -- not to solve the examples precisely but to show the implementation of ICT by using Cabri 3D and Geogebra software.
Keywords for this software
Anything in here will be replaced on browsers that support the canvas element | 677.169 | 1 |
The fun and easy way to learn pre-calculus Getting ready for calculus but still feel a bit confused? Have no fear. Pre-Calculus For Dummies is an un-intimidating, hands-on guide that walks you through all the essential topics, from absolute value and quadratic equations to logarithms and exponential functions to trig identities and matrix operations. With this guide's help you'll quickly and painlessly get a handle on all of the concepts — not just the number crunching — and understand how to perform all pre-calc tasks, from graphing to tackling proofs. You'll also get a new appreciation for how these concepts are used in the real world, and find out that getting a decent grade in pre-calc isn't as impossible as you thought. Updated with fresh example equations and detailed explanations Tracks to a typical pre-calculus class Serves as an excellent supplement to classroom learning If "the fun and easy way to learn pre-calc" seems like a contradiction, get ready for a wealth of surprises in Pre-Calculus For Dummies!
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1 Math Skills for Business- Full Chapters 1 U1-Full Chapter- Algebra Chapter3 Introduction to Algebra 3.1 What is Algebra? Algebra is generalized arithmetic operations that use letters of the alphabet to represent known or unknown quantities. We can use y to represent a company s profit or the costs of labour. The letters used to hold the places for unknown quantities are called variables, while known quantities are called constants. Variables denote a number or quantity that may vary in some circumstances. Algebra now occupies the centre of mathematics, because it could be used to solve a variety of complex problems much faster than using arithmetic methods. Many problems that mathematicians could not solve previously with arithmetic methods can now be solved with algebraic methods. As well, algebra has made it possible to apply mathematics in other areas of human endeavour such as economic planning, pharmacology, medicine, and public health. Consider this, 12 + b = 20. What is the value of b? The only quantity that can take the place of b is 8, because =20. So 8 is the true replacement value for b. What about y + y = 15? The replacement value for the first y could be any number not more than 15. However, the replacement value we pick for the first y will determine the value for the second y. If we say, for example, that the first y is 10, then the second y must be 5. As well, if the first y is 12, the second y must be 3. The reverse is also true. Try it for yourself by picking a replacement value for the second, and determine the value for first y. As we will see later, this simple example is very important for understanding the solution to equations involving two similar variables. Consider another example, 4x + 8 = 40. In this example, we are looking for a number when multiplied by 4 and added 8 to it will give 40. We can try to figure out this number through guessing and checking. Eventually we will find that 8 is the replacement for x. However, with a systematic procedure of solving equations, we can easily solve that problem without going through the throes of guessing and
2 Math Skills for Business- Full Chapters 2 checking. Before we start learning that procedure for solving equations, let us try to understand the meanings of like terms and unlike terms Like Terms and Unlike Terms Consider again, 12 + b = 20. A number or letter separated by the operation sign + is called a term. So 12 is a term; b too is a term. Letters of the same kind in an algebraic statement or expression are called like terms. For example, b + b + c = 2b + c. The two bs are like terms, so we can carry out the operations of addition on them. Look at more examples below. eg2 2b + 3b = 5b, whatever the value of b is. eg3 3y + 5x cannot be simplified two terms are different. eg4 10t - 4t + 12y + 6t = 6t + 6t + 12y (since 10t 4t =6t) = 12t + 12 y (since 6t + 6t = 12t) This is now fully simplified, since the two remaining terms are not like terms. eg 5 3t - 2t = t. Note: traditionally, mathematicians do not write 1t. They just write t. So, t understandably means 1t (one t). The general rule, however, for adding and subtracting like terms does not apply to multiplication or division. In fact, multiplication and division are done as is shown in the following examples. Multiplication/Division examples: eg 1 y y y = y³ eg2 x 2 x 3 = x x x x x = x 5 eg 3 t t = t² eg 4 Note: 4t 6t = (4 6) (t x t) = 24t² t is the same as 26 t yz = 30 y z 3. 48x² = 48 x x
3 Math Skills for Business- Full Chapters t t = = 2, because the two t cancel themselves out x 7 = = 2x, because 7 goes into 14 two times xy, is the same as 20 x y. Again, there is no need to put the multiplication sign between 20 and x or x and y. 7. y y y x x = y³x² 8. 5n²t³ = 5 n n t t t 9. To simplify 3a 2 4a 3, first multiply the numbers together to get 12 from (3 4) 2a and then add the exponents to get a 2+ 3 = a 5 This is the same as a a a a a = a. That is, keep the base a and add the exponents. We now have 12a 2a. Divide the numbers, 12 2 = 6. Then subtract the exponents when dividing, so a 5 a 1 = a 5 4 This equals a. This is the same thing as a x a x a x a x a a = a. The final answer is 6a 10. = = note: We multiplied the top numbers to get 4 ( 2 x 2) and the number letters to get t². 11. = = note that the exponent 3 tells us the number of times the fraction should be multiplied. We multiplied the top numbers to get 1 (1 x 1 x 1) and then the bottom numbers to get 27 (3 x 3 x 3) ³ = 5 x 5 x 5 = Substitution and Formula When we put quantities in place of variables it is called substitution. Consider this, what is b + c + d, where b = 4, c= 2, d= 3? We simply substitute each variable with its corresponding quantity. b + c + d = = 9
4 Math Skills for Business- Full Chapters 4 Example 1 Evaluate, 6x + 8y + 3z, where x = 2, y = 3, z = 4 Example 2 =6(2) + 8(3) + 3(4) = = 48 Evaluate 12a 14b, where a =3, b= 1 Example 3 =12(3) 14 (1) =36 14 = = (dividing by 2, which is the common factor) Evaluate 2(x - 3) = 10, if x=8 Remember the distributive property! 2(x-3) = 10 2(x) - 2(3) = 10 (Distribute 2 over x and -3 in order to remove the bracket) 2x - 6 = 10 2(8) - 6 =10 (substituting 8 for x) 16-6 = 10 A formula describes how one quantity relates to one or more other quantities. A formula is a shorthand form of procedure for doing calculations. Al-Khwarazmi gave the world algebra by using letters of the alphabet to represent unknown numbers. And, then, he gave these symbols all the properties of numbers. Before this, the only way to know a procedure is to see examples of the procedure for specific numbers. For example, procedure for finding the area of a rectangle was taught by showing the procedure for calculating the area for specific rectangles. In a sense, arithmetic reasoning was more of experience rather than deductive reasoning. However, once a formula is constructed, it can be used to calculate unknown quantities. Formulas too are used to calculate quantities in businesses. A common formula is P = R -E, where P= profit, R= revenue (sales), and E = Expenses.
5 Math Skills for Business- Full Chapters 5 The formula says that profit is equal to revenue subtract expenses. Another common formula is E = A - L, where E = owner s equity (the capital the owner has invested in the business), A = assets (Resources such as machines, furniture, buildings, etc.), and L = liabilities are amounts owed to others. As well, when S = VC + FC, this is called the break-even point, where S =sales, VC= variable cost, and FC = fixed cost. The following are other formulas used in businesses: Net sales = Gross sales sales returns and allowances NS = GS SRA Cost of goods sold = Cost of beginning inventory + cost of purchases cost of ending inventory VC = Cost per unit x Quantities produced VC = CPU x Q Total cost = VC + FC, or TC = VQ + FC The Construction of a formula is a straight-forward process. First, you have to know the purpose for which you want to use the formula. Second, you have to understand the information you will be using to construct the formula. Third, with little knowledge of algebra, particularly substitution, you should have no problem constructing a formula. Lastly, test your formula to make sure that it works. Example 1 A technician charges a basic fee of $40.00 for a house visit plus $15 per hour, when repairing central heating system. Construct a formula for calculating three hours of the technician s charge. Solution Let C = charge, and n = number of hours worked. The charge is made up of a fixed cost of $40 and $15 times the number of hours. This translates into the following: Charge = $30 plus $15 times the number of hours worked. C = n
6 Math Skills for Business- Full Chapters 6 Note: no need to write the multiplication sign between 15 and n. To calculate the charge for three hours we substitute 3 in place of the n. C = (3) = C = C = $75 Therefore, the charge for three hours is $ Example 2 A window cleaner charges a fee of $20 for visiting a house and $10 for every window he cleans. a) Write a formula for finding the total cost, C, in dollars, when n windows are cleaned. b) Find C if n = 8 c) If the window cleaner wants to earn an income of $220 a day, how many windows does he have to clean? Solution a) Total cost = fixed charge of $20 plus $10 times the number of windows cleaned. C = $ n b) C = n C = (8) by substituting n=8 C = $100 c) C = $ n is the original equation we constructed. 220 = n (divide each term by 10) 22 = 2 + n 22-2 = n Therefore, the window cleaner must clean 20 windows in order to earn $220 a day. However, this question assumes that the window cleaner will get enough cleaning job to earn the $ Example 3. Julie earns $ x in her first year of work with AP Films Ltd. Her salary is increased by $650 every year. How much will she earn in a) the 4 th year? b) the 6 th year? c) the nth year?
7 Math Skills for Business- Full Chapters 7 Solution a) 1 st Year Salary = $ x + $650(n-1) (x is her earning, and n is number of years) Note that she does not get any raise in the first year that is why we have n 1. 4 th Year Salary = $x + 650(4-1) (Remember: she gets $650 increase every Year, excluding the first year.) = $x (3) = $1, b) 6 th Year Salary = x (6-1) = x + 650(5) = x + 3,250 c) 10th Year Salary = x (10-1) Example 4 = x (9) = x + 5,850 The cost of framing a photograph is calculated using the formula, C =0.55 ( l + b) Where l = length in centimetres b = width in centimetres C = cost in dollars a) Find the cost of framing a photograph whose dimensions are 12cm by 16 cm. b) A photograph with length twice the width was framed for $ How much were the length and the width of the photograph? a) The formula is C = 0.55 (l + b) = 0.55 ( ) = 0.55(28) = $15.40 It costs $15.40 to frame the photograph. b) Let y represent the width and 2y represent the length, since the length is 2 times the width. The formula is C = 0.55 (l + b)
8 Math Skills for Business- Full Chapters 8 By substitution, we get 33 =0.55 (y + 2y) 33 = 0.55 (3y) (since 2y and y add up to 3y) 33 = 1.65y = y 20 = y Therefore, the width is 20 cm. The length is 40cm or 20 x 2. Note that once we have a formula we can use it for evaluating any values we want. However, you should always remember that constructing a formula involves three steps. First, read the information you are going to use to construct the formula carefully. Second, use letters of the alphabet in constructing the formula, bearing in mind its purpose. Finally, test the formula to make sure that it works for the purpose for which you constructed it. 3.4 Equations An equation is a mathematical statement or expression in which two quantities are equal. Similar to formulas, equations also use numbers, letters of the alphabets and operational symbols (+, -., ). In fact, formulas are special form of equations. Example 1 Do you know that, = 7 is an equation? The equality sign (=) separates the statement into left side and right side. The sum of the left side is 2+5 = 7. This is equal to the number on the right side. Example 2 Solve for N if N + 15 = 25. This is an example of an equation, because it is separated by the equality sign (=) and there is a left side and a right side. The equation means a certain number N, plus 15 equals 25. The replacement value for N must be such that when added to 15, it will be equal to 25. By inspection or guessing and checking, you will know that N
9 Math Skills for Business- Full Chapters 9 must be 10. So, = 25. However, in some cases, the replacement value for a variable may not be as simple as in this example. Accordingly, mathematicians have developed a more sophisticated procedure for solving an equation. Solve for N: N+ 15 = 25. When we subtract 15 from both sides of the equation, N = Note that = 0, because they are inverse of each other. This procedure helps to isolate the N on the left side of the equation. The right side becomes 10. So we are left with, N =10 as we figured out before. This procedure leads to the following rule: When the same number is added to or subtracted from both sides of an equation, the equation does not change. Example 3 Solve for x, if 2x + 3 = 15. The above equation reads 2 times a certain number plus 3 equals (or the same as, or equivalent to) 15. The x as a variable is a placeholder. Also, remember that there is no need to write the multiplication sign ( ) between 2 and x. To solve 2x + 3 = 15, our aim is to isolate the variable on the left side of the equation. So we subtract 3 from both sides of the equation. 2x = x + 0 = 12 We now have, 2x = 12. Again, this means 2 times a certain number equals 12. Certainly, we can guess and check to find that 6 is the replacement value for x. However, if we divide both sides of the equation by 2 we have the following situation. = So, x = = 6.
10 Math Skills for Business- Full Chapters 10 This leads us to the next rule of equation. It says that when both sides of an equation is divided or multiplied by the same number the equation does not change. Note that we can check to make sure that 6 is truly the replacement value for x. 2x + 3 = 15 which is the original equation 2(6) + 3 =15 substitute x with 6 (Remember substitution?) =15 Multiply 2 and 6 15 = 15 Add 12 and 3 to get 15. Now you see that 15 equals 15. That means we are right that the replacement value for x is 6. Example 4 Solve for x, 4x 3 +2x = 8x -3 x 6x 3 = 7x -3 We simplified the left side by adding 4x and 2x 6x = 7x x = 7x together to get 6x. We also simplified the right side by adding (8x) and (-x) to get 7x. The basic rule of integer is that when we have a large number that is positive, we always take small negative number from it. We added +3 to both sides. Do you remember the first rule of equation? An equation is like a scale. What you do to one side, you do the same to the other side. This is the result after we added +3 to both sides 6x-7x = 7x -7x We subtract -7x from both sides to isolate x. -x = 0 After subtracting -7x from both sides. -x -1 = 0-1 We multiplied each side by -1, so that we get x. x = 0 Now we are done; x is equal to 0 Let us check that the replacement value for x is 0.
11 Math Skills for Business- Full Chapters 11 4x x = 8x -3 x was the original equation. 4(0) 3 + 2(0) =8 (0) 3 0 ( We substituted 0 in the place of x) = is the result after multiplying. -3 = -3. Since the left side and right side are equal, the replacement value this verifies that x=0 is the correct answer. 3.5 Using Equations to Solve Problems Equations are a powerful tool used to solve many problems in accounting management, marketing, and the sciences. However, before we use this tool, we have to be familiar with its language. The following table will help you to become familiar with algebraic language. Addition Subtraction Multiplication Division Equality The sum of Less than Times divides Equals Plus/total Decreased by Multiplied by Divided by Is/was/are Increased by Subtracted from The product of Divides into The result More than Difference between Twice/two times Exceeds Diminished by Double(two times) Expands Take away Triple (three times) per Quotient Half of What is left What remains The same as Greater than Reduced by Half times Quarter of Gives/giving Gain/profit Less/minus Third of Makes Longer Loss Leaves Heavier Wider Taller Added to Lower shrinks Smaller than slower
12 Math Skills for Business- Full Chapters 12 Apart from learning to understand algebra language in terms of addition, subtraction, division, and multiplication, students also have to learn how to change English sentences to algebraic expressions. Since this is not entirely an algebra course, we will provide only a few simple examples of how to change English phrases into algebraic expressions. Essentially, the skills you learn in this section are what you need for both an elementary and an intermediate algebra course. Study the examples below carefully, so that you can spot these English phrases when used in framing problems. As soon as you spot them, then you have to think about how to convert them into algebraic expressions. English Phrases Eight more than five times a number Ten times the sum of a number and 5 A number subtracted from 50 Four less than a number One-half of a number Five more than twice a number Six less than two-thirds of a number Three-fourths of the sum of a number and 12 Algebraic Expressions 8 + 5n, where n is the unknown. 10 ( y + 5), where y is the unknown 50 x, where x is the unknown number. n 4, where n is the unknown n, where n is the unknown number x n - 6 ( y + 12) Fifty subtracted from three times a number 3z - 50 Now is the time to use equations to solve problems. The first thing we need to do is to read the given information carefully. Second, we have to use a variable to represent the unknowns and then represent one unknown in terms of another. After that, we set up an equation and solve the equation to find the unknown(s). Nine worked examples are provided below to illustrate the power of equations in solving problems.
13 Math Skills for Business- Full Chapters 13 Example 1 Gabby s gross monthly income is 2.5 times that of Maggie income. The sum of their gross monthly income is $7,000. What is each person s gross monthly income? Solution Let y be Maggie s income and 2.5y be Gabby s income. Together their income equals $7,000. So, y +2.5y = 7000 Adding the like terms on the left side 3.5y =7,000 Dividing both sides by 3.5, y = = 2,000 So Maggie s income is $2,000. and Gabby s income is $5,000.(2.5 x 2,000). Check the accuracy of the answer by substituting the values in the original equation in the following way: y + 2.5y = 700 is the original equation 2, (2000) = 7000 Since the value of y is 2000, we substitute it in the place of y. Remember that y is a variable or a placeholder for , ,000 = 7,000 We multiplied 2.5 by 2,000 to get 5,000. 7,000 = 7,000 We added 2,000 and 5,000 to get 7,000. The left side and the right side is the same amount, indicating that our solution is right. In fact, if we calculate 2.5 of $7000 we get $5,000. Again, it shows that our answer is right. So whatever way we check our answer, it indicates that we are right. You must develop this skill of checking the accuracy of your answer after solving an equation. Example 2 A commuter train has four double decker cars and five regular ones. Each double decker has 68 seats more than a regular one. The total number of seats on the train is a) What is the number of seats on each type of car? b) What is the total seats in each type of cars?
14 Math Skills for Business- Full Chapters 14 Solution Let x be the number of seats on a regular car. Then let x + 68 be the number of seats on a double decker. (Note: This is logical since there are 68 seats more on a double decker than a regular car). Since there 4 double decker cars, there must be a total of 4 (x + 68) seats Similarly, since there are 5 regular cars, there must be a total of 5x seats. The given information says that there are a total of 1118 seats on the train The following relationship is given: a) (#of double decker cars) (# of seats on a double decker car) + (# of regular cars) (# of seats on a regular car) = the total # of seats on the commuter train. In terms of algebraic expression, we have the following: 4(x +68) + 5x = x x = x + 5x = x = x = 846 x = = 94 Therefore the number of seats in a regular car is 94. The number of cars in a double decker car is 162 ( ) b) The total seats in the regular cars = 5 (94), since there are 5 regular cars, each with 94 seats. The total number of seats on the four double decker trains = 4(94 +68) =648 Total seats on the commuter train: = The difference is 68 (162-94), indicating that our answer is right. Do you have any other ways to check the answer? Please, note that we could also check the accuracy of our answer by writing 94 in the place of the x and solving the equation. Try this yourself. Example 3 Mr. Martins owns a specialty store. He purchased an equal number of two types of designer phones for a total of $7,200. The top quality phone costs $120 each and the plastic phones costs $80 each.
15 Math Skills for Business- Full Chapters 15 a) How many of each type of phones were purchased? b) What was the total value of each type of phone? Solution a) Let x be the quantity of top quality phones purchased. x can also represent the number of plastic phones purchased. This is the case because the same quantity of each type of phones was purchased. The following relationship is true: (# of top-quality phones) x (the price) + (# of plastic phones) x (the price)= $7200 x $ x 80 = $ x + 80x = 7, x = 7,200 x = = 36 So, 36 quantities of each phone were purchased. We can check the accuracy of our answer by substituting 36 in the place of x and evaluating as we did in substitution. Try that yourself as a practice. b) The dollar value of the Top-Quality phone = Quantity purchased x unit price = 36 x 120 =$4, The dollar value of plastic phones = Quantities purchased x unit price = 36 x 80 = $2,880 Thus, $4,320 + $2,880 = $7,200. This proves the accuracy of our calculations. Example 4 Focus Day Care Centre has 28 children and 7 care-givers. Two of the care-givers work part-time. Children enrolled under 4 years old are 8 more than those over 4 years old. How many children under 4 years old are enrolled in the centre?
16 Math Skills for Business- Full Chapters 16 Solution Let n represent the number of children over 4 years old. Let n + 8 represent the number of children under 4 years old. This so because the number of children under 4 years old are 8 more than those over 4 years old. This relationship is true: (# of children over 4 years old) + (# of children under 4 years old) = 28 n + n + 8 = 28 2n +8 = 28 2n =28-8 = n = 10 The number of children over 4 years old is 10. Thus, the number of children under 4 years old = n + 8 = = 18 Note that we could interpret the above problem in this way. Since there are 8 more children under 4 years old than over 4 years old, we can say that there are 8 less children over 4 years old than under 4 years old. So let n-8 be the number of children over 4 years old, then n can represent those under 4 years old. This will give us the following equation: n + n-8 = 28 Try to solve the above equation for n and compare your answer to the above solution. Did you get the same result? If not, you should check over your work.
17 Math Skills for Business- Full Chapters 17 Example 5 Industrialized nations have 2,017 radios per thousand people. This is about six times the number of radios per thousand in developing nations. What is the number of radios per thousand in developing countries? (Round off to the nearest whole number). Solution Let r represent the number of radios per thousand in developing countries. The following relationship is given in the question: # of radios in developing countries per thousand x 6 =2017 6r =2017 = r= r = 336 (to the nearest whole) So there are 336 radios per thousand in developing countries. We can check the accuracy of our answer in this way: 6r = 2017, then 6 (336.17) = Though this question could be solved using other methods, we used an equation to solve it to show that an equation can be used to solve any question. Its power is limitless! Example 6 Forty acres of land were sold for $810,000. Some were sold for $24,000 per acre and the rest for $18,000 per acre. How much was sold at each price? Solution The unknowns are the number of acres sold at $24,000 per acre and the number of acres sold at $18,000 per acre. Let L represent the number of acres sold at $24,000, then 40- L represent the rest sold at $18,000 per acre.
18 Math Skills for Business- Full Chapters 18 Note the following relationship is implied in the given information. (# of acres sold) x (24,000 per acre) + (# of acres sold) x (18,000 per acre) = $810,000 (L) x (24,000) + (40- L) x (18,000) = 810, L L = ,000L = L = L = = L = 15 Therefore, 15 acres of the land were sold at $24,000. And 25 acres (40-15) were sold at $18,000 per acre. Check the accuracy of the answer by substituting the two values the original equation. Example 7 A bank pays the following interest on an account. Month interest 1 $30 2 $50 3 $70 4 $90 a) Write a formula or equation relating the month to the amount of interest paid. b) How much interest will the bank pay on the 10 th month? Solution a) We have to examine the table for any patterns. If we subtract the first month interest of $30 from the second month interest of $50, we get $20. Again, if we subtract the second month interest of $50 from third month interest of $70, we get $20. If seems that this pattern is true in the table- there is a constant difference of
19 Math Skills for Business- Full Chapters 19 $20. Further, if we multiply the constant difference of $20 by the month and add 10 we get the amount of interest. Let us try this and see if it works. Month x = interest 1 x =30 first month interest 2 x =50 second month interest Therefore, let m represent the month and i the interest. We can write a formula based on the above relationship. 20m + 10 = i b) For the 10 th week, the bank will pay, 20m + 10=i 20 (10) + 10 = = $210 The bank will pay $ Example 8 Modern Restaurant sells pizza, meat pies, and chicken wings. The chart below shows the price of chicken wings. Wings Price 5 wings $ wings $ wings $5.99 a) Based on the information in the above table write a formula to show the relationship between the price and the number of chicken wings. b) How much will 50 chicken wings cost?
20 Math Skills for Business- Full Chapters 20 Solution a) Let s examine the pattern in the table. Though the difference between $3.99 and $1.99 is $2, the pattern does not look like example 7. This is because if we multiply the number of wings by 2 we get 10, the price is far greater than $1.99. This is mainly because the numbers in the left column are not consecutive integers. So, let s divide 1.99 by 5. We get Let s try if this works x 5 = We need to subtract 0.01 from $2 to get $1.99. Let w represent the number of wings, and C the total cost. Therefore, the formula becomes, $0.40w 0.01 = C. Let s test this How much will 15 wings cost? 0.40w (15) 0.01 = C = $5.99, this is the correct price in the chart. b) 50 wings cost, 0.40w 0.01 =C 0.40 (50) 0.01 =C (substitute 50 for w) =$ wings cost $19.99 formula. Example 9 A company s annual profit of $8,400,000 was allocated to reserve fund, dividend distribution, and operation expansion. The amount allocated to dividend distribution was twice the amount allocated to the reserve fund. The amount allocated to operation expansion was $400,000 more than the amount allocated to dividend distribution. Find the amount of each allocation.
22 Math Skills for Business- Full Chapters The Concept of Break-Even Point Break-even point occurs when revenue equals expenses or cost. At that point, a business organization does not make any profit or loss. That is, sales = Fixed expenses + variable expenses. In this case, sales are just enough to cover expenses without incurring a loss. The data below illustrates the concept of break-even. May June July Sales $25,000 $85,000 $125,000 Less: Variable cost 10,000 75, ,000 contribution 15,000 10,000 10,000 Less: fixed costs 10,000 10,000 10,000 Net profit $5,000 $0 $0 From the above table, you will see that the company broke even the month of June and July- it did not make any profit or loss; Sales were just enough to cover variable costs and fixed costs. The calculation of the break-even point is very important for at least two reasons: a) It helps business owners or entrepreneurs to plan for profit or analyze loss. b) It helps business owners to set prices at which they will sell their products. c) It can also be used to predict profit or loss in any business ventures. To calculate the break-even point, we have to be familiar with certain terms, which are defined below: Variable costs or expenses are the costs of labour, utilities, raw materials and any other expenses directly traced to production of goods or services. An important characteristic of variable cost is that it increases as production increases and decreases as production decreases. However, the nature of a business organization will determine what will constitute its variable cost. For example, produce grocery business s variable cost will consist of the cost produce, shipping, storage, spoilage, and wages of cashiers.Free Pre-Algebra Lesson 55! page 1 Lesson 55 Perimeter Problems with Related Variables Take your skill at word problems to a new level in this section. All the problems are the same type, so that you can
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Here are my online notes for my Calculus I course that I teach here at Lamar University. Despite
the fact that these are my "class notes" they should be accessible to anyone wanting to learn
Calculus I or needing a refresher in some of the early topics in calculus.
I've tried to make these notes as self contained as possible and so all the information needed to
read through them is either from an Algebra or Trig class or contained in other sections of the
notes.
Here are a couple of warnings to my students who may be here to get a copy of what happened on
a day that you missed.
1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn
calculus I have included some material that I do not usually have time to cover in class
and because this changes from semester to semester it is not noted here. You will need to
find one of your fellow class mates to see if there is something in these notes that wasn't
covered in class.
2. Because I want these notes to provide some more examples for you to read through, I
don't always work the same problems in class as those given in the notes. Likewise, even
if I do work some of the problems in here I may work fewer problems in class than are
presented here.
3. Sometimes questions in class will lead down paths that are not covered here. I try to
anticipate as many of the questions as possible when writing these up, but the reality is
that I can't anticipate all the questions. Sometimes a very good question gets asked in
class that leads to insights that I've not included here. You should always talk to
someone who was in class on the day you missed and compare these notes to their notes
and see what the differences are.
Here is a listing and brief description of the material in this set of notes.
Review
Review : Functions – Here is a quick review of functions, function notation and
a couple of fairly important ideas about functions.
Review : Inverse Functions – A quick review of inverse functions and the
notation for inverse functions.
Review : Trig Functions – A review of trig functions, evaluation of trig
functions and the unit circle. This section usually gets a quick review in my
class.
Review : Solving Trig Equations – A reminder on how to solve trig equations.
This section is always covered in my class.
Review : Solving Trig Equations with Calculators, Part I – The previous
section worked problem whose answers were always the "standard" angles. In
this section we work some problems whose answers are not "standard" and so a
calculator is needed. This section is always covered in my class as most trig
equations in the remainder will need a calculator.
Review : Solving Trig Equations with Calculators, Part II – Even more trig
equations requiring a calculator to solve.
Review : Exponential Functions – A review of exponential functions. This
section usually gets a quick review in my class.
Review : Logarithm Functions – A review of logarithm functions and
logarithm properties. This section usually gets a quick review in my class.
Review : Exponential and Logarithm Equations – How to solve exponential
and logarithm equations. This section is always covered in my class.
Review : Common Graphs – This section isn't much. It's mostly a collection
of graphs of many of the common functions that are liable to be seen in a
Calculus class.
Computing Limits – Many of the limits we'll be asked to compute will not be
"simple" limits. In other words, we won't be able to just apply the properties and
be done. In this section we will look at several types of limits that require some
work before we can use the limit properties to compute them.
Infinite Limits – Here we will take a look at limits that have a value of infinity
or negative infinity. We'll also take a brief look at vertical asymptotes.
Limits At Infinity, Part I – In this section we'll look at limits at infinity. In
other words, limits in which the variable gets very large in either the positive or
negative sense. We'll also take a brief look at horizontal asymptotes in this
section. We'll be concentrating on polynomials and rational expression
involving polynomials in this section.
Limits At Infinity, Part II – We'll continue to look at limits at infinity in this
section, but this time we'll be looking at exponential, logarithms and inverse
tangents.
Continuity – In this section we will introduce the concept of continuity and how
it relates to limits. We will also see the Mean Value Theorem in this section.
The Definition of the Limit – We will give the exact definition of several of the
limits covered in this section. We'll also give the exact definition of continuity.
Related Rates – In this section we will look at the lone application to derivatives
in this chapter. This topic is here rather than the next chapter because it will help
to cement in our minds one of the more important concepts about derivatives and
because it requires implicit differentiation.
Higher Order Derivatives – Here we will introduce the idea of higher order
derivatives.
Logarithmic Differentiation – The topic of logarithmic differentiation is not
always presented in a standard calculus course. It is presented here for those how
are interested in seeing how it is done and the types of functions on which it can
be used.
Differentials – We will look at differentials in this section as well as an
application for them.
Newton's Method – With this application of derivatives we'll see how to
approximate solutions to an equation.
Business Applications – Here we will take a quick look at some applications of
derivatives to the business field.
Integrals
Indefinite Integrals – In this section we will start with the definition of
indefinite integral. This section will be devoted mostly to the definition and
properties of indefinite integrals and we won't be working many examples in this
section.
Computing Indefinite Integrals – In this section we will compute some
indefinite integrals and take a look at a quick application of indefinite integrals.
Substitution Rule for Indefinite Integrals – Here we will look at the
Substitution Rule as it applies to indefinite integrals. Many of the integrals that
we'll be doing later on in the course and in later courses will require use of the
substitution rule.
More Substitution Rule – Even more substitution rule problems.
Area Problem – In this section we start off with the motivation for definite
integrals and give one of the interpretations of definite integrals.
Definition of the Definite Integral – We will formally define the definite
integral in this section and give many of its properties. We will also take a look
at the first part of the Fundamental Theorem of Calculus.
Computing Definite Integrals – We will take a look at the second part of the
Fundamental Theorem of Calculus in this section and start to compute definite
integrals.
Substitution Rule for Definite Integrals – In this section we will revisit the
substitution rule as it applies to definite integrals.
Introduction
Technically a student coming into a Calculus class is supposed to know both Algebra and
Trigonometry. The reality is often much different however. Most students enter a Calculus class
woefully unprepared for both the algebra and the trig that is in a Calculus class. This is very
unfortunate since good algebra skills are absolutely vital to successfully completing any Calculus
course and if your Calculus course includes trig (as this one does) good trig skills are also
important in many sections.
The intent of this chapter is to do a very cursory review of some algebra and trig skills that are
absolutely vital to a calculus course. This chapter is not inclusive in the algebra and trig skills
that are needed to be successful in a Calculus course. It only includes those topics that most
students are particularly deficient in. For instance factoring is also vital to completing a standard
calculus class but is not included here. For a more in depth review you should visit my
Algebra/Trig review or my full set of Algebra notes at
Note that even though these topics are very important to a Calculus class I rarely cover all of
these in the actual class itself. We simply don't have the time to do that. I do cover certain
portions of this chapter in class, but for the most part I leave it to the students to read this chapter
on their own.
Here is a list of topics that are in this chapter. I've also denoted the sections that I typically cover
during the first couple of days of a Calculus class.
Review : Functions – Here is a quick review of functions, function notation and a couple of
fairly important ideas about functions.
Review : Solving Trig Equations with Calculators, Part I – The previous section worked
problem whose answers were always the "standard" angles. In this section we work some
problems whose answers are not "standard" and so a calculator is needed. This section is always
covered in my class as most trig equations in the remainder will need a calculator.
Review : Solving Trig Equations with Calculators, Part II – Even more trig equations
requiring a calculator to solve.
Review : Exponential Functions – A review of exponential functions. This section usually gets
a quick review in my class.
Review : Logarithm Functions – A review of logarithm functions and logarithm properties.
This section usually gets a quick review in my class.
Review : Exponential and Logarithm Equations – How to solve exponential and logarithm
equations. This section is always covered in my class.
Review : Common Graphs – This section isn't much. It's mostly a collection of graphs of many
of the common functions that are liable to be seen in a Calculus class.
Review : Functions
In this section we're going to make sure that you're familiar with functions and function notation.
Both will appear in almost every section in a Calculus class and so you will need to be able to
deal with them.
First, what exactly is a function? An equation will be a function if for any x in the domain of the
equation (the domain is all the x's that can be plugged into the equation) the equation will yield
exactly one value of y.
This is usually easier to understand with an example.
Example 1 Determine if each of the following are functions.
(a)
2
1 y x = +
(b)
2
1 y x = +
Solution
(a) This first one is a function. Given an x there is only one way to square it and then add 1 to the
result and so no matter what value of x you put into the equation there is only one possible value
of y.
(b) The only difference between this equation and the first is that we moved the exponent off the
x and onto the y. This small change is all that is required, in this case, to change the equation
from a function to something that isn't a function.
To see that this isn't a function is fairly simple. Choose a value of x, say x=3 and plug this into
the equation.
2
3 1 4 y = + =
Now, there are two possible values of y that we could use here. We could use 2 y = or 2 y = − .
Since there are two possible values of y that we get from a single x this equation isn't a function.
Note that this only needs to be the case for a single value of x to make an equation not be a
function. For instance we could have used x=-1 and in this case we would get a single y (y=0).
However, because of what happens at x=3 this equation will not be a function.
Next we need to take a quick look at function notation. Function notation is nothing more than a
fancy way of writing the y in a function that will allow us to simplify notation and some of our
work a little.
Recall that this is NOT a letter times x, this is just a fancy way of writing y.
So, why is this useful? Well let's take the function above and let's get the value of the function at
x=-3. Using function notation we represent the value of the function at x=-3 as f(-3). Function
notation gives us a nice compact way of representing function values.
Now, how do we actually evaluate the function? That's really simple. Everywhere we see an x
on the right side we will substitute whatever is in the parenthesis on the left side. For our
function this gives,
(e) ( ) ( ) ( )
2
2
3 3 6 3 11 12 38 f x x x x x − = − − + − − = − + −
The only difference between this one and the previous one is that I changed the t to an x. Other
than that there is absolutely no difference between the two! Don't get excited if an x appears
inside the parenthesis on the left.
[Return to Problems]
All throughout a calculus course we will be finding roots of functions. A root of a function is
nothing more than a number for which the function is zero. In other words, finding the roots of a
function, g(x), is equivalent to solving
( ) 0 g x =
Example 3 Determine all the roots of ( )
3 2
9 18 6 f t t t t = − +
Solution
So we will need to solve,
3 2
9 18 6 0 t t t − + =
First, we should factor the equation as much as possible. Doing this gives,
( )
2
3 3 6 2 0 t t t − + =
Next recall that if a product of two things are zero then one (or both) of them had to be zero. This
means that,
In order to remind you how to simplify radicals we gave several forms of the answer.
To complete the problem, here is a complete list of all the roots of this function.
3 3 3 3
0, ,
3 3
t t t
+ −
= = =
Note we didn't use the final form for the roots from the quadratic. This is usually where we'll
stop with the simplification for these kinds of roots. Also note that, for the sake of the practice,
we broke up the compact form for the two roots of the quadratic. You will need to be able to do
this so make sure that you can.
This example had a couple of points other than finding roots of functions.
The first was to remind you of the quadratic formula. This won't be the first time that you'll need
it in this class.
The second was to get you used to seeing "messy" answers. In fact, the answers in the above list
are not that messy. However, most students come out of an Algebra class very used to seeing
only integers and the occasional "nice" fraction as answers.
So, here is fair warning. In this class I often will intentionally make the answers look "messy"
just to get you out of the habit of always expecting "nice" answers. In "real life" (whatever that
is) the answer is rarely a simple integer such as two. In most problems the answer will be a
decimal that came about from a messy fraction and/or an answer that involved radicals.
One of the more important ideas about functions is that of the domain and range of a function.
In simplest terms the domain of a function is the set of all values that can be plugged into a
function and have the function exist and have a real number for a value. So, for the domain we
need to avoid division by zero, square roots of negative numbers, logarithms of zero and negative
numbers (if not familiar with logarithms we'll take a look at them in a little later), etc. The range
of a function is simply the set of all possible values that a function can take.
Solution
(a) ( ) 5 3 f x x = −
We know that this is a line and that it's not a horizontal line (because the slope is 5 and not
zero…). This means that this function can take on any value and so the range is all real numbers.
Using "mathematical" notation this is,
( ) Range : , −∞ ∞
This is more generally a polynomial and we know that we can plug any value into a polynomial
and so the domain in this case is also all real numbers or,
( ) Domain : or , x −∞ < < ∞ −∞ ∞
[Return to Problems]
(b) ( ) 4 7 g t t = −
This is a square root and we know that square roots are always positive or zero and because we
can have the square root of zero in this case,
(c) ( )
2
2 12 5 h x x x = − + +
Here we have a quadratic which is a polynomial and so we again know that the domain is all real
numbers or,
( ) Domain : or , x −∞ < < ∞ −∞ ∞
In this case the range requires a little bit of work. From an Algebra class we know that the graph
of this will be a parabola that opens down (because the coefficient of the
2
x is negative) and so
the vertex will be the highest point on the graph. If we know the vertex we can then get the
range. The vertex is then,
So, as discussed, we know that this will be the highest point on the graph or the largest value of
the function and the parabola will take all values less than this so the range is then,
( ]
Range : , 23 −∞
[Return to Problems]
(d) ( ) 6 3 f z z = − −
This function contains an absolute value and we know that absolute value will be either positive
or zero. In this case the absolute value will be zero if 6 z = and so the absolute value portion of
this function will always be greater than or equal to zero. We are subtracting 3 from the absolute
value portion and so we then know that the range will be,
[ ) Range : 3, − ∞
We can plug any value into an absolute value and so the domain is once again all real numbers or,
( ) Domain : or , x −∞ < < ∞ −∞ ∞
[Return to Problems]
In general determining the range of a function can be somewhat difficult. As long as we restrict
ourselves down to "simple" functions, some of which we looked at in the previous example,
finding the range is not too bad, but for most functions it can be a difficult process.
Because of the difficulty in finding the range for a lot of functions we had to keep those in the
previous set somewhat simple, which also meant that we couldn't really look at some of the more
complicated domain examples that are liable to be important in a Calculus course. So, let's take a
look at another set of functions only this time we'll just look for the domain.
So, these are the only values of x that we need to avoid and so the domain is,
Domain : All real numbers except 3 & 5 x x = − =
[Return to Problems]
(b) ( )
2
6 g t t t = + −
In this case we need to avoid square roots of negative numbers and so need to require that,
The first thing that we need to determine where the function is zero and that's not too difficult in
this case.
( )( )
2
6 3 2 0 t t t t − − = − + =
So, the function will be zero at 2 t = − and 3 t = . Recall that these points will be the only place
where the function may change sign. It's not required to change sign at these points, but these
will be the only points where the function can change sign. This means that all we need to do is
break up a number line into the three regions that avoid these two points and test the sign of the
function at a single point in each of the regions. If the function is positive at a single point in the
region it will be positive at all points in that region because it doesn't contain the any of the
points where the function may change sign. We'll have a similar situation if the function is
negative for the test point.
So, here is a number line showing these computations.
From this we can see that in only region in which the quadratic (in its modified form) will be
negative is in the middle region. Recalling that we got to the modified region by multiplying the
quadratic by a -1 this means that the quadratic under the root will only be positive in the middle
region and so the domain for this function is then,
( ) Domain : 2 3 or 2, 3 t − < < −
[Return to Problems]
(c) ( )
2
9
x
h x
x
=
−
In this case we have a mixture of the two previous parts. We have to worry about division by
zero and square roots of negative numbers. We can cover both issues by requiring that,
2
9 0 x − >
Note that we need the inequality here to be strictly greater than zero to avoid the division by zero
issues. We can either solve this by the method from the previous example or, in this case, it is
easy enough to solve by inspection. The domain is this case is,
( ) ( ) Domain : 3 & 3 or , 3 & 3, x x < − > −∞ − ∞
The next topic that we need to discuss here is that of function composition. The composition of
f(x) and g(x) is
( )( ) ( ) ( )
f g x f g x =
In other words, compositions are evaluated by plugging the second function listed into the first
function listed. Note as well that order is important here. Interchanging the order will usually
result in a different answer.
In this case the two compositions where the same and in fact the answer was very simple.
( )( ) ( )( ) f g x g f x x = =
This will usually not happen. However, when the two compositions are the same, or more
specifically when the two compositions are both x there is a very nice relationship between the
two functions. We will take a look at that relationship in the next section.
Review : Inverse Functions
In the last example from the previous section we looked at the two functions ( ) 3 2 f x x = − and
( )
2
3 3
x
g x = + and saw that
( )( ) ( )( ) f g x g f x x = =
and as noted in that section this means that there is a nice relationship between these two
functions. Let's see just what that relationship is. Consider the following evaluations.
In the first case we plugged 1 x = − into ( ) f x and got a value of -5. We then turned around and
plugged 5 x = − into ( ) g x and got a value of -1, the number that we started off with.
In the second case we did something similar. Here we plugged 2 x = into ( ) g x and got a value
of
4
3
, we turned around and plugged this into ( ) f x and got a value of 2, which is again the
number that we started with.
Note as well that these both agree with the formula for the compositions that we found in the
previous section. We get back out of the function evaluation the number that we originally
plugged into the composition.
So, just what is going on here? In some way we can think of these two functions as undoing what
the other did to a number. In the first case we plugged 1 x = − into ( ) f x and then plugged the
result from this function evaluation back into ( ) g x and in some way ( ) g x undid what ( ) f x
had done to 1 x = − and gave us back the original x that we started with.
Function pairs that exhibit this behavior are called inverse functions. Before formally defining
inverse functions and the notation that we're going to use for them we need to get a definition out
of the way.
A function is called one-to-one if no two values of x produce the same y. Mathematically this is
the same as saying,
( ) ( )
1 2 1 2
whenever f x f x x x ≠ ≠
So, a function is one-to-one if whenever we plug different values into the function we get
different function values.
Sometimes it is easier to understand this definition if we see a function that isn't one-to-one.
Let's take a look at a function that isn't one-to-one. The function ( )
2
f x x = is not one-to-one
because both ( ) 2 4 f − = and ( ) 2 4 f = . In other words there are two different values of x that
produce the same value of y. Note that we can turn ( )
2
f x x = into a one-to-one function if we
restrict ourselves to 0 x ≤ < ∞. This can sometimes be done with functions.
Showing that a function is one-to-one is often tedious and/or difficult. For the most part we are
going to assume that the functions that we're going to be dealing with in this course are either
one-to-one or we have restricted the domain of the function to get it to be a one-to-one function.
Now, be careful with the notation for inverses. The "-1" is NOT an exponent despite the fact that
is sure does look like one! When dealing with inverse functions we've got to remember that
( )
( )
1
1
f x
f x
−
≠
This is one of the more common mistakes that students make when first studying inverse
functions.
The process for finding the inverse of a function is a fairly simple one although there are a couple
of steps that can on occasion be somewhat messy. Here is the process
Finding the Inverse of a Function
Given the function ( ) f x we want to find the inverse function, ( )
1
f x
−
.
1. First, replace ( ) f x with y. This is done to make the rest of the process easier.
2. Replace every x with a y and replace every y with an x.
3. Solve the equation from Step 2 for y. This is the step where mistakes are most often
made so be careful with this step.
4. Replace y with ( )
1
f x
−
. In other words, we've managed to find the inverse at this point!
5. Verify your work by checking that
( )( )
1
f f x x
−
= and
( )( )
1
f f x x
−
= are both
true. This work can sometimes be messy making it easy to make mistakes so again be
careful.
That's the process. Most of the steps are not all that bad but as mentioned in the process there are
a couple of steps that we really need to be careful with since it is easy to make mistakes in those
steps.
In the verification step we technically really do need to check that both
( )( )
1
f f x x
−
= and
( )( )
1
f f x x
−
= are true. For all the functions that we are going to be looking at in this course
if one is true then the other will also be true. However, there are functions (they are beyond the
scope of this course however) for which it is possible for only one of these to be true. This is
brought up because in all the problems here we will be just checking one of them. We just need
to always remember that technically we should check both.
Now, we need to verify the results. We already took care of this in the previous section, however,
we really should follow the process so we'll do that here. It doesn't matter which of the two that
we check we just need to check one of them. This time we'll check that
( )( )
1
f f x x
−
= is
true.
Review : Trig Functions
The intent of this section is to remind you of some of the more important (from a Calculus
standpoint…) topics from a trig class. One of the most important (but not the first) of these topics
will be how to use the unit circle. We will actually leave the most important topic to the next
section.
First let's start with the six trig functions and how they relate to each other.
true in many science classes. However, in a calculus course almost everything is done in radians.
The following table gives some of the basic angles in both degrees and radians.
Degree 0 30 45 60 90 180 270 360
Radians 0
6
π
4
π
3
π
2
π
π
3
2
π
2π
Know this table! We may not see these specific angles all that much when we get into the
Calculus portion of these notes, but knowing these can help us to visualize each angle. Now, one
more time just make sure this is clear.
Be forewarned, everything in most calculus classes will be done in radians!
Let's next take a look at one of the most overlooked ideas from a trig class. The unit circle is one
of the more useful tools to come out of a trig class. Unfortunately, most people don't learn it as
well as they should in their trig class.
Below is the unit circle with just the first quadrant filled in. The way the unit circle works is to
draw a line from the center of the circle outwards corresponding to a given angle. Then look at
the coordinates of the point where the line and the circle intersect. The first coordinate is the
cosine of that angle and the second coordinate is the sine of that angle. We've put some of the
basic angles along with the coordinates of their intersections on the unit circle. So, from the unit
circle below we can see that
3
cos
6 2
π ⎛ ⎞
=
⎜ ⎟
⎝ ⎠
and
1
sin
6 2
π ⎛ ⎞
=
⎜ ⎟
⎝ ⎠
.
Remember how the signs of angles work. If you rotate in a counter clockwise direction the angle
is positive and if you rotate in a clockwise direction the angle is negative.
Recall as well that one complete revolution is 2π , so the positive x-axis can correspond to either
an angle of 0 or 2π (or 4π , or 6π , or 2π − , or 4π − , etc. depending on the direction of
rotation). Likewise, the angle
6
π
(to pick an angle completely at random) can also be any of the
following angles:
In fact
6
π
can be any of the following angles
6
2 , 0, 1, 2, 3, n n
π
π + = ± ± ± … In this case n is
the number of complete revolutions you make around the unit circle starting at
6
π
. Positive
values of n correspond to counter clockwise rotations and negative values of n correspond to
clockwise rotations.
So, why did I only put in the first quadrant? The answer is simple. If you know the first quadrant
then you can get all the other quadrants from the first with a small application of geometry.
You'll see how this is done in the following set of examples.
Solution
(a) The first evaluation in this part uses the angle
2
3
π
. That's not on our unit circle above,
however notice that
2
3 3
π π
π = − . So
2
3
π
is found by rotating up
3
π
from the negative x-axis.
This means that the line for
2
3
π
will be a mirror image of the line for
3
π
only in the second
quadrant. The coordinates for
2
3
π
will be the coordinates for
3
π
except the x coordinate will be
negative.
Likewise for
2
3
π
− we can notice that
2
3 3
π π
π − = − + , so this angle can be found by rotating
down
3
π
from the negative x-axis. This means that the line for
2
3
π
− will be a mirror image of
the line for
3
π
only in the third quadrant and the coordinates will be the same as the coordinates
for
3
π
except both will be negative.
Both of these angles along with their coordinates are shown on the following unit circle.
This leads to a nice fact about the sine function. The sine function is called an odd function and
so for ANY angle we have
( ) ( ) sin sin θ θ − = −
[Return to Problems]
(b) For this example notice that
7
6 6
π π
π = + so this means we would rotate down
6
π
from the
negative x-axis to get to this angle. Also
7
6 6
π π
π − = − − so this means we would rotate up
6
π
from the negative x-axis to get to this angle. So, as with the last part, both of these angles will be
mirror images of
6
π
in the third and second quadrants respectively and we can use this to
determine the coordinates for both of these new angles.
Both of these angles are shown on the following unit circle along with appropriate coordinates for
the intersection points.
(c) Here we should note that
7
2
4 4
π π
π = − so
7
4
π
and
4
π
− are in fact the same angle! Also
note that this angle will be the mirror image of
4
π
in the fourth quadrant. The unit circle for this
angle is
Now, if we remember that ( )
( )
( )
sin
tan
cos
x
x
x
= we can use the unit circle to find the values the
tangent function. So,
On a side note, notice that tan 1
4
π ⎛ ⎞
=
⎜ ⎟
⎝ ⎠
and we can see that the tangent function is also called an
odd function and so for ANY angle we will have
( ) ( ) tan tan θ θ − = − .
[Return to Problems]
(d) Here we need to notice that
25
4
6 6
π π
π = + . In other words, we've started at
6
π
and rotated
around twice to end back up at the same point on the unit circle. This means that
So, in the last example we saw how the unit circle can be used to determine the value of the trig
functions at any of the "common" angles. It's important to notice that all of these examples used
the fact that if you know the first quadrant of the unit circle and can relate all the other angles to
"mirror images" of one of the first quadrant angles you don't really need to know whole unit
circle. If you'd like to see a complete unit circle I've got one on my Trig Cheat Sheet that is
available at
Another important idea from the last example is that when it comes to evaluating trig functions all
that you really need to know is how to evaluate sine and cosine. The other four trig functions are
defined in terms of these two so if you know how to evaluate sine and cosine you can also
evaluate the remaining four trig functions.
We've not covered many of the topics from a trig class in this section, but we did cover some of
the more important ones from a calculus standpoint. There are many important trig formulas that
you will use occasionally in a calculus class. Most notably are the half-angle and double-angle
formulas. If you need reminded of what these are, you might want to download my Trig Cheat
Sheet as most of the important facts and formulas from a trig class are listed there.
circle there is another angle which will also be a solution. We need to determine what this angle
is. When we look for these angles we typically want positive angles that lie between 0 and 2π .
This angle will not be the only possibility of course, but by convention we typically look for
angles that meet these conditions.
To find this angle for this problem all we need to do is use a little geometry. The angle in the first
quadrant makes an angle of
6
π
with the positive x-axis, then so must the angle in the fourth
quadrant. So we could use
6
π
− , but again, it's more common to use positive angles so, we'll use
11
2
6 6
t
π π
π = − = .
We aren't done with this problem. As the discussion about finding the second angle has shown
there are many ways to write any given angle on the unit circle. Sometimes it will be
6
π
− that
we want for the solution and sometimes we will want both (or neither) of the listed angles.
Therefore, since there isn't anything in this problem (contrast this with the next problem) to tell
us which is the correct solution we will need to list ALL possible solutions.
This is very easy to do. Recall from the previous section and you'll see there that I used
2 , 0, 1, 2, 3,
6
n n
π
π + = ± ± ± …
to represent all the possible angles that can end at the same location on the unit circle, i.e. angles
that end at
6
π
. Remember that all this says is that we start at
6
π
then rotate around in the
counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete
rotations. The same thing can be done for the second solution.
Solution
In a calculus class we are often more interested in only the solutions to a trig equation that fall in
a certain interval. The first step in this kind of problem is to first find all possible solutions. We
did this in the first example.
Now, to find the solutions in the interval all we need to do is start picking values of n, plugging
them in and getting the solutions that will fall into the interval that we've been given.
n=0.
( )
( )
2 0 2
6 6
11 11
2 0 2
6 6
π π
π π
π π
π π
+ = <
+ = <
Now, notice that if we take any positive value of n we will be adding on positive multiples of 2π
onto a positive quantity and this will take us past the upper bound of our interval and so we don't
need to take any positive value of n.
However, just because we aren't going to take any positive value of n doesn't mean that we
shouldn't also look at negative values of n.
These are both greater than 2π − and so are solutions, but if we subtract another 2π off (i.e use
2 n = − ) we will once again be outside of the interval so we've found all the possible solutions
that lie inside the interval [ 2 , 2 ] π π − .
Solution
This problem is very similar to the other problems in this section with a very important
difference. We'll start this problem in exactly the same way. We first need to find all possible
solutions.
2sin(5 ) 3
3
sin(5 )
2
x
x
= −
−
=
So, we are looking for angles that will give
3
2
− out of the sine function. Let's again go to our
trusty unit circle.
This is not the set of solutions because we are NOT looking for values of x for which
( )
3
sin
2
x = − , but instead we are looking for values of x for which ( )
3
sin 5
2
x = − . Note the
difference in the arguments of the sine function! One is x and the other is 5x . This makes all the
difference in the world in finding the solution! Therefore, the set of solutions is
And we're now past the left endpoint of the interval. Sometimes, there will be many solutions as
there were in this example. Putting all of this together gives the following set of solutions that lie
in the given interval.
Solution
This problem is a little different from the previous ones. First, we need to do some rearranging
and simplification.
( )
sin(2 ) cos(2 )
sin(2 )
1
cos(2 )
tan 2 1
x x
x
x
x
= −
= −
= −
So, solving sin(2 ) cos(2 ) x x = − is the same as solving tan(2 ) 1 x = − . At some level we didn't
need to do this for this problem as all we're looking for is angles in which sine and cosine have
the same value, but opposite signs. However, for other problems this won't be the case and we'll
want to convert to tangent.
Unlike the previous example only one of these will be in the interval. This will happen
occasionally so don't always expect both answers from a particular n to work. Also, we should
now check n=2 for the first to see if it will be in or out of the interval. I'll leave it to you to check
that it's out of the interval.
Let's work one more example so that I can make a point that needs to be understood when
solving some trig equations.
Example 5 Solve ( ) cos 3 2 x = .
Solution
This example is designed to remind you of certain properties about sine and cosine. Recall that
( ) 1 cos 1 θ − ≤ ≤ and ( ) 1 sin 1 θ − ≤ ≤ . Therefore, since cosine will never be greater that 1 it
definitely can't be 2. So THERE ARE NO SOLUTIONS to this equation!
It is important to remember that not all trig equations will have solutions.
In this section we solved some simple trig equations. There are more complicated trig equations
that we can solve so don't leave this section with the feeling that there is nothing harder out there
in the world to solve. In fact, we'll see at least one of the more complicated problems in the next
section. Also, every one of these problems came down to solutions involving one of the
"common" or "standard" angles. Most trig equations won't come down to one of those and will
in fact need a calculator to solve. The next section is devoted to this kind of problem.
Review : Solving Trig Equations with Calculators, Part I
In the previous section we started solving trig equations. The only problem with the equations we
solved in there is that they pretty much all had solutions that came from a handful of "standard"
angles and of course there are many equations out there that simply don't. So, in this section we
are going to take a look at some more trig equations, the majority of which will require the use of
a calculator to solve (a couple won't need a calculator).
The fact that we are using calculators in this section does not however mean that the problems in
the previous section aren't important. It is going to be assumed in this section that the basic ideas
of solving trig equations are known and that we don't need to go back over them here. In
particular, it is assumed that you can use a unit circle to help you find all answers to the equation
(although the process here is a little different as we'll see) and it is assumed that you can find
answers in a given interval. If you are unfamiliar with these ideas you should first go to the
previous section and go over those problems.
Before proceeding with the problems we need to go over how our calculators work so that we can
get the correct answers. Calculators are great tools but if you don't know how they work and
how to interpret their answers you can get in serious trouble.
First, as already pointed out in previous sections, everything we are going to be doing here will be
in radians so make sure that your calculator is set to radians before attempting the problems in
this section. Also, we are going to use 4 decimal places of accuracy in the work here. You can
use more if you want, but in this class we'll always use at least 4 decimal places of accuracy.
Next, and somewhat more importantly, we need to understand how calculators give answers to
inverse trig functions. We didn't cover inverse trig functions in this review, but they are just
inverse functions and we have talked a little bit about inverse functions in a review section. The
only real difference is that we are now using trig functions. We'll only be looking at three of
them and they are:
As shown there are two different notations that are commonly used. In these notes we'll be using
the first form since it is a little more compact. Most calculators these days will have buttons on
them for these three so make sure that yours does as well.
From the previous section we know that there should in fact be an infinite number of answers to
this including a second angle that is in the interval
[ ]
0, 2π . However, our calculator only gave us
a single answer. How to determine what the other angles are will be covered in the following
examples so we won't go into detail here about that. We did need to point out however, that the
calculators will only give a single answer and that we're going to have more work to do than just
plugging a number into a calculator.
Since we know that there are supposed to be an infinite number of solutions to ( )
3
4
cos x = the
next question we should ask then is just how did the calculator decide to return the answer that it
did? Why this one and not one of the others? Will it give the same answer every time?
There are rules that determine just what answer the calculator gives. All calculators will give
answers in the following ranges.
If you think back to the unit circle and recall that we think of cosine as the horizontal axis the we
can see that we'll cover all possible values of cosine in the upper half of the circle and this is
exactly the range give above for the inverse cosine. Likewise, since we think of sine as the
vertical axis in the unit circle we can see that we'll cover all possible values of sine in the right
half of the unit circle and that is the range given above.
For the tangent range look back to the graph of the tangent function itself and we'll see that one
branch of the tangent is covered in the range given above and so that is the range we'll use for
inverse tangent. Note as well that we don't include the endpoints in the range for inverse tangent
since tangent does not exist there.
So, if we can remember these rules we will be able to determine the remaining angle in
[ ]
0, 2π
that also works for each solution.
As a final quick topic let's note that it will, on occasion, be useful to remember the decimal
representations of some basic angles. So here they are,
3
1.5708 3.1416 4.7124 2 6.2832
2 2
π π
π π = = = =
Using these we can quickly see that
( )
1
3
4
cos
−
must be in the first quadrant since 0.7227 is
between 0 and 1.5708. This will be of great help when we go to determine the remaining angles
So, once again, we can't stress enough that calculators are great tools that can be of tremendous
help to us, but it you don't understand how they work you will often get the answers to problems
wrong.
So, with all that out of the way let's take a look at our first problem.
Example 1 Solve ( ) 4cos 3 t = on[-8,10].
Solution
Okay, the first step here is identical to the problems in the previous section. We first need to
isolate the cosine on one side by itself and then use our calculator to get the first answer.
( )
1
3 3
cos cos 0.7227
4 4
t t
−
⎛ ⎞
= ⇒ = =
⎜ ⎟
⎝ ⎠
So, this is the one we were using above in the opening discussion of this section. At the time we
mentioned that there were infinite number of answers and that we'd be seeing how to find them
later. Well that time is now.
line segment in the first quadrant forms an angle of 0.7227 radians with the positive x-axis then so
does the line segment in the fourth quadrant. This means that we can use either -0.7227 as the
second angle or 2 0.7227 5.5605 π − = . Which you use depends on which you prefer. We'll
pretty much always use the positive angle to avoid the possibility that we'll lose the minus sign.
So, all possible solutions, ignoring the interval for a second, are then,
0.7227 2
0, 1, 2,
5.5605 2
t n
n
t n
π
π
= +
= ± ±
= +
…
Now, all we need to do is plug in values of n to determine the angle that are actually in the
interval. Here's the work for that.
Note that we had a choice of angles to use for the second angle in the previous example. The
choice of angles there will also affect the value(s) of n that we'll need to use to get all the
solutions. In the end, regardless of the angle chosen, we'll get the same list of solutions, but the
value(s) of n that give the solutions will be different depending on our choice.
Also, in the above example we put in a little more explanation than we'll show in the remaining
examples in this section to remind you how these work.
Example 3 Solve 6sin 1
2
x ⎛ ⎞
=
⎜ ⎟
⎝ ⎠
on [-20,30]
Solution
Let's first get the calculator work out of the way since that isn't where the difference comes into
play.
1
1 1
sin sin 0.1674
2 6 2 6
x x
−
⎛ ⎞ ⎛ ⎞
= ⇒ = =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Here's a unit circle for this example.
To find the second angle in this case we can notice that the line in the first quadrant makes an
angle of 0.1674 with the positive x-axis and so the angle in the second quadrant will then make an
angle of 0.1674 with the negative x-axis and so the angle that we're after is then,
0.1674 2.9742 π − = .
Here's the rest of the solution for this example. We're going to assume from this point on that
you can do this work without much explanation.
Solution
You should be getting pretty good at these by now, so we won't be putting much explanation in
for this one. Here we go.
( )
1
2 2
sin 5 5 sin 0.7297
3 3
z z
−
⎛ ⎞
= − ⇒ = − = −
⎜ ⎟
⎝ ⎠
Okay, with this one we're going to do a little more work than with the others. For the first angle
we could use the answer our calculator gave us. However, it's easy to lose minus signs so we'll
instead use 2 0.7297 5.5535 π − = . Again, there is no reason to this other than a worry about
losing the minus sign in the calculator answer. If you'd like to use the calculator answer you are
more than welcome to. For the second angle we'll note that the lines in the third and fourth
quadrant make an angle of 0.7297 with the x-axis and so the second angle is
0.7297 3.8713 π + = .
we'd just relied on our calculator without worrying about other angles we would not have gotten
this solution. Again, it can't be stressed enough that while calculators are a great tool if we don't
understand how to correctly interpret/use the result we can (and often will) get the solution
wrong.
To this point we've only worked examples involving sine and cosine. Let's no work a couple of
examples that involve other trig functions to see how they work.
Example 5 Solve ( ) ( ) 9sin 2 5cos 2 x x = − on[-10,0].
Solution
At first glance this problem seems to be at odds with the sentence preceding the example.
However, it really isn't.
First, when we have more than one trig function in an equation we need a way to get equations
that only involve one trig function. There are many ways of doing this that depend on the type of
equation we're starting with. In this case we can simply divide both sides by a cosine and we'll
get a single tangent in the equation. We can now see that this really is an equation that doesn't
involve a sine or a cosine.
value of tangent. So, the second angle will always be the first angle plus π .
Before getting the second angle let's also note that, like the previous example, we'll use the
2 0.5071 5.7761 π − = for the first angle. Again, this is only because of a concern about losing
track of the minus sign in our calculator answer. We could just as easily do the work with the
original angle our calculator gave us.
Now, this is where is seems like we're just randomly making changes and doing things for no
reason. The second angle that we're going to use is,
( ) 0.5071 0.5071 2.6345 π π + − = − =
The fact that we used the calculator answer here seems to contradict the fact that we used a
different angle for the first above. The reason for doing this here is to give a second angle that is
in the range
[ ]
0, 2π . Had we used 5.7761 to find the second angle we'd get
5.7761 8.9177 π + = . This is a perfectly acceptable answer, however it is larger than 2π
(6.2832) and the general rule of thumb is to keep the initial angles as small as possible.
Solution
We'll start this one in exactly the same way we've done all the others.
( )
1
10 10
sec 3 3 sec
7 7
t t
−
⎛ ⎞
= − ⇒ = −
⎜ ⎟
⎝ ⎠
Now we reach the problem. As noted above, most calculators can't handle inverse secant so
we're going to need a different solution method for this one. To finish the solution here we'll
simply recall the definition of secant in terms of cosine and convert this into an equation
involving cosine instead and we already know how to solve those kinds of trig equations.
( )
( ) ( )
1 10 7
sec 3 cos 3
cos 3 7 10
t t
t
= = − ⇒ = −
Now, we solved this equation in the second example above so we won't redo our work here. The
solution is,
2
0.7821
3
0, 1, 2,
2
1.3123
3
n
t
n
n
t
π
π
= +
= ± ±
= +
…
We weren't given an interval in this problem so here is nothing else to do here.
For the remainder of the examples in this section we're not going to be finding solutions in an
interval to save some space. If you followed the work from the first few examples in which we
were given intervals you should be able to do any of the remaining examples if given an interval.
Also, we will no longer be including sketches of unit circles in the remaining solutions. We are
going to assume that you can use the above sketches as guides for sketching unit circles to verify
our claims in the following examples.
The next three examples don't require a calculator but are important enough or cause enough
problems for students to include in this section in case you run across them and haven't seen them
anywhere else.
Solution
There really isn't too much to do with this problem. It is, however, different from all the others
done to this point. All the others done to this point have had two angles in the interval
[ ] 0, 2π
that were solutions to the equation. This only has one. Here is the solution to this equation.
4 2 0, 1, 2,
4 2
n
n n
π π
θ π π θ = + ⇒ = + = ± ± …
Example 8 Solve sin 0
7
α ⎛ ⎞
=
⎜ ⎟
⎝ ⎠
.
Solution
Again, not much to this problem. Using a unit circle it isn't too hard to see that the solutions to
this equation are,
This next example has an important point that needs to be understood when solving some trig
equations.
Example 9 Solve ( ) sin 3 2 t = .
Solution
This example is designed to remind you of certain properties about sine and cosine. Recall that
( ) 1 sin 1 θ − ≤ ≤ and ( ) 1 cos 1 θ − ≤ ≤ . Therefore, since sine will never be greater that 1 it
definitely can't be 2. So THERE ARE NO SOLUTIONS to this equation!
It is important to remember that not all trig equations will have solutions.
Because this document is also being prepared for viewing on the web we're going to split this
section in two in order to keep the page size (and hence load time in a browser) to a minimum. In
the next section we're going to take a look at some slightly more "complicated" equations.
Although, as you'll see, they aren't as complicated as they may at first seem.
Review : Solving Trig Equations with Calculators, Part II
Because this document is also being prepared for viewing on the web we split this section into
two parts to keep the size of the pages to a minimum.
Also, as with the last few examples in the previous part of this section we are not going to be
looking for solutions in an interval in order to save space. The important part of these examples
is to find the solutions to the equation. If we'd been given an interval it would be easy enough to
find the solutions that actually fall in the interval.
In all the examples in the previous section all the arguments, the 3t ,
7
α
, etc., were fairly simple.
Let's take a look at an example that has a slightly more complicated argument.
Example 1 Solve ( ) 5cos 2 1 3 x − = − .
Solution
Note that the argument here is not really all that complicated but the addition of the "-1" often
seems to confuse people so we need to a quick example with this kind of argument. The solution
process is identical to all the problems we've done to this point so we won't be putting in much
explanation. Here is the solution.
This angle is in the second quadrant and so we can use either -2.2143 or 2 2.2143 4.0689 π − =
for the second angle. As usual for these notes we'll use the positive one. Therefore the two
angles are,
2 1 2.2143 2
0, 1, 2,
2 1 4.0689 2
x n
n
x n
π
π
− = +
= ± ±
− = +
…
Now, we still need to find the actual values of x that are the solutions. These are found in the
same manner as all the problems above. We'll first add 1 to both sides and then divide by 2.
Doing this gives,
We now need to move into a different type of trig equation. All of the trig equations solved to
this point (the previous example as well as the previous section) were, in some way, more or less
the "standard" trig equation that is usually solved in a trig class. There are other types of
equations involving trig functions however that we need to take a quick look at. The remaining
examples show some of these different kinds of trig equations.
Example 2 Solve ( ) ( ) ( ) 2cos 6 11cos 6 sin 3 0 y y y + = .
Solution
So, this definitely doesn't look like any of the equations we've solved to this point and initially
the process is different as well. First, notice that there is a ( ) cos 6y in each term, so let's factor
that out and see what we have.
( ) ( ) ( )
cos 6 2 11sin 3 0 y y + =
We now have a product of two terms that is zero and so we know that we must have,
( ) ( ) cos 6 0 OR 2 11sin 3 0 y y = + =
Now, at this point we have two trig equations to solve and each is identical to the type of equation
we were solving earlier. Because of this we won't put in much detail about solving these two
equations.
This next example also involves "factoring" trig equations but in a slightly different manner than
the previous example.
Example 3 Solve
2
4sin 3sin 1
3 3
t t ⎛ ⎞ ⎛ ⎞
− =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
.
Solution
Before solving this equation let's solve an apparently unrelated equation.
( )( )
2 2
1
4 3 1 4 3 1 4 1 1 0 ,1
4
x x x x x x x − = ⇒ − − = + − = ⇒ = −
This is an easy (or at least I hope it's easy as this point) equation to solve. The obvious question
then is, why did we do this? We'll, if you compare the two equations you'll see that the only real
difference is that the one we just solved has an x everywhere the equation we want to solve has a
sine. What this tells us is that we can work the two equations in exactly the same way.
Now, set each of the two factors equal to zero and solve for the sine,
1
sin sin 1
3 4 3
t t ⎛ ⎞ ⎛ ⎞
= − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
We now have two trig equations that we can easily (hopefully…) solve at this point. We'll leave
the details to you to verify that the solutions to each of these and hence the solutions to the
original equation are,
Let's work one more trig equation that involves solving a quadratic equation. However, this time,
unlike the previous example this one won't factor and so we'll need to use the quadratic formula.
Example 4 Solve ( ) ( )
2
8cos 1 13cos 1 5 0 x x − + − − = .
Solution
Now, as mentioned prior to starting the example this quadratic does not factor. However, that
doesn't mean all is lost. We can solve the following equation with the quadratic formula (you do
remember this and how to use it right?),
2
13 329
8 13 5 0 0.3211, 1.9461
16
t t t
− ±
+ − = ⇒ = = −
So, if we can use the quadratic formula on this then we can also use it on the equation we're
asked to solve. Doing this gives us,
( ) ( ) cos 1 0.3211 OR cos 1 1.9461 x x − = − = −
Now, recall Example 9 from the previous section. In that example we noted that
( ) 1 cos 1 θ − ≤ ≤ and so the second equation will have no solutions. Therefore, the solutions to
the first equation will yield the only solutions to our original equation. Solving this gives the
following set of solutions,
0.2439 2
0, 1, 2,
4.0393 2
x n
n
x n
π
π
= − −
= ± ±
= − −
…
Note that we did get some negative numbers here and that does seem to violate the general form
that we've been using in most of these examples. However, in this case the "-" are coming about
when we solved for x after computing the inverse cosine in our calculator.
There is one more example in this section that we need to work that illustrates another way in
which factoring can arise in solving trig equations. This equation is also the only one where the
variable appears both inside and outside of the trig equation. Not all equations in this form can be
easily solved, however some can so we want to do a quick example of one.
Solution
First, before we even start solving we need to make one thing clear. DO NOT CANCEL AN x
FROM BOTH SIDES!!! While this may seem like a natural thing to do it WILL cause us to
lose a solution here.
So, to solve this equation we'll first get all the terms on one side of the equation and then factor
an x out of the equation. If we can cancel an x from all terms then it can be factored out. Doing
this gives,
( ) ( ) ( )
5 tan 8 3 5tan 8 3 0 x x x x x − = − =
Upon factoring we can see that we must have either,
( )
3
0 OR tan 8
5
x x = =
Note that if we'd canceled the x we would have missed the first solution. Now, we solved an
equation with a tangent in it in Example 5 of the previous section so we'll not go into the details
of this solution here. Here is the solution to the trig equation.
Review : Exponential Functions
In this section we're going to review one of the more common functions in both calculus and the
sciences. However, before getting to this function let's take a much more general approach to
things.
These will all be very useful properties to recall at times as we move throughout this course (and
later Calculus courses for that matter…).
There is a very important exponential function that arises naturally in many places. This function
is called the natural exponential function. However, for must people this is simply the
exponential function.
The main point behind this problem is to make sure you can do this type of evaluation so make
sure that you can get the values that we graphed in this example. You will be asked to do this
kind of evaluation on occasion in this class.
You will be seeing exponential functions in pretty much every chapter in this class so make sure
that you are comfortable with them.
Review : Logarithm Functions
In this section we'll take a look at a function that is related to the exponential functions we looked
at in the last section. We will look logarithms in this section. Logarithms are one of the
functions that students fear the most. The main reason for this seems to be that they simply have
never really had to work with them. Once they start working with them, students come to realize
that they aren't as bad as they first thought.
We'll start with 0 b > , 1 b ≠ just as we did in the last section. Then we have
log is equivalent to
y
b
y x x b = =
The first is called logarithmic form and the second is called the exponential form. Remembering
this equivalence is the key to evaluating logarithms. The number, b, is called the base.
(a)
3 4 5
ln x y z
Property 6 above can be extended to products of more than two functions. Once we've used
Property 6 we can then use Property 8.
3 4 5 3 4 5
ln ln ln ln
3ln 4ln 5ln
x y z x y z
x y z
= + +
= + +
[Return to Problems]
(b)
4
3
9
log
x
y
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
When using property 7 above make sure that the logarithm that you subtract is the one that
contains the denominator as its argument. Also, note that that we'll be converting the root to
fractional exponents in the first step.
You can use Property 8 on the second term because the WHOLE term was raised to the 3, but in
the first logarithm, only the individual terms were squared and not the term as a whole so the 2's
must stay where they are!
[Return to Problems]
The last topic that we need to look at in this section is the change of base formula for logarithms.
The change of base formula is,
log
log
log
a
b
a
x
x
b
=
This is the most general change of base formula and will convert from base b to base a.
However, the usual reason for using the change of base formula is to compute the value of a
logarithm that is in a base that you can't easily deal with. Using the change of base formula
means that you can write the logarithm in terms of a logarithm that you can deal with. The two
most common change of base formulas are
b
ln log
log and log
ln log
b
x x
x x
b b
= =
In fact, often you will see one or the other listed as THE change of base formula!
In the first part of this section we computed the value of a few logarithms, but we could do these
without the change of base formula because all the arguments could be written in terms of the
base to a power. For instance,
2
7
log 49 2 because 7 49 = =
However, this only works because 49 can be written as a power of 7! We would need the change
of base formula to compute
7
log 50.
So, it doesn't matter which we use, we will get the same answer regardless of the logarithm that
we use in the change of base formula.
Note as well that we could use the change of base formula on
7
log 49 if we wanted to as well.
7
ln 49 3.89182029811
log 49 2
ln 7 1.94591014906
= = =
This is a lot of work however, and is probably not the best way to deal with this.
So, in this section we saw how logarithms work and took a look at some of the properties of
logarithms. We will run into logarithms on occasion so make sure that you can deal with them
when we do run into them.
Solution
The first step is to get the exponential all by itself on one side of the equation with a coefficient of
one.
1 3
1 3
1 3
7 15 10
15 3
1
5
z
z
z
−
−
−
+ =
=
=
e
e
e
Now, we need to get the z out of the exponent so we can solve for it. To do this we will use the
property above. Since we have an e in the equation we'll use the natural logarithm. First we take
the logarithm of both sides and then use the property to simplify the equation.
Now that we've seen a couple of equations where the variable only appears in the exponent we
need to see an example with variables both in the exponent and out of it.
Example 3 Solve
5 2
0
x
x x
+
− = e .
Solution
The first step is to factor an x out of both terms.
DO NOT DIVIDE AN x FROM BOTH TERMS!!!!
Note that it is very tempting to "simplify" the equation by dividing an x out of both terms.
However, if you do that you'll miss a solution as we'll see.
( )
5 2
5 2
0
1 0
x
x
x x
x
+
+
− =
− =
e
e
So, it's now a little easier to deal with. From this we can see that we get one of two possibilities.
5 2
0 OR
1 0
x
x
+
=
− = e
The first possibility has nothing more to do, except notice that if we had divided both sides by an
x we would have missed this one so be careful. In the second possibility we've got a little more
to do. This is an equation similar to the first two that we did in this section.
The next equation is a more complicated (looking at least…) example similar to the previous one.
Example 4 Solve
( ) ( )
2 2 7
5 4 4
x
x x
−
− = − e .
Solution
As with the previous problem do NOT divide an
2
4 x − out of both sides. Doing this will lose
solutions even though it "simplifies" the equation. Note however, that if you can divide a term
out then you can also factor it out if the equation is written properly.
So, the first step here is to move everything to one side of the equation and then to factor out the
2
4 x − .
As a final example let's take a look at an equation that contains two different logarithms.
Example 5 Solve
1 3 5 2
4 9 0
x x + −
− = e e .
Solution
The first step here is to get one exponential on each side and then we'll divide both sides by one
of them (which doesn't matter for the most part) so we'll have a quotient of two exponentials.
The quotient can then be simplified and we'll finally get both coefficients on the other side.
Doing all of this gives,
Note that while we said that it doesn't really matter which exponential we divide out by doing it
the way we did here we'll avoid a negative coefficient on the x. Not a major issue, but those
minus signs on coefficients are really easy to lose on occasion.
This is now in a form that we can deal with so here's the rest of the solution.
At this point we might be tempted to say that we're done and move on. However, we do need to
be careful. Recall from the previous section that we can't plug a negative number into a
logarithm. This, by itself, doesn't mean that our answer won't work since its negative. What we
need to do is plug it into the logarithm and make sure that 3
7
x
+ will not be negative. I'll leave it
to you to verify that this is in fact positive upon plugging our solution into the logarithm and so
20.78861832 x = − is in fact a solution to the equation.
Let's now take a look at a more complicated equation. Often there will be more than one
logarithm in the equation. When this happens we will need to use on or more of the following
properties to combine all the logarithms into a single logarithm. Once this has been done we can
proceed as we did in the previous example.
Finally, we just need to make sure that the solution, 0.8807970780 x = , doesn't produce
negative numbers in both of the original logarithms. It doesn't, so this is in fact our solution to
this problem.
So, potential solutions are 5 x = and 2 x = − . Note, however that if we plug 2 x = − into either
of the two original logarithms we would get negative numbers so this can't be a solution. We can
however, use 5 x = .
Therefore, the solution to this equation is 5 x = .
When solving equations with logarithms it is important to check your potential solutions to make
sure that they don't generate logarithms of negative numbers or zero. It is also important to make
sure that you do the checks in the original equation. If you check them in the second logarithm
above (after we've combined the two logs) both solutions will appear to work! This is because in
combining the two logarithms we've actually changed the problem. In fact, it is this change that
introduces the extra solution that we couldn't use!
Also be careful in solving equations containing logarithms to not get locked into the idea that you
will get two potential solutions and only one of these will work. It is possible to have problems
where both are solutions and where neither are solutions.
Review : Common Graphs
The purpose of this section is to make sure that you're familiar with the graphs of many of the
basic functions that you're liable to run across in a calculus class.
Example 1 Graph
2
3
5
y x = − + .
Solution
This is a line in the slope intercept form
y mx b = +
In this case the line has a y intercept of (0,b) and a slope of m. Recall that slope can be thought of
as
rise
run
m =
Note that if the slope is negative we tend to think of the rise as a fall.
The slope allows us to get a second point on the line. Once we have any point on the line and the
slope we move right by run and up/down by rise depending on the sign. This will be a second
point on the line.
In this case we know (0,3) is a point on the line and the slope is
2
5
− . So starting at (0,3) we'll
move 5 to the right (i.e. 0 5 → ) and down 2 (i.e. 3 1 → ) to get (5,1) as a second point on the
line. Once we've got two points on a line all we need to do is plot the two points and connect
them with a line.
Solution
This is a parabola in the general form.
( )
2
f x ax bx c = + +
In this form, the x-coordinate of the vertex (the highest or lowest point on the parabola) is
2
b
x
a
= − and we get the y-coordinate is
2
b
y f
a
⎛ ⎞
= −
⎜ ⎟
⎝ ⎠
. So, for our parabola the coordinates of
the vertex will be.
( )
( ) ( ) ( )
2
2
1
2 1
1 1 2 1 3 4
x
y f
= − =
−
= = − + + =
So, the vertex for this parabola is (1,4).
We can also determine which direction the parabola opens from the sign of a. If a is positive the
parabola opens up and if a is negative the parabola opens down. In our case the parabola opens
down.
Now, because the vertex is above the x-axis and the parabola opens down we know that we'll
have x-intercepts (i.e. values of x for which we'll have ( ) 0 f x = ) on this graph. So, we'll solve
the following.
( )( )
2
2
2 3 0
2 3 0
3 1 0
x x
x x
x x
− + + =
− − =
− + =
So, we will have x-intercepts at 1 x = − and 3 x = . Notice that to make our life easier in the
solution process we multiplied everything by -1 to get the coefficient of the
2
x positive. This
made the factoring easier.
Here's a sketch of this parabola.
Example 4 Graph ( )
2
6 5 f y y y = − +
Solution
Most people come out of an Algebra class capable of dealing with functions in the form
( ) y f x = . However, many functions that you will have to deal with in a Calculus class are in
the form ( ) x f y = and can only be easily worked with in that form. So, you need to get used to
working with functions in this form.
The nice thing about these kinds of function is that if you can deal with functions in the form
( ) y f x = then you can deal with functions in the form ( ) x f y = even if you aren't that familiar
with them.
For our function we have essentially the same equation except the x and y's are switched around.
In other words, we have a parabola in the form,
2
x ay by c = + +
This is the general form of this kind of parabola and this will be a parabola that opens left or right
depending on the sign of a. The y-coordinate of the vertex is given by
2
b
y
a
= − and we find the
x-coordinate by plugging this into the equation. So, you can see that this is very similar to the
type of parabola that you're already used to dealing with.
Now, let's get back to the example. Our function is a parabola that opens to the right (a is
positive) and has a vertex at (-4,3). The vertex is to the left of the y-axis and opens to the right so
we'll need the y-intercepts (i.e. values of y for which we'll have ( ) 0 f y = )). We find these just
like we found x-intercepts in the previous problem.
( )( )
2
6 5 0
5 1 0
y y
y y
− + =
− − =
So, our parabola will have y-intercepts at 1 y = and 5 y = . Here's a sketch of the graph.
Upon doing this we see that we have a circle and it's now written in standard form.
( ) ( )
2 2
2
x h y k r − + − =
When circles are in this form we can easily identify the center : (h, k) and radius : r. Once we
have these we can graph the circle simply by starting at the center and moving right, left, up and
down by r to get the rightmost, leftmost, top most and bottom most points respectively.
Our circle has a center at (-1, 4) and a radius of 3. Here's a sketch of this circle.
Example 6 Graph
( )
( )
2
2
2
4 2 1
9
x
y
−
+ + =
Solution
This is an ellipse. The standard form of the ellipse is
( ) ( )
2 2
2 2
1
x h y k
a b
− −
+ =
This is an ellipse with center (h, k) and the right most and left most points are a distance of a
away from the center and the top most and bottom most points are a distance of b away from the
center.
The ellipse for this problem has center (2, -2) and has 3 a = and
1
2
b = . Note that to get the b
we're really rewriting the equation as,
Vertices
a units right and left
of center.
b units up and down
from center.
Slope of Asymptotes
b
a
±
b
a
±
So, what does all this mean? First, notice that one of the terms is positive and the other is
negative. This will determine which direction the two parts of the hyperbola open. If the x term
is positive the hyperbola opens left and right. Likewise, if the y term is positive the parabola
opens up and down.
Both have the same "center". Note that hyperbolas don't really have a center in the sense that
circles and ellipses have centers. The center is the starting point in graphing a hyperbola. It tells
up how to get to the vertices and how to get the asymptotes set up.
Let's also note here that we can put all values of x into cosine (which won't be the case for most
of the trig functions) and so the domain is all real numbers. Also note that
( ) 1 cos 1 x − ≤ ≤
It is important to notice that cosine will never be larger than 1 or smaller than -1. This will be
useful on occasion in a calculus class. In general we can say that
( ) cos R R x R ω − ≤ ≤
Example 13 Graph ( ) sin y x =
Solution
As with the first problem in this section there really isn't a lot to do other than graph it. Here is
the graph.
From this graph we can see that sine has the same range that cosine does. In general
( ) sin R R x R ω − ≤ ≤
As with cosine, sine itself will never be larger than 1 and never smaller than -1. Also the domain
of sine is all real numbers.
Example 14 Graph ( ) tan y x = .
Solution
In the case of tangent we have to be careful when plugging x's in since tangent doesn't exist
wherever cosine is zero (remember that
sin
tan
cos
x
x
x
= ). Tangent will not exist at
Notice that the graph is always greater than 1 and less than -1. This should not be terribly
surprising. Recall that ( ) 1 cos 1 x − ≤ ≤ . So, one divided by something less than one will be
greater than 1. Also,
1
1
1
= ±
±
and so we get the following ranges for secant.
( ) ( ) sec 1 and sec 1 x x ω ω ≥ ≤ −
Introduction
The topic that we will be examining in this chapter is that of Limits. This is the first of three
major topics that we will be covering in this course. While we will be spending the least amount
of time on limits in comparison to the other two topics limits are very important in the study of
Calculus. We will be seeing limits in a variety of places once we move out of this chapter. In
particular we will see that limits are part of the formal definition of the other two major topics.
Here is a quick listing of the material that will be covered in this chapter.
Tangent Lines and Rates of Change – In this section we will take a look at two problems that
we will see time and again in this course. These problems will be used to introduce the topic of
limits.
The Limit – Here we will take a conceptual look at limits and try to get a grasp on just what they
are and what they can tell us.
One-Sided Limits – A brief introduction to one-sided limits.
Limit Properties – Properties of limits that we'll need to use in computing limits. We will also
compute some basic limits in this section
Computing Limits – Many of the limits we'll be asked to compute will not be "simple" limits.
In other words, we won't be able to just apply the properties and be done. In this section we will
look at several types of limits that require some work before we can use the limit properties to
compute them.
Infinite Limits – Here we will take a look at limits that have a value of infinity or negative
infinity. We'll also take a brief look at vertical asymptotes.
Limits At Infinity, Part I – In this section we'll look at limits at infinity. In other words, limits
in which the variable gets very large in either the positive or negative sense. We'll also take a
brief look at horizontal asymptotes in this section. We'll be concentrating on polynomials and
rational expression involving polynomials in this section.
Rates of Change and Tangent Lines
In this section we are going to take a look at two fairly important problems in the study of
calculus. There are two reasons for looking at these problems now.
First, both of these problems will lead us into the study of limits, which is the topic of this chapter
after all. Looking at these problems here will allow us to start to understand just what a limit is
and what it can tell us about a function.
Secondly, the rate of change problem that we're going to be looking at is one of the most
important concepts that we'll encounter in the second chapter of this course. In fact, it's probably
one of the most important concepts that we'll encounter in the whole course. So looking at it now
will get us to start thinking about it from the very beginning.
Tangent Lines
The first problem that we're going to take a look at is the tangent line problem. Before getting
into this problem it would probably be best to define a tangent line.
A tangent line to the function f(x) at the point x a = is a line that just touches the graph of the
function at the point in question and is "parallel" (in some way) to the graph at that point. Take a
look at the graph below.
In this graph the line is a tangent line at the indicated point because it just touches the graph at
that point and is also "parallel" to the graph at that point. Likewise, at the second point shown,
the line does just touch the graph at that point, but it is not "parallel" to the graph at that point and
so it's not a tangent line to the graph at that point.
At the second point shown (the point where the line isn't a tangent line) we will sometimes call
the line a secant line.
moving in the same direction at that point. So, in the first point above the graph and the line are
moving in the same direction and so we will say they are parallel at that point. At the second
point, on the other hand, the line and the graph are not moving in the same direction and so they
aren't parallel at that point.
Okay, now that we've gotten the definition of a tangent line out of the way let's move on to the
tangent line problem. That's probably best done with an example.
Example 1 Find the tangent line to ( )
2
15 2 f x x = − at x = 1.
Solution
We know from algebra that to find the equation of a line we need either two points on the line or
a single point on the line and the slope of the line. Since we know that we are after a tangent line
we do have a point that is on the line. The tangent line and the graph of the function must touch
at x = 1 so the point ( ) ( ) ( ) 1, 1 1,13 f = must be on the line.
Now we reach the problem. This is all that we know about the tangent line. In order to find the
tangent line we need either a second point or the slope of the tangent line. Since the only reason
for needing a second point is to allow us to find the slope of the tangent line let's just concentrate
on seeing if we can determine the slope of the tangent line.
At this point in time all that we're going to be able to do is to get an estimate for the slope of the
tangent line, but if we do it correctly we should be able to get an estimate that is in fact the actual
slope of the tangent line. We'll do this by starting with the point that we're after, let's call it
( ) 1,13 P = . We will then pick another point that lies on the graph of the function, let's call that
point ( ) ( )
, Q x f x = .
For the sake of argument let's take choose 2 x = and so the second point will be ( ) 2, 7 Q = .
Below is a graph of the function, the tangent line and the secant line that connects P and Q.
We can see from this graph that the secant and tangent lines are somewhat similar and so the
slope of the secant line should be somewhat close to the actual slope of the tangent line. So, as an
estimate of the slope of the tangent line we can use the slope of the secant line, let's call it
PQ
m ,
which is,
Now, if we weren't too interested in accuracy we could say this is good enough and use this as an
estimate of the slope of the tangent line. However, we would like an estimate that is at least
somewhat close the actual value. So, to get a better estimate we can take an x that is closer to
1 x = and redo the work above to get a new estimate on the slope. We could then take a third
value of x even closer yet and get an even better estimate.
In other words, as we take Q closer and closer to P the slope of the secant line connecting Q and
P should be getting closer and closer to the slope of the tangent line. If you are viewing this on
the web, the image below shows this process.
worry about how I got the exact or approximate slopes. We'll be computing the approximate
slopes shortly and we'll be able to compute the exact slope in a few sections.
In this figure we only looked at Q's that were to the right of P, but we could have just as easily
used Q's that were to the left of P and we would have received the same results. In fact, we
should always take a look at Q's that are on both sides of P. In this case the same thing is
happening on both sides of P. However, we will eventually see that doesn't have to happen.
Therefore we should always take a look at what is happening on both sides of the point in
question when doing this kind of process.
So, let's see if we can come up with the approximate slopes I showed above, and hence an
estimation of the slope of the tangent line. In order to simplify the process a little let's get a
formula for the slope of the line between P and Q,
PQ
m , that will work for any x that we choose
to work with. We can get a formula by finding the slope between P and Q using the "general"
form of ( ) ( )
, Q x f x = .
Now, let's pick some values of x getting closer and closer to 1 x = , plug in and get some
slopes.
x
PQ
m x
PQ
m
2 -6 0 -2
1.5 -5 0.5 -3
1.1 -4.2 0.9 -3.8
1.01 -4.02 0.99 -3.98
1.001 -4.002 0.999 -3.998
1.0001 -4.0002 0.9999 -3.9998
So, if we take x's to the right of 1 and move them in very close to 1 it appears that the slope of the
secant lines appears to be approaching -4. Likewise, if we take x's to the left of 1 and move them
in very close to 1 the slope of the secant lines again appears to be approaching -4.
Based on this evidence it seems that the slopes of the secant lines are approaching -4 as we move
in towards 1 x = , so we will estimate that the slope of the tangent line is also -4. As noted above,
this is the correct value and we will be able to prove this eventually.
There are a couple of important points to note about our work above. First, we looked at points
that were on both sides of 1 x = . In this kind of process it is important to never assume that what
is happening on one side of a point will also be happening on the other side as well. We should
always look at what is happening on both sides of the point. In this example we could sketch a
graph and from that guess that what is happening on one side will also be happening on the other,
but we will usually not have the graphs in front of us or be able to easily get them.
Next, notice that when we say we're going to move in close to the point in question we do mean
that we're going to move in very close and we also used more than just a couple of points. We
should never try to determine a trend based on a couple of points that aren't really all that close to
the point in question.
The next thing to notice is really a warning more than anything. The values of
PQ
m in this
example were fairly "nice" and it was pretty clear what value they were approaching after a
couple of computations. In most cases this will not be the case. Most values will be far
"messier" and you'll often need quite a few computations to be able to get an estimate.
Last, we were after something that was happening at 1 x = and we couldn't actually plug 1 x =
into our formula for the slope. Despite this limitation we were able to determine some
information about what was happening at 1 x = simply by looking at what was happening around
1 x = . This is more important than you might at first realize and we will be discussing this point
in detail in later sections.
Before moving on let's do a quick review of just what we did in the above example. We wanted
the tangent line to ( ) f x at a point x a = . First, we know that the point ( ) ( )
, P a f a = will be
on the tangent line. Next, we'll take a second point that is on the graph of the function, call it
( ) ( )
, Q x f x = and compute the slope of the line connecting P and Q as follows,
( ) ( )
PQ
f x f a
m
x a
−
=
−
We then take values of x that get closer and closer to x a = (making sure to look at x's on both
sides of x a = and use this list of values to estimate the slope of the tangent line, m.
The tangent line will then be,
( ) ( ) y f a m x a = + −
Rates of Change
The next problem that we need to look at is the rate of change problem. This will turn out to be
one of the most important concepts that we will look at throughout this course.
Here we are going to consider a function, f(x), that represents some quantity that varies as x
varies. For instance, maybe f(x) represents the amount of water in a holding tank after x minutes.
Or maybe f(x) is the distance traveled by a car after x hours. In both of these example we used x
to represent time. Of course x doesn't have to represent time, but it makes for examples that are
easy to visualize.
What we want to do here is determine just how fast f(x) is changing at some point, say x a = .
This is called the instantaneous rate of change or sometimes just rate of change of f(x) at
x a = .
As with the tangent line problem all that we're going to be able to do at this point is to estimate
the rate of change. So let's continue with the examples above and think of f(x) as something that
is changing in time and x being the time measurement. Again x doesn't have to represent time
but it will make the explanation a little easier. While we can't compute the instantaneous rate of
change at this point we can find the average rate of change.
To compute the average rate of change of f(x) at x a = all we need to do is to choose another
point, say x, and then the average rate of change will be,
( )
( ) ( )
change in
. . .
change in
f x
ARC
x
f x f a
x a
=
−
=
−
Then to estimate the instantaneous rate of change at x a = all we need to do is to choose values of
x getting closer and closer to x a = (don't forget to chose them on both sides of x a = ) and
compute values of A.R.C. We can then estimate the instantaneous rate of change form that.
Let's take a look at an example.
Example 2 Suppose that the amount of air in a balloon after t hours is given by
( )
3 2
6 35 V t t t = − +
Estimate the instantaneous rate of change of the volume after 5 hours.
Solution
Okay. The first thing that we need to do is get a formula for the average rate of change of the
volume. In this case this is,
So, from this table it looks like the average rate of change is approaching 15 and so we can
estimate that the instantaneous rate of change is 15 at this point.
So, just what does this tell us about the volume at this point? Let's put some units on the answer
from above. This might help us to see what is happening to the volume at this point. Let's
suppose that the units on the volume were in cm
3
. The units on the rate of change (both average
and instantaneous) are then cm
3
/hr.
We have estimated that at 5 t = the volume is changing at a rate of 15 cm
3
/hr. This means that
at 5 t = the volume is changing in such a way that, if the rate were constant, then an hour later
there would be 15 cm
3
more air in the balloon than there was at 5 t = .
We do need to be careful here however. In reality there probably won't be 15 cm
3
more air in the
balloon after an hour. The rate at which the volume is changing is generally not constant and so
we can't make any real determination as to what the volume will be in another hour. What we
can say is that the volume is increasing, since the instantaneous rate of change is positive, and if
we had rates of change for other values of t we could compare the numbers and see if the rate of
change is faster or slower at the other points.
For instance, at 4 t = the instantaneous rate of change is 0 cm
3
/hr and at 3 t = the instantaneous
rate of change is -9 cm
3
/hr. I'll leave it to you to check these rates of change. In fact, that would
be a good exercise to see if you can build a table of values that will support my claims on these
rates of change.
Anyway, back to the example. At 4 t = the rate of change is zero and so at this point in time the
volume is not changing at all. That doesn't mean that it will not change in the future. It just
means that exactly at 4 t = the volume isn't changing. Likewise at 3 t = the volume is
decreasing since the rate of change at that point is negative. We can also say that, regardless of
the increasing/decreasing aspects of the rate of change, the volume of the balloon is changing
faster at 5 t = than it is at 3 t = since 15 is larger than 9.
We will be talking a lot more about rates of change when we get into the next chapter.
Velocity Problem
Let's briefly look at the velocity problem. Many calculus books will treat this as its own
problem. I however, like to think of this as a special case of the rate of change problem. In the
velocity problem we are given a position function of an object, f(t), that gives the position of an
object at time t. Then to compute the instantaneous velocity of the object we just need to recall
that the velocity is nothing more than the rate at which the position is changing.
In other words, to estimate the instantaneous velocity we would first compute the average
velocity,
( ) ( )
change in position
. .
time traveled
AV
f t f a
t a
=
−
=
−
and then take values of t closer and closer to t a = and use these values to estimate the
instantaneous velocity.
Change of Notation
There is one last thing that we need to do in this section before we move on. The main point of
this section was to introduce us to a couple of key concepts and ideas that we will see throughout
the first portion of this course as well as get us started down the path towards limits.
Before we move into limits officially let's go back and do a little work that will relate both (or all
three if you include velocity as a separate problem) problems to a more general concept.
First, notice that whether we wanted the tangent line, instantaneous rate of change, or
instantaneous velocity each of these came down to using exactly the same formula. Namely,
( ) ( ) f x f a
x a
−
−
(1)
This should suggest that all three of these problems are then really the same problem. In fact this
is the case as we will see in the next chapter. We are really working the same problem in each of
these cases the only difference is the interpretation of the results.
In preparation for the next section where we will discuss this in much more detail we need to do a
quick change of notation. It's easier to do here since we've already invested a fair amount of
time into these problems.
In all of these problems we wanted to determine what was happening at x a = . To do this we
chose another value of x and plugged into (1). For what we were doing here that is probably most
intuitive way of doing it. However, when we start looking at these problems as a single problem
(1) will not be the best formula to work with.
What we'll do instead is to first determine how far from x a = we want to move and then define
our new point based on that decision. So, if we want to move a distance of h from x a = the new
point would be x a h = + .
As we saw in our work above it is important to take values of x that are both sides of x a = . This
way of choosing new value of x will do this for us. If h>0 we will get value of x that are to the
right of x a = and if h<0 we will get values of x that are to the left of x a = .
On the surface it might seem that (2) is going to be an overly complicated way of dealing with
this stuff. However, as we will see it will often be easier to deal with (2) than it will be to deal
with (1).
The Limit
In the previous section we looked at a couple of problems and in both problems we had a function
(slope in the tangent problem case and average rate of change in the rate of change problem) and
we wanted to know how that function was behaving at some point x a = . At this stage of the
game we no longer care where the functions came from and we no longer care if we're going to
see them down the road again or not. All that we need to know or worry about is that we've got
these functions and we want to know something about them.
To answer the questions in the last section we choose values of x that got closer and closer to
x a = and we plugged these into the function. We also made sure that we looked at values of x
that were on both the left and the right of x a = . Once we did this we looked at our table of
function values and saw what the function values were approaching as x got closer and closer to
x a = and used this to guess the value that we were after.
This process is called taking a limit and we have some notation for this. The limit notation for
the two problems from the last section is,
2 3 2
1 5
2 2 6 25
lim 4 lim 15
1 5
x t
x t t
x t
→ →
− − +
= − =
− −
In this notation we will note that we always give the function that we're working with and we
also give the value of x (or t) that we are moving in towards.
In this section we are going to take an intuitive approach to limits and try to get a feel for what
they are and what they can tell us about a function. With that goal in mind we are not going to
get into how we actually compute limits yet. We will instead rely on what we did in the previous
section as well as another approach to guess the value of the limits.
Both of the approaches that we are going to use in this section are designed to help us understand
just what limits are. In general we don't typically use the methods in this section to compute
limits and in many cases can be very difficult to use to even estimate the value of a limit and/or
will give the wrong value on occasion. We will look at actually computing limits in a couple of
sections.
Let's first start off with the following "definition" of a limit.
Definition
We say that the limit of f(x) is L as x approaches a and write this as
( ) lim
x a
f x L
→
=
provided we can make f(x) as close to L as we want for all x sufficiently close to a, from both
sides, without actually letting x be a.
This is not the exact, precise definition of a limit. If you would like to see the more precise and
mathematical definition of a limit you should check out the The Definition of a Limit section at
the end of this chapter. The definition given above is more of a "working" definition. This
definition helps us to get an idea of just what limits are and what they can tell us about functions.
So just what does this definition mean? Well let's suppose that we know that the limit does in
fact exist. According to our "working" definition we can then decide how close to L that we'd
like to make f(x). For sake of argument let's suppose that we want to make f(x) no more that
0.001 away from L. This means that we want one of the following
Now according to the "working" definition this means that if we get x sufficiently close to we can
make one of the above true. However, it actually says a little more. It actually says that
somewhere out there in the world is a value of x, say X, so that for all x's that are closer to a than
X then one of the above statements will be true.
This is actually a fairly important idea. There are many functions out there in the work that we
can make as close to L for specific values of x that are close to a, but there will other values of x
closer to a that give functions values that are nowhere near close to L. In order for a limit to exist
once we get f(x) as close to L as we want for some x then it will need to stay in that close to L (or
get closer) for all values of x that are closer to a. We'll see an example of this later in this
section.
In somewhat simpler terms the definition says that as x gets closer and closer to x=a (from both
sides of course…) then f(x) must be getting closer and closer to L. Or, as we move in towards
x=a then f(x) must be moving in towards L.
It is important to note once again that we must look at values of x that are on both sides of x=a.
We should also note that we are not allowed to use x=a in the definition. We will often use the
information that limits give us to get some information about what is going on right at x=a, but
the limit itself is not concerned with what is actually going on at x=a. The limit is only
concerned with what is going on around the point x=a. This is an important concept about limits
that we need to keep in mind.
An alternative notation that we will occasionally use in denoting limits is
( ) as f x L x a → →
Solution
Notice that I did say estimate the value of the limit. Again, we are not going to directly compute
limits in this section. The point of this section is to give us a better idea of how limits work and
what they can tell us about the function.
So, with that in mind we are going to work this in pretty much the same way that we did in the
last section. We will choose values of x that get closer and closer to x=2 and plug these values
into the function. Doing this gives the following table of values.
Note that we made sure and picked values of x that were on both sides of 2 x = and that
we moved in very close to 2 x = to make sure that any trends that we might be seeing are
in fact correct.
Also notice that we can't actually plug in 2 x = into the function as this would give us a
division by zero error. This is not a problem since the limit doesn't care what is
happening at the point in question.
From this table it appears that the function is going to 4 as x approaches 2, so
2
2
2
4 12
lim 4
2
x
x x
x x
→
+ −
=
−
.
Let's think a little bit more about what's going on here. Let's graph the function from the
last example. The graph of the function in the range of x's that were interested in is
shown below.
First, notice that there is a rather large open dot at 2 x = . This is there to remind us that
the function (and hence the graph) doesn't exist at 2 x = .
When we are computing limits the question that we are really asking is what y value is
our graph approaching as we move in towards x a = on our graph. We are NOT asking
what y value the graph takes at the point in question. In other words, we are asking what
the graph is doing around the point x a = . In our case we can see that as x moves in
towards 2 (from both sides) the function is approaching 4 y = even though the function
itself doesn't even exist at 2 x = . Therefore we can say that the limit is in fact 4.
So what have we learned about limits? Limits are asking what the function is doing around
x a = and are not concerned with what the function is actually doing at x a = . This is a good
thing as many of the functions that we'll be looking at won't even exist at x a = as we saw in our
last example.
Solution
The first thing to note here is that this is exactly the same function as the first example with the
exception that we've now given it a value for 2 x = . So, let's first note that
( ) 2 6 g =
As far as estimating the value of this limit goes, nothing has changed in comparison to the first
example. We could build up a table of values as we did in the first example or we could take a
quick look at the graph of the function. Either method will give us the value of the limit.
that are getting closer to 2 x = but we never take 2 x = . In other words the table of values that
we used in the first example will be exactly the same table that we'll use here. So, since we've
already got it down once there is no reason to redo it here.
From this table it is again clear that the limit is,
( )
2
lim 4
x
g x
→
=
The limit is NOT 6! Remember from the discussion after the first example that limits do not care
what the function is actually doing at the point in question. Limits are only concerned with what
is going on around the point. Since the only thing about the function that we actually changed
was its behavior at 2 x = this will not change the limit.
Let's also take a quick look at this functions graph to see if this says the same thing.
Again, we can see that as we move in towards 2 x = on our graph the function is still
approaching a y value of 4. Remember that we are only asking what the function is doing
around 2 x = and we don't care what the function is actually doing at 2 x = . The graph then
also supports the conclusion that the limit is,
( )
2
lim 4
x
g x
→
=
Let's make the point one more time just to make sure we've got it. Limits are not concerned with
what is going on at x a = . Limits are only concerned with what is going on around x a = . We
keep saying this, but it is a very important concept about limits that we must always keep in mind.
So, we will take every opportunity to remind ourselves of this idea.
value at a point. It happens sometimes and so we will need to be able to deal with those cases
when they arise.
Let's take a look another example to try and beat this idea into the ground.
Example 3 Estimate the value of the following limit.
( )
0
1 cos
lim
θ
θ
θ
→
−
Solution
First don't get excited about the θ in function. It's just a letter, just like x is a letter! It's a Greek
letter, but it's a letter and you will be asked to deal with Greek letters on occasion so it's a good
idea to start getting used to them at this point.
Now, also notice that if we plug in θ =0 that we will get division by zero and so the function
doesn't exist at this point. Actually, we get 0/0 at this point, but because of the division by zero
this function does not exist at θ =0.
So, as we did in the first example let's get a table of values and see what if we can guess what
value the function is heading in towards.
In all three of these function evaluations we evaluated the function at a number that is less that
0.001 and got three totally different numbers. Recall that the definition of the limit that we're
working with requires that the function be approaching a single value (our guess) as t gets closer
and closer to the point in question. It doesn't say that only some of the function values must be
getting closer to the guess. It says that all the function values must be getting closer and closer to
our guess.
To see what's happening here a graph of the function would be convenient.
From this graph we can see that as we move in towards 0 t = the function starts oscillating
wildly and in fact the oscillations increases in speed the closer to 0 t = that we get. Recall from
our definition of the limit that in order for a limit to exist the function must be settling down in
towards a single value as we get closer to the point in question.
This function clearly does not settle in towards a single number and so this limit does not exist!
This last example points out the drawback of just picking values of x using a table of function
values to estimate the value of a limit. The values of x that we chose in the previous example
were valid and in fact were probably values that many would have picked. In fact they were
exactly the same values we used in the problem before this one and they worked in that problem!
When using a table of values there will always be the possibility that we aren't choosing the
correct values and that we will guess incorrectly for our limit. This is something that we should
always keep in mind when doing this to guess the value of limits. In fact, this is such a problem
that after this section we will never use a table of values to guess the value of a limit again.
This last example also has shown us that limits do not have to exist. To this point we've only
seen limits that have existed, but that just doesn't always have to be the case.
Solution
This function is often called either the Heaviside or step function. We could use a table of values
to estimate the limit, but it's probably just as quick in this case to use the graph so let's do that.
Below is the graph of this function.
We can see from the graph that if we approach 0 t = from the right side the function is moving in
towards a y value of 1. Well actually it's just staying at 1, but in the terminology that we've been
using in this section it's moving in towards 1…
Also, if we move in towards 0 t = from the left the function is moving in towards a y value of 0.
we move in towards t a = (from both sides). This isn't happening in this case and so in this
example we will also say that the limit doesn't exist.
Note that the limit in this example is a little different from the previous example. In the previous
example the function did not settle down to a single number as we moved in towards 0 t = . In
this example however, the function does settle down to a single number as 0 t = on either side.
The problem is that the number is different on each side of 0 t = . This is an idea that we'll look
at in a little more detail in the next section.
Let's summarize what we (hopefully) learned in this section. In the first three examples we saw
that limits do not care what the function is actually doing at the point in question. They only are
concerned with what is happening around the point. In fact, we can have limits at x a = even if
the function itself does not exist at that point. Likewise, even if a function exists at a point there
is no reason (at this point) to think that the limit will have the same value as the function at that
point. Sometimes the limit and the function will have the same value at a point and other times
they won't have the same value.
Next, in the third and fourth examples we saw the main reason for not using a table of values to
guess the value of a limit. In those examples we used exactly the same set of values, however
they only worked in one of the examples. Using tables of values to guess the value of limits is
simply not a good way to get the value of a limit. This is the only section in which we will do
this. Tables of values should always be your last choice in finding values of limits.
The last two examples showed us that not all limits will in fact exist. We should not get locked
into the idea that limits will always exist. In most calculus courses we work with limits that
almost always exist and so it's easy to start thinking that limits always exist. Limits don't always
exist and so don't get into the habit of assuming that they will.
Finally, we saw in the fourth example that the only way to deal with the limit was to graph the
function. Sometimes this is the only way, however this example also illustrated the drawback of
using graphs. In order to use a graph to guess the value of the limit you need to be able to
actually sketch the graph. For many functions this is not that easy to do.
There is another drawback in using graphs. Even if you actually have the graph it's only going to
be useful if the y value is approaching an integer. If the y value is approaching say
15
123
−
there is
no way that you're going to be able to guess that value from the graph and we are usually going
to want exact values for our limits.
So while graphs of functions can, on occasion, make your life easier in guessing values of limits
they are again probably not the best way to get values of limits. They are only going to be useful
if you can get your hands on it and the value of the limit is a "nice" number.
The natural question then is why did we even talk about using tables and/or graphs to estimate
limits if they aren't the best way. There were a couple of reasons.
First, they can help us get a better understanding of what limits are and what they can tell us. If
we don't do at least a couple of limits in this way we might not get all that good of an idea on just
what limits are.
The second reason for doing limits in this way is to point out their drawback so that we aren't
tempted to use them all the time!
We will eventually talk about how we really do limits. However, there is one more topic that we
need to discuss before doing that. Since this section has already gone on for a while we will talk
about this in the next section.
OneSided Limits
In the final two examples in the previous section we saw two limits that did not exist. However,
the reason for each of the limits not existing was different for each of the examples.
We saw that
0
lim cos
t
t
π
→
⎛ ⎞
⎜ ⎟
⎝ ⎠
did not exist because the function did not settle down to a single value as t approached 0 t = .
The closer to 0 t = we moved the more wildly the function oscillated and in order for a limit to
exist the function must settle down to a single value.
did not exist not because the function didn't settle down to a single number as we moved in
towards 0 t = , but instead because it settled into two different numbers depending on which side
of 0 t = we were on.
In this case the function was a very well behaved function, unlike the first function. The only
problem was that, as we approached 0 t = , the function was moving in towards different numbers
on each side. We would like a way to differentiate between these two examples.
We do this with one-sided limits. As the name implies, with one-sided limits we will only be
looking at one side of the point in question. Here are the definitions for the two one sided limits.
Right-handed limit
We say
( ) lim
x a
f x L
+
→
=
provided we can make f(x) as close to L as we want for all x sufficiently close to a and x>a
without actually letting x be a.
Left-handed limit
We say
( ) lim
x a
f x L
−
→
=
provided we can make f(x) as close to L as we want for all x sufficiently close to a and x<a
without actually letting x be a.
x>a. Likewise for the left-handed limit we have x a
−
→ (note the "-") which means that we will
only be looking at x<a.
Also, note that as with the "normal" limit (i.e. the limits from the previous section) we still need
the function to settle down to a single number in order for the limit to exist. The only difference
this time is that the function only needs to settle down to a single number on either the right side
of x a = or the left side of x a = depending on the one-sided limit we're dealing with.
So when we are looking at limits it's now important to pay very close attention to see whether we
are doing a normal limit or one of the one-sided limits. Let's now take a look at the some of the
problems from the last section and look at one-sided limits instead of the normal limit.
So, we can see that if we stay to the right of 0 t = (i.e. 0 t > ) then the function is moving in
towards a value of 1 as we get closer and closer to 0 t = , but staying to the right. We can
therefore say that the right-handed limit is,
( )
0
lim 1
t
H t
+
→
=
Likewise, if we stay to the left of 0 t = (i.e 0 t < ) the function is moving in towards a value of 0
as we get closer and closer to 0 t = , but staying to the left. Therefore the left-handed limit is,
( )
0
lim 0
t
H t
−
→
=
we can see that both of the one-sided limits suffer the same problem that the normal limit did in
the previous section. The function does not settle down to a single number on either side of
0 t = . Therefore, neither the left-handed nor the right-handed limit will exist in this case.
So, one-sided limits don't have to exist just as normal limits aren't guaranteed to exist.
In this case regardless of which side of 2 x = we are on the function is always approaching a
value of 4 and so we get,
( ) ( )
2 2
lim 4 lim 4
x x
g x g x
+ −
→ →
= =
Note that one-sided limits do not care about what's happening at the point any more than normal
limits do. They are still only concerned with what is going on around the point. The only real
difference between one-sided limits and normal limits is the range of x's that we look at when
determining the value of the limit.
Now let's take a look at the first and last example in this section to get a very nice fact about the
relationship between one-sided limits and normal limits. In the last example the one-sided limits
as well as the normal limit existed and all three had a value of 4. In the first example the two
one-sided limits both existed, but did not have the same value and the normal limit did not exist.
The relationship between one-sided limits and normal limits can be summarized by the following
fact.
This fact can be turned around to also say that if the two one-sided limits have different values,
i.e.,
( ) ( ) lim lim
x a x a
f x f x
+ −
→ →
≠
then the normal limit will not exist.
This should make some sense. If the normal limit did exist then by the fact the two one-sided
limits would have to exist and have the same value by the above fact. So, if the two one-sided
limits have different values (or don't even exist) then the normal limit simply can't exist.
Let's take a look at one more example to make sure that we've got all the ideas about limits down
that we've looked at in the last couple of sections.
(d) ( )
4
lim 2
x
f x
→−
= We can do this one of two ways. Either we can use the fact here and notice
that the two one-sided limits are the same and so the normal limit must exist and have the same
value as the one-sided limits or just get the answer from the graph.
Also recall that a limit can exist at a point even if the function doesn't exist at that point.
(e) ( ) 1 4 f = . The function will take on the y value where the closed dot is.
(g) ( )
1
lim 2
x
f x
+
→
= − The function is approaching a value of -2 as x moves in towards 1 from the
right. Remember that the limit does NOT care about what the function is actually doing at the
point, it only cares about what the function is doing around the point. In this case, always staying
to the right of 1 x = , the function is approaching a value of -2 and so the limit is -2. The limit is
not 4, as that is value of the function at the point and again the limit doesn't care about that!
(h) ( )
1
lim
x
f x
→
doesn't exist. The two one-sided limits both exist, however they are different and
so the normal limit doesn't exist.
(i) ( ) 6 2 f = . The function will take on the y value where the closed dot is.
(l) ( )
6
lim 5
x
f x
→
= Again, we can use either the graph or the fact to get this. Also, once more
remember that the limit doesn't care what is happening at the point and so it's possible for the
limit to have a different value than the function at a point. When dealing with limits we've
always got to remember that limits simply do not care about what the function is doing at the
point in question. Limits are only concerned with what the function is doing around the point.
Hopefully over the last couple of sections you've gotten an idea on how limits work and what
they can tell us about functions. Some of these ideas will be important in later sections so it's
important that you have a good grasp on them.
Limit Properties
The time has almost come for us to actually compute some limits. However, before we do that
we will need some properties of limits that will make our life somewhat easier. So, let's take a
look at those first. The proof of some of these properties can be found in the Proof of Various
Limit Properties section of the Extras chapter.
So to take the limit of a sum or difference all we need to do is take the limit of the
individual parts and then put them back together with the appropriate sign. This is also
not limited to two functions. This fact will work no matter how many functions we've
got separated by "+" or "-".
We take the limits of products in the same way that we can take the limit of sums or
differences. Just take the limit of the pieces and then put them back together. Also, as
with sums or differences, this fact is not limited to just two functions.
As noted in the statement we only need to worry about the limit in the denominator being
zero when we do the limit of a quotient. If it were zero we would end up with a division
by zero error and we need to avoid that.
In other words, in this case we were the limit is the same value that we'd get by just evaluating
the function at the point in question. This seems to violate one of the main concepts about limits
that we've seen to this point.
In the previous two sections we made a big deal about the fact that limits do not care about what
is happening at the point in question. They only care about what is happening around the point.
So how does the previous example fit into this since it appears to violate this main idea about
limits?
Despite appearances the limit still doesn't care about what the function is doing at 2 x = − . In
this case the function that we've got is simply "nice enough" so that what is happening around the
point is exactly the same as what is happening at the point. Eventually we will formalize up just
what is meant by "nice enough". At this point let's not worry too much about what "nice
enough" is. Let's just take advantage of the fact that some functions will be "nice enough",
whatever that means.
The function in the last example was a polynomial. It turns out that all polynomials are "nice
enough" so that what is happening around the point is exactly the same as what is happening at
the point. This leads to the following fact.
Notice that the limit of the denominator wasn't zero and so our use of property 4 was legitimate.
Notice in this last example that again all we really did was evaluate the function at the point in
question. So it appears that there is a fairly large class of functions for which this can be done.
Let's generalize the fact from above a little.
Again, we will formalize up just what we mean by "nice enough" eventually. At this point all we
want to do is worry about which functions are "nice enough". Some functions are "nice enough"
for all x while others will only be "nice enough" for certain values of x. It will all depend on the
function.
As noted in the statement, this fact also holds for the two one-sided limits as well as the normal
limit.
Here is a list of some of the more common functions that are "nice enough".
The last bullet is important. This means that for any combination of these functions all we need
to do is evaluate the function at the point in question, making sure that none of the restrictions are
violated. This means that we can now do a large number of limits.
Computing Limits
In the previous section we saw that there is a large class of function that allows us to use
( ) ( ) lim
x a
f x f a
→
=
to compute limits. However, there are also many limits for which this won't work easily. The
purpose of this section is to develop techniques for dealing with some of these limits that will not
allow us to just use this fact.
Let's first got back and take a look at one of the first limits that we looked at and compute its
exact value and verify our guess for the limit.
Example 1 Evaluate the following limit.
2
2
2
4 12
lim
2
x
x x
x x
→
+ −
−
Solution
First let's notice that if we try to plug in 2 x = we get,
2
2
2
4 12 0
lim
2 0
x
x x
x x
→
+ −
=
−
So, we can't just plug in 2 x = to evaluate the limit. So, we're going to have to do something
else.
The first thing that we should always do when evaluating limits is to simplify the function as
much as possible. In this case that means factoring both the numerator and denominator. Doing
this gives,
So, upon factoring we saw that we could cancel an 2 x − from both the numerator and the
denominator. Upon doing this we now have a new rational expression that we can plug 2 x =
into because we lost the division by zero problem. Therefore, the limit is,
denominator is also zero. Likewise anything divided by itself is 1, unless we're talking about
zero.
So, there are really three competing "rules" here and it's not clear which one will win out. It's
also possible that none of them will win out and we will get something totally different from
undefined, zero, or one. We might, for instance, get a value of 4 out of this, to pick a number
completely at random.
There are many more kinds of indeterminate forms and we will be discussing indeterminate forms
at length in the next chapter.
Let's take a look at a couple of more examples.
Example 2 Evaluate the following limit.
( )
2
0
2 3 18
lim
h
h
h
→
− + −
Solution
In this case we also get 0/0 and factoring is not really an option. However, there is still some
simplification that we can do.
So, upon multiplying out the first term we get a little cancellation and now notice that we can
factor an h out of both terms in the numerator which will cancel against the h in the denominator
and the division by zero problem goes away and we can then evaluate the limit.
two examples will be of any help here, at least initially. We can't factor and we can't just
multiply something out to get the function to simplify.
When there is a square root in the numerator or denominator we can try to rationalize and see if
that helps. Recall that rationalizing makes use of the fact that
( )( )
2 2
a b a b a b + − = −
So, if either the first and/or the second term have a square root in them the rationalizing will
eliminate the root(s). This might help in evaluating the limit.
Notice that we didn't multiply the denominator out as well. Most students come out of an
Algebra class having it beaten into their heads to always multiply this stuff out. However, in this
case multiplying out will make the problem very difficult and in the end you'll just end up
factoring it back out anyway.
At this stage we are almost done. Notice that we can factor the numerator so let's do that.
In this case there really isn't a whole lot to do. In doing limits recall that we must always look at
what's happening on both sides of the point in question as we move in towards it. In this case
6 y = is completely inside the second interval for the function and so there are values of y on
both sides of 6 y = that are also inside this interval. This means that we can just use the fact to
evaluate this limit.
portion because this interval does not contain values of y to the left of 2 y = − and we need to
know what is happening on both sides of the point.
To do this part we are going to have to remember the fact from the section on one-sided limits
that says that if the two one-sided limits exist and are the same then the normal limit will also
exist and have the same value.
Notice that both of the one sided limits can be done here since we are only going to be looking at
one side of the point in question. So let's do the two one-sided limits and see what we get.
Note that this fact should make some sense to you if we assume that both functions are nice
enough. If both of the functions are "nice enough" to use the limit evaluation fact then we have,
( ) ( ) ( ) ( ) lim lim
x c x c
f x f c g c g x
→ →
= ≤ =
The inequality is true because we know that c is somewhere between a and b and in that range we
also know ( ) ( ) f x g x ≤ .
Note that we don't really need the two functions to be nice enough for the fact to be true, but it
does provide a nice way to give a quick "justification" for the fact.
Also, note that we said that we assumed that ( ) ( ) f x g x ≤ for all x on [a, b] (except possibly at
x c = ). Because limits do not care what is actually happening at x c = we don't really need the
inequality to hold at that specific point. We only need it to hold around x c = since that is what
the limit is concerned about.
As with the previous fact we only need to know that ( ) ( ) ( ) f x h x g x ≤ ≤ is true around x c =
because we are working with limits and they are only concerned with what is going on around
x c = and not what is actually happening at x c = .
sketch of what the Squeeze Theorem is telling us. The following figure illustrates what is
happening in this theorem.
From the figure we can see that if the limits of f(x) and g(x) are equal at x c = then the function
values must also be equal at x c = (this is where we're using the fact that we assumed the
functions where "nice enough", which isn't really required for the Theorem). However, because
h(x) is "squeezed" between f(x) and g(x) at this point then h(x) must have the same value.
Therefore, the limit of h(x) at this point must also be the same.
The Squeeze theorem is also known as the Sandwich Theorem and the Pinching Theorem.
So, how do we use this theorem to help us with limits? Let's take a look at the following
example to see the theorem in action.
Example 6 Evaluate the following limit.
2
0
1
lim cos
x
x
x
→
⎛ ⎞
⎜ ⎟
⎝ ⎠
Solution
In this example none of the previous examples can help us. There's no factoring or simplifying to
do. We can't rationalize and one-sided limits won't work. There's even a question as to whether
this limit will exist since we have division by zero inside the cosine at x=0.
The first thing to notice is that we know the following fact about cosine.
( ) 1 cos 1 x − ≤ ≤
Our function doesn't have just an x in the cosine, but as long as we avoid 0 x = we can say the
same thing for our cosine.
Now if we have the above inequality for our cosine we can just multiply everything by an x
2
and
get the following.
2 2 2
1
cos x x x
x
⎛ ⎞
− ≤ ≤
⎜ ⎟
⎝ ⎠
In other words we've managed to squeeze the function that we were interested in between two
other functions that are very easy to deal with. So, the limits of the two outer functions are.
( )
2 2
0 0
lim 0 lim 0
x x
x x
→ →
= − =
These are the same and so by the Squeeze theorem we must also have,
2
0
1
lim cos 0
x
x
x
→
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
We can verify this with the graph of the three functions. This is shown below.
In this section we've seen several tools that we can use to help us to compute limits in which we
can't just evaluate the function at the point in question. As we will see many of the limits that
we'll be doing in later sections will require one or more of these tools.
Infinite Limits
In this section we will take a look at limits whose value is infinity or minus infinity. These kinds
of limit will show up fairly regularly in later sections and in other courses and so you'll need to
be able to deal with them when you run across them.
The first thing we should probably do here is to define just what we mean when we sat that a limit
has a value of infinity or minus infinity.
Definition
We say
( ) lim
x a
f x
→
= ∞
if we can make f(x) arbitrarily large for all x sufficiently close to x=a, from both sides, without
actually letting x a = .
We say
( ) lim
x a
f x
→
= −∞
if we can make f(x) arbitrarily large and negative for all x sufficiently close to x=a, from both
sides, without actually letting x a = .
These definitions can be appropriately modified for the one-sided limits as well. To see a more
precise and mathematical definition of this kind of limit see the The Definition of the Limit
section at the end of this chapter.
Solution
So we're going to be taking a look at a couple of one-sided limits as well as the normal limit here.
In all three cases notice that we can't just plug in 0 x = . If we did we would get division by
zero. Also recall that the definitions above can be easily modified to give similar definitions for
the two one-sided limits which we'll be needing here.
Now, there are several ways we could proceed here to get values for these limits. One way is to
plug in some points and see what value the function is approaching. In the proceeding section we
said that we were no longer going to do this, but in this case it is a good way to illustrate just
what's going on with this function.
From this table we can see that as we make x smaller and smaller the function
1
x
gets larger and
larger and will retain the same sign that x originally had. It should make sense that this trend will
continue for any smaller value of x that we chose to use. The function is a constant (one in this
case) divided by an increasingly small number. The resulting fraction should be an increasingly
large number and as noted above the fraction will retain the same sign as x.
We can make the function as large and positive as we want for all x's sufficiently close to zero
while staying positive (i.e. on the right). Likewise, we can make the function as large and
negative as we want for all x's sufficiently close to zero while staying negative (i.e. on the left).
So, from our definition above it looks like we should have the following values for the two one
sided limits.
0 0
1 1
lim lim
x x
x x
+ −
→ →
= ∞ = −∞
Another way to see the values of the two one sided limits here is to graph the function. Again, in
the previous section we mentioned that we won't do this too often as most functions are not
something we can just quickly sketch out as well as the problems with accuracy in reading values
off the graph. In this case however, it's not too hard to sketch a graph of the function and, in this
case as we'll see accuracy is not really going to be an issue. So, here is a quick sketch of the
graph.
So, we can see from this graph that the function does behave much as we predicted that it would
from our table values. The closer x gets to zero from the right the larger (in the positive sense)
the function gets, while the closer x gets to zero from the left the larger (in the negative sense) the
function gets.
Finally, the normal limit, in this case, will not exist since the two one-sided have different values.
So, in summary here are the values of the three limits for this example.
0 0 0
1 1 1
lim lim lim doesn't exist
x x x
x x x
+ −
→ → →
= ∞ = −∞
For most of the remaining examples in this section we'll attempt to "talk our way through" each
limit. This means that we'll see if we can analyze what should happen to the function as we get
very close to the point in question without actually plugging in any values into the function. For
most of the following examples this kind of analysis shouldn't be all that difficult to do. We'll
also verify our analysis with a quick graph.
So, let's do a couple more examples.
Example 2 Evaluate each of the following limits.
2 2 2
0 0 0
6 6 6
lim lim lim
x x x
x x x
+ −
→ → →
Solution
As with the previous example let's start off by looking at the two one-sided limits. Once we have
those we'll be able to determine a value for the normal limit.
So, let's take a look at the right-hand limit first and as noted above let's see if we can see if we
can figure out what each limit will be doing without actually plugging in any values of x into the
function. As we take smaller and smaller values of x, while staying positive, squaring them will
only make them smaller (recall squaring a number between zero and one will make it smaller)
and of course it will stay positive. So we have a positive constant divided by an increasingly
small positive number. The result should then be an increasingly large positive number. It looks
like we should have the following value for the right-hand limit in this case,
2
0
6
lim
x
x
+
→
= ∞
Now, let's take a look at the left hand limit. In this case we're going to take smaller and smaller
values of x, while staying negative this time. When we square them we'll get smaller, but upon
squaring the result is now positive. So, we have a positive constant divided by an increasingly
small positive number. The result, as with the right hand limit, will be an increasingly large
positive number and so the left-hand limit will be,
Now, in this example, unlike the first one, the normal limit will exist and be infinity since the two
one-sided limits both exist and have the same value. So, in summary here are all the limits for
this example as well as a quick graph verifying the limits.
2 2 2
0 0 0
6 6 6
lim lim lim
x x x
x x x
+ −
→ → →
= ∞ = ∞ = ∞
With this next example we'll move away from just an x in the denominator, but as we'll see in the
next couple of examples they work pretty much the same way.
Example 3 Evaluate each of the following limits.
2 2 2
4 4 4
lim lim lim
2 2 2
x x x
x x x
+ −
→− →− →−
− − −
+ + +
Solution
Let's again start with the right-hand limit. With the right hand limit we know that we have,
2 2 0 x x > − ⇒ + >
Also, as x gets closer and closer to -2 then 2 x + will be getting closer and closer to zero, while
staying positive as noted above. So, for the right-hand limit, we'll have a negative constant
divided by an increasingly small positive number. The result will be an increasingly large and
negative number. So, it looks like the right-hand limit will be negative infinity.
For the left hand limit we have,
2 2 0 x x < − ⇒ + <
and 2 x + will get closer and closer to zero (and be negative) as x gets closer and closer to -2. In
this case then we'll have a negative constant divided by an increasingly small negative number.
The result will then be an increasingly large positive number and so it looks like the left-hand
limit will be positive infinity.
At this point we should briefly acknowledge the idea of vertical asymptotes. Each of the three
previous graphs have had one. Recall from an Algebra class that a vertical asymptote is a vertical
line (the dashed line at 2 x = − in the previous example) in which the graph will go towards
infinity and/or minus infinity on one or both sides of the line.
In an Algebra class they are a little difficult to define other than to say pretty much what we just
said. Now that we have infinite limits under our belt we can easily define a vertical asymptote as
follows,
Definition
The function f(x) will have a vertical asymptote at x a = if we have any of the following limits at
x a = .
( ) ( ) ( ) lim lim lim
x a x a x a
f x f x f x
− +
→ → →
= ±∞ = ±∞ = ±∞
Note that it only requires one of the above limits for a function to have a vertical asymptote at
x a = .
Using this definition we can see that the first two examples had vertical asymptotes at 0 x =
while the third example had a vertical asymptote at 2 x = − .
Solution
Let's start with the right-hand limit. For this limit we have,
( )
3
4 4 0 4 0 x x x > ⇒ − < ⇒ − <
also, 4 0 x − → as 4 x → . So, we have a positive constant divided by an increasingly small
negative number. The results will be an increasingly large negative number and so it looks like
the right-hand limit will be negative infinity.
For the left-handed limit we have,
( )
3
4 4 0 4 0 x x x < ⇒ − > ⇒ − >
and we still have, 4 0 x − → as 4 x → . In this case we have a positive constant divided by an
increasingly small positive number. The results will be an increasingly large positive number and
so it looks like the right-hand limit will be positive infinity.
The normal limit will not exist since the two one-sided limits are not the same. The official
answers to this example are then,
The main difference here with this example is the behavior of the numerator as we let x get closer
and closer to 3. In this case we have the following behavior for both the numerator and
denominator.
3 0 and 2 6 as 3 x x x − → → →
So, as we let x get closer and closer to 3 (always staying on the right of course) the numerator,
while not a constant, is getting closer and closer to a positive constant while the denominator is
getting closer and closer to zero, and will be positive since we are on the right side.
This means that we'll have a numerator that is getting closer and closer to a non-zero and positive
constant divided by an increasingly smaller positive number and so the result should be an
increasingly larger positive number. The right-hand limit should then be positive infinity.
For the left-hand limit we'll have,
3 3 0 x x < ⇒ − <
As with the right-hand limit we'll have the following behaviors for the numerator and the
denominator,
3 0 and 2 6 as 3 x x x − → → →
The main difference in this case is that the denominator will now be negative. So, we'll have a
numerator that is approaching a positive, non-zero constant divided by an increasingly small
negative number. The result will be an increasingly large and negative number.
So far all we've done is look at limits of rational expressions, let's do a couple of quick examples
with some different functions.
Example 6 Evaluate ( )
0
lim ln
x
x
+
→
Solution
First, notice that we can only evaluate the right-handed limit here. We know that the domain of
any logarithm is only the positive numbers and so we can't even talk about the left-handed limit
because that would necessitate the use of negative numbers. Likewise, since we can't deal with
the left-handed limit then we can't talk about the normal limit.
From this it's easy to see that we have the following values for each of these limits,
( ) ( )
2 2
lim tan lim tan
x x
x x
π π
+ −
→ →
= −∞ = ∞
Note that the normal limit will not exist because the two one-sided limits are not the same.
Limits At Infinity, Part I
In the previous section we saw limits that were infinity and it's now time to take a look at limits
at infinity. By limits at infinity we mean one of the following two limits.
( ) ( ) lim lim
x x
f x f x
→∞ →−∞
In other words, we are going to be looking at what happens to a function if we let x get very large
in either the positive or negative sense. Also, as well soon see, these limits may also have infinity
as a value.
For many of the limits that we're going to be looking at we will need the following facts.
Fact 1
1. If r is a positive rational number and c is any real number then,
lim 0
r
x
c
x
→∞
=
The first part of this fact should make sense if you think about it. Because we are requiring
0 r > we know that x
r
will stay in the denominator. Next as we increase x then x
r
will also
increase. So, we have a constant divided by an increasingly large number and so the result will
be increasingly small. Or, in the limit we will get zero.
The second part is nearly identical except we need to worry about x
r
being defined for negative x.
This condition is here to avoid cases such as
1
2
r = . If this r were allowed then we'd be taking
the square root of negative numbers which would be complex and we want to avoid that at this
level.
Note as well that the sign of c will not affect the answer. Regardless of the sign of c we'll still
have a constant divided by a very large number which will result in a very small number and the
larger x get the smaller the fraction gets. The sign of c will affect which direction the fraction
approaches zero (i.e. from the positive or negative side) but it still approaches zero.
To see the proof of this fact see the Proof of Various Limit Properties section in the Extras
chapter.
Let's start the off the examples with one that will lead us to a nice idea that we'll use on a regular
basis about limits at infinity for polynomials.
Solution
(a)
4 2
lim2 8
x
x x x
→∞
− −
Our first thought here is probably to just "plug" infinity into the polynomial and "evaluate" each
term to determine the value of the limit. It is pretty simple to see what each term will do in the
limit and so this seems like an obvious step, especially since we've been doing that for other
limits in previous sections.
So, let's see what we get if we do that. As x approaches infinity, then x to a power can only get
larger and the coefficient on each term (the first and third) will only make the term even larger.
So, if we look at what each term is doing in the limit we get the following,
4 2
lim2 8
x
x x x
→∞
− − = ∞−∞−∞
Now, we've got a small, but easily fixed, problem to deal with. We are probably tempted to say
that the answer is zero (because we have an infinity minus an infinity) or maybe −∞(because
we're subtracting two infinities off of one infinity). However, in both cases we'd be wrong. This
is one of those indeterminate forms that we first started seeing in a previous section.
Infinities just don't always behave as real numbers do when it comes to arithmetic. Without more
work there is simply no way to know what ∞−∞ will be and so we really need to be careful
with this kind of problem. To read a little more about this see the Types of Infinity section in the
Extras chapter.
So, we need a way to get around this problem. What we'll do here is factor the largest power of x
out of the whole polynomial as follows,
If you're not sure you agree with the factoring above (there's a chance you haven't really been
asked to do this kind of factoring prior to this) then recall that to check all you need to do is
multiply the
4
x back through the parenthesis to verify it was done correctly. Also, an easy way
to remember how to do this kind of factoring is to note that the second term is just the original
polynomial divided by
4
x . This will always work when factoring a power of x out of a
polynomial.
The first limit is clearly infinity and for the second limit we'll use the fact above on the last two
terms and so we'll arrive at the following value of the limit,
( )( )
4 2
lim2 8 2
x
x x x
→∞
− − = ∞ = ∞
Note that while we can't give a value for ∞−∞, if we multiply an infinity by a constant we will
still have an infinity no matter how large or small the constant is. The only thing that we need to
be careful of is signs. If the constant had been negative then we'd have gotten negative infinity
for a value.
[Return to Problems]
(b)
5 3 2
1
3
lim 2 8
t
t t t
→−∞
+ − +
We'll work this part much quicker than the previous part. All we need to do is factor out the
largest power of t to get the following,
Remember that all you need to do to get the factoring correct is divide the original polynomial by
the power of t we're factoring out,
5
t in this case.
Now all we need to do is take the limit of the two terms. In the first don't forget that since we're
going out towards −∞ and we're raising t to the 5
th
power that the limit will be negative
(negative number raised to an odd power is still negative). In the second term well again make
heavy use of the fact above.
What this fact is really saying is that when we go to take a limit at infinity for a polynomial then
all we need to really do is look at the term with the largest power and ask what that term is doing
in the limit since the polynomial will have the same behavior.
You can see the proof in the Proof of Various Limit Properties section in the Extras chapter.
Solution
First, the only difference between these two is that one is going to positive infinity and the other
is going to negative infinity. Sometimes this small difference will affect then value of the limit
and at other times it won't.
Let's start with the first limit and as with our first set of examples it might be tempting to just
"plug" in the infinity. Since both the numerator and denominator are polynomials we can use the
above fact to determine the behavior of each. Doing this gives,
4 2
4
2 8
lim
5 7
x
x x x
x
→∞
− + ∞
=
− + −∞
This is yet another indeterminate form. In this case we might be tempted to say that the limit is
infinity (because of the infinity in the numerator), zero (because of the infinity in the
denominator) or -1 (because something divided by itself is one). There are three separate
arithmetic "rules" at work here and without work there is no way to know which "rule" will be
correct and to make matters worse it's possible that none of them may work and we might get a
completely different answer, say
2
5
− to pick a number completely at random.
In this case the indeterminate form was neither of the "obvious" choices of infinity, zero, or -1 so
be careful with make these kinds of assumptions with this kind of indeterminate forms.
The second limit is done in a similar fashion. Notice however, that nowhere in the work for the
first limit did we actually use the fact that the limit was going to plus infinity. In this case it
doesn't matter which infinity we are going towards we will get the same value for the limit.
4 2
4
2 8 2
lim
5 7 5
x
x x x
x
→−∞
− +
= −
− +
In the previous example the infinity that we were using in the limit didn't change the answer.
This will not always be the case so don't make the assumption that this will always be the case.
Let's take a look at an example where we get different answers for each limit.
Example 3 Evaluate each of the following limits.
2 2
3 6 3 6
lim lim
5 2 5 2
x x
x x
x x
→∞ →−∞
+ +
− −
Solution
The square root in this problem won't change our work, but it will make the work a little messier.
Let's start with the first limit. In this case the largest power of x in the denominator is just an x.
So we need to factor an x out of the numerator and the denominator. When we are done factoring
the x out we will need an x in both of the numerator and the denominator. To get this in the
numerator we will have to factor an x
2
out of the square root so that after we take the square root
we will get an x.
This is probably not something you're used to doing, but just remember that when it comes out of
the square root it needs to be an x and the only way have an x come out of a square is to take the
square root of x
2
and so that is what we'll need to factor out of the term under the radical. Here's
the factoring work for this part,
This is where we need to be really careful with the square root in the problem. Don't forget that
2
x x =
Square roots are ALWAYS positive and so we need the absolute value bars on the x to make sure
that it will give a positive answer. This is not something that most people every remember seeing
in an Algebra class and in fact it's not always given in an Algebra class. However, at this point it
becomes absolutely vital that we know and use this fact. Using this fact the limit becomes,
Let's now take a look at the second limit (the one with negative infinity). In this case we will
need to pay attention to the limit that we are using. The initial work will be the same up until we
reach the following step.
In this limit we are going to minus infinity so in this case we can assume that x is negative. So, in
order to drop the absolute value bars in this case we will need to tack on a minus sign as well.
The limit is then,
So, as we saw in the last two examples sometimes the infinity in the limit will affect the answer
and other times it won't. Note as well that it doesn't always just change the sign of the number.
It can on occasion completely change the value. We'll see an example of this later in this section.
Before moving on to a couple of more examples let's revisit the idea of asymptotes that we first
saw in the previous section. Just as we can have vertical asymptotes defined in terms of limits we
can also have horizontal asymptotes defined in terms of limits.
Definition
The function f(x) will have a horizontal asymptote at y=L if either of the following are true.
( ) ( ) lim lim
x x
f x L f x L
→∞ →−∞
= =
Let's work another couple of examples involving of rational expressions.
Example 4 Evaluate each of the following limits.
2 6 2 6
3 3
4 4
lim lim
1 5 1 5
z z
z z z z
z z
→∞ →−∞
+ +
− −
Solution
Let's do the first limit and in this case it looks like we will factor a z
3
out of both the numerator
and denominator. Remember that we only look at the denominator when determining the largest
power of z here. There is a larger power of z in the numerator but we ignore it. We ONLY look
at the denominator when doing this! So doing the factoring gives,
When we take the limit we'll need to be a little careful. The first term in the numerator and
denominator will both be zero. However, the z
3
in the numerator will be going to plus infinity in
the limit and so the limit is,
2 6
3
4
lim
1 5 5
z
z z
z
→−∞
+ ∞
= = −∞
− −
The final limit is negative because we have a quotient of positive quantity and a negative
quantity.
In this case the z
3
in the numerator gives negative infinity in the limit since we are going out to
minus infinity and the power is odd. The answer is positive since we have a quotient of two
negative numbers.
Example 5 Evaluate the following limit.
2
4 3
5 9
lim
2 3
t
t t
t t
→−∞
− −
+
Solution
In this case it looks like we will factor a
4
t out of both the numerator and denominator. Doing
this gives,
In this case using Fact 1 we can see that the numerator is zero and so since the denominator is
also not zero the fraction, and hence the limit, will be zero.
In this section we concentrated on limits at infinity with functions that only involved polynomials
and/or rational expression involving polynomials. There are many more types of functions that
we could use here. That is the subject of the next section.
To see a precise and mathematical definition of this kind of limit see the The Definition of the
Limit section at the end of this chapter.
Limits At Infinity, Part II
In the previous section we look at limit at infinity of polynomials and/or rational expression
involving polynomials. In this section we want to take a look at some other types of functions
that often show up in limits at infinity. The functions we'll be looking at here are exponentials,
natural logarithms and inverse tangents.
Let's start by taking a look at a some of very basic examples involving exponential functions.
The main point of this example was to point out that if the exponent of an exponential goes to
infinity in the limit then the exponential function will also go to infinity in the limit. Likewise, if
the exponent goes to minus infinity in the limit then the exponential will go to zero in the limit.
The exponent goes to infinity in the limit and so the exponential will also need to go to infinity in
the limit. Or,
4 2
5 1
lim
t
t t − +
→−∞
= ∞ e
[Return to Problems]
(c)
1
0
lim
z
z
+
→
e
On the surface this part doesn't appear to belong in this section since it isn't a limit at infinity.
However, it does fit into the ideas we're examining in this set of examples.
So, let's first note that using the idea from the previous section we have,
0
1
lim
z
z
+
→
= ∞
Remember that in order to do this limit here we do need to do a right hand limit.
So, the exponent goes to infinity in the limit and so the exponential must also go to infinity.
Here's the answer to this part.
1
0
lim
z
z
+
→
= ∞ e
[Return to Problems]
Let's work some more complicated examples involving exponentials. In the following set of
examples it won't be that the exponents are more complicated, but instead that there will be more
than one exponential function to deal with.
The last two terms aren't any problem (they will be in the next part however, do you see that?).
The first three are a problem however as they present us with another indeterminate form.
When dealing with polynomials we factored out the term with the largest exponent in it. Let's do
the same thing here. However, we now have to deal with both positive and negative exponents
and just what do we mean by the "largest" exponent. When dealing with these here we look at
the terms that are causing the problems and ask which is the largest exponent in those terms. So,
since only the first three terms are causing us problems (i.e. they all evaluate to an infinity in the
limit) we'll look only at those.
So, since 10x is the largest of the three exponents there we'll "factor" an
10x
e out of the whole
thing. Just as with polynomials we do the factoring by, in essence, dividing each term by
10x
e
and remembering that to simply the division all we need to do is subtract the exponents. For
example, let's just take a look at the last term,
Notice that in doing this factoring all the remaining exponentials now have negative exponents
and we know that for this limit (i.e. going out to positive infinity) these will all be zero in the
limit and so will no longer cause problems.
(b)
10 6 2 15
lim 4 3 2 9
x
x x x x x − −
→−∞
− + + − e e e e e
Let's start this one off in the same manner as the first part. Let's take the limit of each of the
pieces. This time note that because our limit is going to negative infinity the first three
exponentials will in fact go to zero (because their exponents go to minus infinity in the limit).
The final two exponentials will go to infinity in the limit (because their exponents go to plus
infinity in the limit).
So, the last two terms are the problem here as they once again leave us with an indeterminate
form. As with the first example we're going to factor out the "largest" exponent in the last two
terms. This time however, "largest" doesn't refer to the bigger of the two numbers (-2 is bigger
than -15). Instead we're going to use "largest" to refer to the exponent that is farther away from
zero. Using this definition of "largest" means that we're going to factor an
15x −
e out.
Again, remember that to factor this out all we really are doing is dividing each term by
15x −
e and
then subtracting exponents. Here's the work for the first term as an example,
( )
10
10 15 25
15
x
x x x
x
− −
−
= =
e
e e
e
As with the first part we can either factor it out of only the "problem" terms (i.e. the last two
terms), or all the terms. For the practice we'll factor it out of all the terms. Here is the factoring
work for this limit,
So, when dealing with sums and/or differences of exponential functions we look for the
exponential with the "largest" exponent and remember here that "largest" means the exponent
farthest from zero. Also remember that if we're looking at a limit at plus infinity only the
exponentials with positive exponents are going to cause problems so those are the only terms we
look at in determining the largest exponent. Likewise, if we are looking at a limit at minus
infinity then only exponentials with negative exponents are going to cause problems and so only
those are looked at in determining the largest exponent.
Finally, as you might have been able to guess from the previous example when dealing with a
sum and/or difference of exponentials all we need to do is look at the largest exponent to
determine the behavior of the whole expression. Again, remembering that if the limit is at plus
infinity we only look at exponentials with positive exponents and if we're looking at a limit at
minus infinity we only look at exponentials with negative exponents.
Let's next take a look at some rational functions involving exponentials.
Solution
As with the previous example, the only difference between the first two parts is that one of the
limits is going to plus infinity and the other is going to minus infinity and just as with the
previous example each will need to be worked differently.
(a)
4 2
4 2
6
lim
8 3
x
x x
x x x
−
−
→∞
−
− +
e e
e e e
The basic concept involved in working this problem is the same as with rational expressions in
the previous section. We look at the denominator and determine the exponential function with
the "largest" exponent which we will then factor out from both numerator and denominator. We
will use the same reasoning as we did with the previous example to determine the "largest"
exponent. In the case since we are looking at a limit at plus infinity we only look at exponentials
with positive exponents.
So, we'll factor an
4x
e out of both then numerator and denominator. Once that is done we can
cancel the
4x
e and then take the limit of the remaining terms. Here is the work for this limit,
In this case we're going to minus infinity in the limit and so we'll look at exponentials in the
denominator with negative exponents in determining the "largest" exponent. There's only one
however in this problem so that is what we'll use.
Again, remember to only look at the denominator. Do NOT use the exponential from the
numerator, even though that one is "larger" than the exponential in then denominator. We always
look only at the denominator when determining what term to factor out regardless of what is
going on in the numerator.
Note that we had to do a right-handed limit for the first one since we can't plug negative x's into a
logarithm. This means that the normal limit won't exist since we must look at x's from both sides
of the point in question and x's to the left of zero are negative.
From the previous example we can see that if the argument of a log (the stuff we're taking the log
of) goes to zero from the right (i.e. always positive) then the log goes to negative infinity in the
limit while if the argument goes to infinity then the log also goes to infinity in the limit.
Note as well that we can't look at a limit of a logarithm as x approaches minus infinity since we
can't plug negative numbers into the logarithm.
The argument of the log is going to infinity and so the log must also be going to infinity in the
limit. The answer to this part is then,
( )
3 2
limln 7 1
x
x x
→∞
− + = ∞
[Return to Problems]
(b)
2
1
lim ln
5
t
t t
→−∞
⎛ ⎞
⎜ ⎟
−
⎝ ⎠
First, note that the limit going to negative infinity here isn't a violation (necessarily) of the fact
that we can't plug negative numbers into the logarithm. The real issue is whether or not the
argument of the log will be negative or not.
to minus infinity in the limit) the denominator will always be positive and so the quotient will
also always be positive. Therefore, not only does the argument go to zero, it goes to zero from
the right. This is exactly what we need to do this limit.
So, the answer here is,
2
1
lim ln
5
t
t t
→−∞
⎛ ⎞
= −∞
⎜ ⎟
−
⎝ ⎠
[Return to Problems]
As a final set of examples let's take a look at some limits involving inverse tangents.
Solution
The first two parts here are really just the basic limits involving inverse tangents and can easily be
found by examining the following sketch of inverse tangents. The remaining two parts are more
involved but as with the exponential and logarithm limits really just refer back to the first two
parts as we'll see.
Again, not much to do here other than examine the graph of the inverse tangent.
1
lim tan
2
x
x
π
−
→−∞
= −
[Return to Problems]
(c)
( )
1 3
limtan 5 6
x
x x
−
→∞
− +
Okay, in part (a) above we saw that if the argument of the inverse tangent function (the stuff
inside the parenthesis) goes to plus infinity then we know the value of the limit. In this case
(using the techniques from the previous section) we have,
3
lim 5 6
x
x x
→∞
− + = ∞
So, this limit is,
( )
1 3
limtan 5 6
2
x
x x
π
−
→∞
− + =
[Return to Problems]
(d)
1
0
1
lim tan
x
x
−
−
→
⎛ ⎞
⎜ ⎟
⎝ ⎠
Even though this limit is not a limit at infinity we're still looking at the same basic idea here.
We'll use part (b) from above as a guide for this limit. We know from the Infinite Limits section
that we have the following limit for the argument of this inverse tangent,
0
1
lim
x
x
−
→
= −∞
So, since the argument goes to minus infinity in the limit we know that this limit must be,
1
0
1
lim tan
2
x
x
π
−
−
→
⎛ ⎞
= −
⎜ ⎟
⎝ ⎠
[Return to Problems]
To see a precise and mathematical definition of this kind of limit see the The Definition of the
Limit section at the end of this chapter.
Continuity
Over the last few sections we've been using the term "nice enough" to define those functions that
we could evaluate limits by just evaluating the function at the point in question. It's now time to
formally define what we mean by "nice enough".
Definition
A function ( ) f x is said to be continuous at x a = if
( ) ( ) lim
x a
f x f a
→
=
A function is said to be continuous on the interval [a, b] if it is continuous at each point in the
interval.
This is exactly the same fact that we first put down back when we started looking at limits with
the exception that we have replaced the phrase "nice enough" with continuous.
It's nice to finally know what we mean by "nice enough", however, the definition doesn't really
tell us just what it means for a function to be continuous. Let's take a look at an example to help
us understand just what it means for a function to be continuous.
function value at that point. If they are equal the function is continuous at that point and if they
aren't equal the function isn't continuous at that point.
First 2 x = − .
( ) ( )
2
2 2 lim doesn't exist
x
f f x
→−
− =
The function value and the limit aren't the same and so the function is not continuous at this
point. This kind of discontinuity in a graph is called a jump discontinuity. Jump discontinuities
occur where the graph has a break in it is as this graph does.
Now 0 x = .
( ) ( )
0
0 1 lim 1
x
f f x
→
= =
The function is continuous at this point since the function and limit have the same value.
Finally 3 x = .
( ) ( )
3
3 1 lim 0
x
f f x
→
= − =
The function is not continuous at this point. This kind of discontinuity is called a removable
discontinuity. Removable discontinuities are those where there is a hole in the graph as there is
in this case.
From this example we can get a quick "working" definition of continuity. A function is
continuous on an interval if we can draw the graph from start to finish without ever once picking
up our pencil. The graph in the last example has only two discontinuities since there are only two
places where we would have to pick up our pencil in sketching it.
In other words, a function is continuous if its graph has no holes or breaks in it.
For many functions it's easy to determine where it won't be continuous. Functions won't be
continuous where we have things like division by zero or logarithms of zero. Let's take a quick
look at an example of determining where a function is not continuous.
Solution
Rational functions are continuous everywhere except where we have division by zero. So all that
we need to is determine where the denominator is zero. That's easy enough to determine by
setting the denominator equal to zero and solving.
( )( )
2
2 15 5 3 0 t t t t − − = − + =
To see a proof of this fact see the Proof of Various Limit Properties section in the Extras chapter.
With this fact we can now do limits like the following example.
Example 3 Evaluate the following limit.
0
sin
lim
x
x
→
e
Solution
Since we know that exponentials are continuous everywhere we can use the fact above.
0
0
0
limsin
sin
lim 1
x
x
x
x
→
→
= = = e e e
Another very nice consequence of continuity is the Intermediate Value Theorem.
Intermediate Value Theorem
Suppose that f(x) is continuous on [a, b] and let M be any number between f(a) and f(b). Then
there exists a number c such that,
1. a c b < <
2. ( ) f c M =
All the Intermediate Value Theorem is really saying is that a continuous function will take on all
values between f(a) and f(b). Below is a graph of a continuous function that illustrates the
Intermediate Value Theorem.
As we can see from this image if we pick any value, M, that is between the value of f(a) and the
value of f(b) and draw a line straight out from this point the line will hit the graph in at least one
point. In other words somewhere between a and b the function will take on the value of M. Also,
as the figure shows the function may take on the value at more than one place.
It's also important to note that the Intermediate Value Theorem only says that the function will
take on the value of M somewhere between a and b. It doesn't say just what that value will be. It
only says that it exists.
So, the Intermediate Value Theorem tells us that a function will take the value of M somewhere
between a and b but it doesn't tell us where it will take the value nor does it tell us how many
times it will take the value. There are important idea to remember about the Intermediate Value
Theorem.
A nice use of the Intermediate Value Theorem is to prove the existence of roots of equations as
the following example shows.
Solution
What we're really asking here is whether or not the function will take on the value
( ) 0 p x =
somewhere between -1 and 2. In other words, we want to show that there is a number c such that
1 2 c − < < and ( ) 0 p c = . However if we define 0 M = and acknowledge that 1 a = − and
2 b = we can see that these two condition on c are exactly the conclusions of the Intermediate
Value Theorem.
So, this problem is set up to use the Intermediate Value Theorem and in fact, all we need to do is
to show that the function is continuous and that 0 M = is between ( ) 1 p − and ( ) 2 p (i.e.
( ) ( ) 1 0 2 p p − < < or ( ) ( ) 2 0 1 p p < < − and we'll be done.
For the sake of completeness here is a graph showing the root that we just proved existed. Note
that we used a computer program to actually find the root and that the Intermediate Value
Theorem did not tell us what this value was.
Let's take a look at another example of the Intermediate Value Theorem.
Solution
Okay, so much as the previous example we're being asked to determine, if possible, if the
function takes on either of the two values above in the interval [0,5]. First, let's notice that this is
a continuous function and so we know that we can use the Intermediate Value Theorem to do this
problem.
Now, for each part we will let M be the given value for that part and then we'll need to show that
M lives between ( ) 0 f and ( ) 5 f . If it does then we can use the Intermediate Value Theorem to
prove that the function will take the given value.
So, since we'll need the two function evaluations for each part let's give them here,
So, by the Intermediate Value Theorem there must be a number 0 5 c ≤ ≤ such that
( ) 10 f c =
[Return to Problems]
(b) In this part we'll define 10 M = − . We now have a problem. In this part M does not live
between ( ) 0 f and ( ) 5 f . So, what does this mean for us? Does this mean that ( ) 10 f x ≠ −
in [0,5]?
Unfortunately for us, this doesn't mean anything. It is possible that ( ) 10 f x ≠ − in [0,5], but is
it also possible that ( ) 10 f x = − in [0,5]. The Intermediate Value Theorem will only tell us that
c's will exist. The theorem will NOT tell us that c's don't exist.
In this case it is not possible to determine if ( ) 10 f x = − in [0,5] using the Intermediate Value
Theorem.
[Return to Problems]
Okay, as the previous example has shown, the Intermediate Value Theorem will not always be
able to tell us what we want to know. Sometimes we can use it to verify that a function will take
some value in a given interval and in other cases we won't be able to use it.
From this graph we can see that not only does ( ) 10 f x = − in [0,5] it does so a total of 4 times!
Also note that as we verified in the first part of the previous example ( ) 10 f x = in [0,5] and in
fact it does so a total of 3 times.
So, remember that the Intermediate Value Theorem will only verify that a function will take on a
given value. It will never exclude a value from being taken by the function. Also, if we can use
the Intermediate Value Theorem to verify that a function will take on a value it never tells us how
many times the function will the value, it only tells us that it does take the value.
The Definition of the Limit
In this section we're going to be taking a look at the precise, mathematical definition of the three
kinds of limits we looked at in this chapter. We'll be looking at the precise definition of limits at
finite points that have finite values, limits that are infinity and limits at infinity. We'll also give
the precise, mathematical definition of continuity.
Let's start this section out with the definition of a limit at a finite point that has a finite value.
Definition 1 Let f(x) be a function defined on an interval that contains x a = , except possibly at
x a = . Then we say that,
( ) lim
x a
f x L
→
=
if for every number 0 ε > there is some number 0 δ > such that
( ) whenever 0 f x L x a ε δ − < < − <
Wow. That's a mouth full. Now that it's written down, just what does this mean?
Let's take a look at the following graph and let's also assume that the limit does exist.
What the definition is telling us is that for any number 0 ε > that we pick we can go to our graph
and sketch two horizontal lines at L ε + and L ε − as shown on the graph above. Then
somewhere out there in the world is another number 0 δ > , which we will need to determine,
that will allow us to add in two vertical lines to our graph at a δ + and a δ − .
Now, if we take any x in the pink region, i.e. between a δ + and a δ − , then this x will be closer
to a than either of a δ + and a δ − . Or,
x a δ − <
If we now identify the point on the graph that our choice of x gives then this point on the graph
will lie in the intersection of the pink and yellow region. This means that this function value f(x)
will be closer to L than either of L ε + and L ε − . Or,
( ) f x L ε − <
So, if we take any value of x in the pink region then the graph for those values of x will lie in the
yellow region.
Notice that there are actually an infinite number of possible δ 's that we can choose. In fact, if we
go back and look at the graph above it looks like we could have taken a slightly larger δ and still
gotten the graph from that pink region to be completely contained in the yellow region.
Also, notice that as the definition points out we only need to make sure that the function is
defined in some interval around x a = but we don't really care if it is defined at x a = .
Remember that limits do not care what is happening at the point, they only care what is
happening around the point in question.
Okay, now that we've gotten the definition out of the way and made an attempt to understand it
let's see how it's actually used in practice.
These are a little tricky sometimes and it can take a lot of practice to get good at these so don't
feel too bad if you don't pick up on this stuff right away. We're going to be looking a couple of
examples that work out fairly easily.
Example 1 Use the definition of the limit to prove the following limit.
2
0
lim 0
x
x
→
=
Solution
In this case both L and a are zero. So, let 0 ε > be any number. Don't worry about what the
number is, ε is just some arbitrary number. Now according to the definition of the limit, if this
limit is to be true we will need to find some other number 0 δ > so that the following will be
true.
Now, the results of this simplification looks an awful lot like 0 x δ < < with the exception of
the " 0 < " part. Missing that however isn't a problem, it is just telling us that we can't take
0 x = . So, it looks like if we choose δ ε = we should get what we want.
Verification is in fact pretty much the same work that we did to get our guess. First, let's again
let 0 ε > be any number and then choose δ ε = . Now, assume that 0 x ε < < . We need to
show that by choosing x to satisfy this we will get,
2
x ε <
To start the verification process we'll start with
2
x and then first strip out the exponent from the
absolute values. Once this is done we'll use our assumption on x, namely that x ε < . Doing
all this gives,
inequality to get our guess for δ and then seemingly went through exactly the same work to then
prove that our guess was correct. This is often who these work, although we will see an example
here in a bit where things don't work out quite so nicely.
So, having said that let's take a look at a slightly more complicated limit, although this one will
still be fairly similar to the first example.
Example 2 Use the definition of the limit to prove the following limit.
2
lim5 4 6
x
x
→
− =
Solution
We'll start this one out the same way that we did the first one. We won't be putting in quite the
same amount of explanation however.
Let's start off by letting 0 ε > be any number then we need to find a number 0 δ > so that the
following will be true.
So, as with the first example it looks like if we do enough simplification on the left inequality we
get something that looks an awful lot like the right inequality and this leads us to choose
5
ε
δ = .
Let's now verify this guess. So, again let 0 ε > be any number and then choose
5
ε
δ = . Next,
assume that 0 2
5
x
ε
δ < − < = and we get the following,
Okay, so again the process seems to suggest that we have to essentially redo all our work twice,
once to make the guess for δ and then another time to prove our guess. Let's do an example that
doesn't work out quite so nicely.
Example 3 Use the definition of the limit to prove the following limit.
2
4
lim 11 9
x
x x
→
+ − =
Solution
So, let's get started. Let 0 ε > be any number then we need to find a number 0 δ > so that the
following will be true.
( )
2
11 9 whenever 0 4 x x x ε δ + − − < < − <
We'll start the guess process in the same manner as the previous two examples.
Okay, we've managed to show that
( )
2
11 9 x x ε + − − < is equivalent to 5 4 x x ε + − < .
However, unlike the previous two examples, we've got an extra term in here that doesn't show up
in the right inequality above. If we have any hope of proceeding here we're going need to find
some way to get rid of the 5 x + .
To do this let's just note that if, by some chance, we can show that 5 x K + < for some number
K then, we'll have the following,
5 4 4 x x K x + − < −
If we now assume that what we really want to show is 4 K x ε − < instead of 5 4 x x ε + − <
we get the following,
4 x
K
ε
− <
This is starting to seem familiar isn't it?
All this work however, is based on the assumption that we can show that 5 x K + < for some K.
Without this assumption we can't do anything so let's see if we can do this.
Why choose 1 here? There is no reason other than it's a nice number to work with. We could
just have easily chosen 2, or 5, or
1
3
. The only difference our choice will make is on the actual
value of K that we end up with. You might want to go through this process with another choice
of K and see if you can do it.
It may not seem like it, but we're now ready to chose a δ . In the previous examples we had only
a single assumption and we used that to give us δ . In this case we've got two and they BOTH
need to be true. So, we'll let δ be the smaller of the two assumptions, 1 and
10
ε
.
Mathematically, this is written as,
min 1,
10
δ
3 ⎧ ⎫
=
⎨ ⎬
⎩ ⎭
Okay, that was a lot more work that the first two examples and unfortunately, it wasn't all that
difficult of a problem. Well, maybe we should say that in comparison to some of the other limits
we could have tried to prove it wasn't all that difficult. When first faced with these kinds of
proofs using the precise definition of a limit they can all seem pretty difficult.
Do not feel bad if you don't get this stuff right away. It's very common to not understand this
right away and to have to struggle a little to fully start to understand how these kinds of limit
definition proofs work.
Next, let's give the precise definitions for the right- and left-handed limits.
Note that with both of these definitions there are two ways to deal with the restriction on x and
the one in parenthesis is probably the easier to use, although the main one given more closely
matches the definition of the normal limit above.
Let's work a quick example of one of these, although as you'll see they work in much the same
manner as the normal limit problems do.
Example 4 Use the definition of the limit to prove the following limit.
0
lim 0
x
x
+
→
=
Solution
Let 0 ε > be any number then we need to find a number 0 δ > so that the following will be true.
0 whenever 0 0 x x ε δ − < < − <
As with the previous problems let's start with the left hand inequality and see if we can't use that
to get a guess for δ . The only simplification that we really need to do here is to square both
sides.
and so by the definition of the right-hand limit we have,
0
lim 0
x
x
+
→
=
Let's now move onto the definition of infinite limits. Here are the two definitions that we need to
cover both possibilities, limits that are positive infinity and limits that are negative infinity.
Definition 4 Let f(x) be a function defined on an interval that contains x a = , except possibly at
x a = . Then we say that,
( ) lim
x a
f x
→
= ∞
if for every number 0 M > there is some number 0 δ > such that
( ) whenever 0 f x M x a δ > < − <
Definition 5 Let f(x) be a function defined on an interval that contains x a = , except possibly at
x a = . Then we say that,
( ) lim
x a
f x
→
= −∞
if for every number 0 N < there is some number 0 δ > such that
( ) whenever 0 f x N x a δ < < − <
In these two definitions note that M must be a positive number and that N must be a negative
number. That's an easy distinction to miss if you aren't paying close attention.
Also note that we could also write down definitions for one-sided limits that are infinity if we
wanted to. We'll leave that to you to do if you'd like to.
What Definition 4 is telling us is that no matter how large we choose M to be we can always find
an interval around x a = , given by 0 x a δ < − < for some number δ , so that as long as we
stay within that interval the graph of the function will be above the line y M = as shown in the
graph. Also note that we don't need the function to actually exist at x a = in order for the
definition to hold. This is also illustrated in the sketch above.
Note as well that the larger M is the smaller we're probably going to need to make δ .
To see an illustration of Definition 5 reflect the above graph about the x-axis and you'll see a
sketch of Definition 5.
Let's work a quick example of one of these to see how these differ from the previous examples.
Example 5 Use the definition of the limit to prove the following limit.
2
0
1
lim
x
x
→
= ∞
Solution
These work in pretty much the same manner as the previous set of examples do. The main
difference is that we're working with an M now instead of an ε . So, let's get going.
Let 0 M > be any number and we'll need to choose a 0 δ > so that,
2
1
whenever 0 0 M x x
x
δ > < − = <
As with the all the previous problems we'll start with the left inequality and try to get something
in the end that looks like the right inequality. To do this we'll basically solve the left inequality
for x and we'll need to recall that
2
x x = . So, here's that work.
In the previous examples we tried to show that our assumptions satisfied the left inequality by
working with it directly. However, in this, the function and our assumption on x that we've got
actually will make this easier to start with the assumption on x and show that we can get the left
inequality out of that. Note that this is being done this way mostly because of the function that
we're working with and not because of the type of limit that we've got.
For our next set of limit definitions let's take a look at the two definitions for limits at infinity.
Again, we need one for a limit at plus infinity and another for negative infinity.
Definition 6 Let f(x) be a function defined on an interval that contains x a = , except possibly at
x a = . Then we say that,
( ) lim
x
f x L
→∞
=
if for every number 0 ε > there is some number 0 M > such that
( ) whenever f x L x M ε − < >
Definition 7 Let f(x) be a function defined on an interval that contains x a = , except possibly at
x a = . Then we say that,
( ) lim
x
f x L
→−∞
=
if for every number 0 ε > there is some number 0 N < such that
( ) whenever f x L x N ε − < <
To see what these definitions are telling us here is a quick sketch illustrating Definition 6.
Definition 6 tells us is that no matter how close to L we want to get, mathematically this is given
by ( ) f x L ε − < for any chosen ε , we can find another number M such that provided we take
any x bigger than M, then the graph of the function for that x will be closer to L than either L ε −
and L ε + . Or, in other words, the graph will be in the shaded region as shown in the sketch
below.
Since we're heading out towards negative infinity it looks like we can choose
1
N
ε
= − . Note
that we need the "-" to make sure that N is negative (recall that 0 ε > ).
Let's verify that our guess will work. Let 0 ε > and choose
1
N
ε
= − and assume that
1
x
ε
< − .
As with the previous example the function that we're working with here suggests that it will be
easier to start with this assumption and show that we can get the left inequality out of that.
For our final limit definition let's look at limits at infinity that are also infinite in value. There are
four possible limits to define here. We'll do one of them and leave the other three to you to write
down if you'd like to.
Definition 8 Let f(x) be a function defined on an interval that contains x a = , except possibly at
x a = . Then we say that,
( ) lim
x
f x
→∞
= ∞
if for every number 0 N > there is some number 0 M > such that
( ) whenever f x N x M > >
The other three definitions are almost identical. The only differences are the signs of M and/or N
and the corresponding inequality directions.
As a final definition in this section let's recall that we previously said that a function was
continuous if,
( ) ( ) lim
x a
f x f a
→
=
Definition 9 Let f(x) be a function defined on an interval that contains x a = . Then we say that
f(x) is continuous at x a = if for every number 0 ε > there is some number 0 δ > such that
( ) ( ) whenever 0 f x f a x a ε δ − < < − <
This definition is very similar to the first definition in this section and of course that should make
some sense since that is exactly the kind of limit that we're doing to show that a function is
continuous. The only real difference is that here we need to make sure that the function is
actually defined at x a = , while we didn't need to worry about that for the first definition since
limits don't really care what is happening at the point.
We won't do any examples of proving a function is continuous at a point here mostly because
we've already done some examples. Go back and look at the first three examples. In each of
these examples the value of the limit was the value of the function evaluated at x a = and so in
each of these examples not only did we prove the value of the limit we also managed to prove
that each of these functions are continuous at the point in question.
Introduction
In this chapter we will start looking at the next major topic in a calculus class. We will be
looking at derivatives in this chapter (as well as the next chapter). This chapter is devoted almost
exclusively to finding derivatives. We will be looking at one application of them in this chapter.
We will be leaving most of the applications of derivatives to the next chapter.
Here is a listing of the topics covered in this chapter.
The Definition of the Derivative – In this section we will be looking at the definition of the
derivative.
Interpretation of the Derivative – Here we will take a quick look at some interpretations of the
derivative.
Differentiation Formulas – Here we will start introducing some of the differentiation formulas
used in a calculus course.
Product and Quotient Rule – In this section we will took at differentiating products and
quotients of functions.
Derivatives of Trig Functions – We'll give the derivatives of the trig functions in this section.
Derivatives of Exponential and Logarithm Functions – In this section we will get the
derivatives of the exponential and logarithm functions.
Derivatives of Inverse Trig Functions – Here we will look at the derivatives of inverse trig
functions.
Derivatives of Hyperbolic Functions – Here we will look at the derivatives of hyperbolic
functions.
Chain Rule – The Chain Rule is one of the more important differentiation rules and will allow us
to differentiate a wider variety of functions. In this section we will take a look at it.
Related Rates – In this section we will look at the lone application to derivatives in this chapter.
This topic is here rather than the next chapter because it will help to cement in our minds one of
the more important concepts about derivatives and because it requires implicit differentiation.
Higher Order Derivatives – Here we will introduce the idea of higher order derivatives.
Logarithmic Differentiation – The topic of logarithmic differentiation is not always presented in
a standard calculus course. It is presented here for those how are interested in seeing how it is
done and the types of functions on which it can be used.
The Definition of the Derivative
In the first section of the last chapter we saw that the computation of the slope of a tangent line,
the instantaneous rate of change of a function, and the instantaneous velocity of an object at
x a = all required us to compute the following limit.
( ) ( )
lim
x a
f x f a
x a
→
−
−
We also saw that with a small change of notation this limit could also be written as,
( ) ( )
0
lim
h
f a h f a
h
→
+ −
(3)
This is such an important limit and it arises in so many places that we give it a name. We call it a
derivative. Here is the official definition of the derivative.
Note that we replaced all the a's in (3) with x's to acknowledge the fact that the derivative is
really a function as well. We often "read" ( ) f x ′ as "f prime of x".
Let's compute a couple of derivatives using the definition.
Example 1 Find the derivative of the following function using the definition of the derivative.
( )
2
2 16 35 f x x x = − +
Solution
So, all we really need to do is to plug this function into the definition of the derivative, (3), and do
some algebra. While, admittedly, the algebra will get somewhat unpleasant at times, but it's just
algebra so don't get excited about the fact that we're now computing derivatives.
Notice that every term in the numerator that didn't have an h in it canceled out and we can now
factor an h out of the numerator which will cancel against the h in the denominator. After that we
can compute the limit.
Example 2 Find the derivative of the following function using the definition of the derivative.
( )
1
t
g t
t
=
+
Solution
This one is going to be a little messier as far as the algebra goes. However, outside of that it will
work in exactly the same manner as the previous examples. First, we plug the function into the
definition of the derivative,
Before finishing this let's note a couple of things. First, we didn't multiply out the denominator.
Multiplying out the denominator will just overly complicate things so let's keep it simple. Next,
as with the first example, after the simplification we only have terms with h's in them left in the
numerator and so we can now cancel an h out.
So, upon canceling the h we can evaluate the limit and get the derivative.
doesn't exist. However, this is the limit that gives us the derivative that we're after.
If the limit doesn't exist then the derivative doesn't exist either.
In this example we have finally seen a function for which the derivative doesn't exist at a point.
This is a fact of life that we've got to be aware of. Derivatives will not always exist. Note as
well that this doesn't say anything about whether or not the derivative exists anywhere else. In
fact, the derivative of the absolute value function exists at every point except the one we just
looked at, 0 x = .
The preceding discussion leads to the following definition.
Definition
A function ( ) f x is called differentiable at x a = if ( ) f x ′ exists and ( ) f x is called
differentiable on an interval if the derivative exists for each point in that interval.
The next theorem shows us a very nice relationship between functions that are continuous and
those that are differentiable.
Theorem
If ( ) f x is differentiable at x a = then ( ) f x is continuous at x a = .
Alternate Notation
Next we need to discuss some alternate notation for the derivative. The typical derivative
notation is the "prime" notation. However, there is another notation that is used on occasion so
let's cover that.
Because we also need to evaluate derivatives on occasion we also need a notation for evaluating
derivatives when using the fractional notation. So if we want to evaluate the derivative at x=a all
of the following are equivalent.
( )
x a
x a x a
df dy
f a y
dx dx
=
= =
′ ′ = = =
Note as well that on occasion we will drop the (x) part on the function to simplify the notation
somewhat. In these cases the following are equivalent.
( ) f x f ′ ′ =
As a final note in this section we'll acknowledge that computing most derivatives directly from
the definition is a fairly complex (and sometimes painful) process filled with opportunities to
make mistakes. In a couple of section we'll start developing formulas and/or properties that will
help us to take the derivative of many of the common functions so we won't need to resort to the
definition of the derivative too often.
This does not mean however that it isn't important to know the definition of the derivative! It is
an important definition that we should always know and keep in the back of our minds. It is just
something that we're not going to be working with all that much.
Interpretations of the Derivative
Before moving on to the section where we learn how to compute derivatives by avoiding the
limits we were evaluating in the previous section we need to take a quick look at some of the
interpretations of the derivative. All of these interpretations arise from recalling how our
definition of the derivative came about. The definition came about by noticing that all the
problems that we worked in the first section in the chapter on limits required us to evaluate the
same limit.
Rate of Change
The first interpretation of a derivative is rate of change. This was not the first problem that we
looked at in the limit chapter, but it is the most important interpretation of the derivative. If
( ) f x represents a quantity at any x then the derivative ( ) f a ′ represents the instantaneous rate
of change of ( ) f x at x a = .
Example 1 Suppose that the amount of water in a holding tank at t minutes is given by
( )
2
2 16 35 V t t t = − + . Determine each of the following.
(a) Is the volume of water in the tank increasing or decreasing at 1 t = minute?
[Solution]
(b) Is the volume of water in the tank increasing or decreasing at 5 t = minutes?
[Solution]
(c) Is the volume of water in the tank changing faster at 1 t = or 5 t = minutes?
[Solution]
(d) Is the volume of water in the tank ever not changing? If so, when? [Solution]
Solution
In the solution to this example we will use both notations for the derivative just to get you
familiar with the different notations.
We are going to need the rate of change of the volume to answer these questions. This means that
we will need the derivative of this function since that will give us a formula for the rate of change
at any time t. Now, notice that the function giving the volume of water in the tank is the same
function that we saw in Example 1 in the last section except the letters have changed. The change
in letters between the function in this example versus the function in the example from the last
section won't affect the work and so we can just use the answer from that example with an
appropriate change in letters.
Recall from our work in the first limits section that we determined that if the rate of change was
positive then the quantity was increasing and if the rate of change was negative then the quantity
was decreasing.
We can now work the problem.
(a) Is the volume of water in the tank increasing or decreasing at 1 t = minute?
In this case all that we need is the rate of change of the volume at 1 t = or,
( )
1
1 12 OR 12
t
dV
V
dt
=
′ = − = −
So, at 1 t = the rate of change is negative and so the volume must be decreasing at this time.
[Return to Problems]
(b) Is the volume of water in the tank increasing or decreasing at 5 t = minutes?
Again, we will need the rate of change at 5 t = .
( )
5
5 4 OR 4
t
dV
V
dt
=
′ = =
In this case the rate of change is positive and so the volume must be increasing at 5 t = .
[Return to Problems]
(c) Is the volume of water in the tank changing faster at 1 t = or 5 t = minutes?
To answer this question all that we look at is the size of the rate of change and we don't worry
about the sign of the rate of change. All that we need to know here is that the larger the number
the faster the rate of change. So, in this case the volume is changing faster at 1 t = than at 5 t = .
[Return to Problems]
(d) Is the volume of water in the tank ever not changing? If so, when?
The volume will not be changing if it has a rate of change of zero. In order to have a rate of
change of zero this means that the derivative must be zero. So, to answer this question we will
then need to solve
( ) 0 OR 0
dV
V t
dt
′ = =
This is easy enough to do.
4 16 0 4 t t − = ⇒ =
So at 4 t = the volume isn't changing. Note that all this is saying is that for a brief instant the
volume isn't changing. It doesn't say that at this point the volume will quit changing
permanently.
If we go back to our answers from parts (a) and (b) we can get an idea about what is going on. At
1 t = the volume is decreasing and at 5 t = the volume is increasing. So at some point in time
the volume needs to switch from decreasing to increasing. That time is 4 t = .
This is the time in which the volume goes from decreasing to increasing and so for the briefest
instant in time the volume will quit changing as it changes from decreasing to increasing.
[Return to Problems]
Note that one of the more common mistakes that students make in these kinds of problems is to
try and determine increasing/decreasing from the function values rather than the derivatives. In
this case if we took the function values at 0 t = , 1 t = and 5 t = we would get,
( ) ( ) ( ) 0 35 1 21 5 5 V V V = = =
Clearly as we go from 0 t = to 1 t = the volume has decreased. This might lead us to decide that
AT 1 t = the volume is decreasing. However, we just can't say that. All we can say is that
between 0 t = and 1 t = the volume has decreased at some point in time. The only way to know
what is happening right at 1 t = is to compute ( ) 1 V′ and look at its sign to determine
increasing/decreasing. In this case ( ) 1 V′ is negative and so the volume really is decreasing at
1 t = .
Now, if we'd plugged into the function rather than the derivative we would have been gotten the
correct answer for 1 t = even though our reasoning would have been wrong. It's important to not
let this give you the idea that this will always be the case. It just happened to work out in the case
of 1 t = .
To see that this won't always work let's now look at 5 t = . If we plug 1 t = and 5 t = into the
volume we can see that again as we go from 1 t = to 5 t = the volume has decreases. Again,
however all this says is that the volume HAS decreased somewhere between 1 t = and 5 t = . It
does NOT say that the volume is decreasing at 5 t = . The only way to know what is going on
right at 5 t = is to compute ( ) 5 V′ and in this case ( ) 5 V′ is positive and so the volume is
actually increasing at 5 t = .
So, be careful. When asked to determine if a function is increasing or decreasing at a point make
sure and look at the derivative. It is the only sure way to get the correct answer. We are not
looking to determine is the function has increased/decreased by the time we reach a particular
point. We are looking to determine if the function is increasing/decreasing at that point in
question.
Slope of Tangent Line
This is the next major interpretation of the derivative. The slope of the tangent line to ( ) f x at
x a = is ( ) f a ′ . The tangent line then is given by,
( ) ( )( ) y f a f a x a ′ = + −
Example 2 Find the tangent line to the following function at 3 z = .
( ) 5 8 R z z = −
Solution
We first need the derivative of the function and we found that in Example 3 in the last section.
The derivative is,
( )
5
2 5 8
R z
z
′ =
−
Velocity
Recall that this can be thought of as a special case of the rate of change interpretation. If the
position of an object is given by ( ) f t after t units of time the velocity of the object at t a = is
given by ( ) f a ′ .
Example 3 Suppose that the position of an object after t hours is given by,
( )
1
t
g t
t
=
+
Answer both of the following about this object.
(a) Is the object moving to the right or the left at 10 t = hours? [Solution]
(b) Does the object ever stop moving? [Solution]
Solution
Once again we need the derivative and we found that in Example 2 in the last section. The
derivative is,
( )
( )
2
1
1
g t
t
′ =
+
So the velocity at 10 t = is positive and so the object is moving to the right at 10 t = .
[Return to Problems]
(b) Does the object ever stop moving?
The object will stop moving if the velocity is ever zero. However, note that the only way a
rational expression will ever be zero is if the numerator is zero. Since the numerator of the
derivative (and hence the speed) is a constant it can't be zero.
Therefore, the velocity will never stop moving.
In fact, we can say a little more here. The object will always be moving to the right since the
velocity is always positive.
[Return to Problems]
We've seen three major interpretations of the derivative here. You will need to remember these,
especially the rate of change, as they will show up continually throughout this course.
Differentiation Formulas
In the first section of this chapter we saw the definition of the derivative and we computed a
couple of derivatives using the definition. As we saw in those examples there was a fair amount
of work involved in computing the limits and the functions that we worked with were not terribly
complicated.
For more complex functions using the definition of the derivative would be an almost impossible
task. Luckily for us we won't have to use the definition terribly often. We will have to use it on
occasion, however we have a large collection of formulas and properties that we can use to
simplify our life considerably and will allow us to avoid using the definition whenever possible.
We will introduce most of these formulas over the course of the next several sections. We will
start in this section with some of the basic properties and formulas. We will give the properties
and formulas in this section in both "prime" notation and "fraction" notation.
Properties
1) ( ) ( ) ( ) ( ) ( ) f x g x f x g x
′
′ ′ ± = ± OR ( ) ( ) ( )
d df dg
f x g x
dx dx dx
± = ±
In other words, to differentiate a sum or difference all we need to do is differentiate the
individual terms and then put them back together with the appropriate signs. Note as well
that this property is not limited to two functions.
See the Proof of Various Derivative Formulas section of the Extras chapter to see the
proof of this property. It's a very simple proof using the definition of the derivative.
2) ( ) ( ) ( ) cf x cf x
′
′ = OR ( ) ( )
d df
cf x c
dx dx
= , c is any number
In other words, we can "factor" a multiplicative constant out of a derivative if we need to.
See the Proof of Various Derivative Formulas section of the Extras chapter to see the
proof of this property.
Note that we have not included formulas for the derivative of products or quotients of two
functions here. The derivative of a product or quotient of two functions is not the product or
quotient of the derivatives of the individual pieces. We will take a look at these in the next
section.
Next, let's take a quick look at a couple of basic "computation" formulas that will allow us to
actually compute some derivatives.
Formulas
1) If ( ) f x c = then ( ) 0 f x ′ = OR ( ) 0
d
c
dx
=
The derivative of a constant is zero. See the Proof of Various Derivative Formulas
section of the Extras chapter to see the proof of this formula.
2) If ( )
n
f x x = then ( )
1 n
f x nx
−
′ = OR
( )
1 n n
d
x nx
dx
−
= , n is any number.
This formula is sometimes called the power rule. All we are doing here is bringing the
original exponent down in front and multiplying and then subtracting one from the
original exponent.
Note as well that in order to use this formula n must be a number, it can't be a variable.
Also note that the base, the x, must be a variable, it can't be a number. It will be tempting
in some later sections to misuse the Power Rule when we run in some functions where
the exponent isn't a number and/or the base isn't a variable.
See the Proof of Various Derivative Formulas section of the Extras chapter to see the
proof of this formula. There are actually three different proofs in this section. The first
two restrict the formula to n being an integer because at this point that is all that we can
do at this point. The third proof is for the general rule, but does suppose that you've read
most of this chapter.
These are the only properties and formulas that we'll give in this section. Let's compute some
derivatives using these properties.
Notice that in the third term the exponent was a one and so upon subtracting 1 from the original
exponent we get a new exponent of zero. Now recall that
0
1 x = . Don't forget to do any basic
arithmetic that needs to be done such as any multiplication and/or division in the coefficients.
[Return to Problems]
(b) ( )
6 6
2 7 g t t t
−
= +
The point of this problem is to make sure that you deal with negative exponents correctly. Here
is the derivative.
( ) ( ) ( )
5 7
5 7
2 6 7 6
12 42
g t t t
t t
−
−
′ = + −
= −
Make sure that you correctly deal with the exponents in these cases, especially the negative
exponents. It is an easy mistake to "go the other way" when subtracting one off from a negative
exponent and get
5
6t
−
− instead of the correct
7
6t
−
− .
[Return to Problems]
(c)
3
5
1
8 23
3
y z z
z
= − + −
Now in this function the second term is not correctly set up for us to use the power rule. The
power rule requires that the term be a variable to a power only and the term must be in the
numerator. So, prior to differentiating we first need to rewrite the second term into a form that
we can deal with.
3 5
1
8 23
3
y z z z
−
= − + −
Note that we left the 3 in the denominator and only moved the variable up to the numerator.
Remember that the only thing that gets an exponent is the term that is immediately to the left of
the exponent. If we'd wanted the three to come up as well we'd have written,
( )
5
1
3z
so be careful with this! It's a very common mistake to bring the 3 up into the numerator as well
at this stage.
Now that we've gotten the function rewritten into a proper form that allows us to use the Power
Rule we can differentiate the function. Here is the derivative for this part.
(d) ( )
3 7
5 2
2
9 T x x x
x
= + −
All of the terms in this function have roots in them. In order to use the power rule we need to
first convert all the roots to fractional exponents. Again, remember that the Power Rule requires
us to have a variable to a number and that it must be in the numerator of the term. Here is the
function written in "proper" form.
In the last two terms we combined the exponents. You should always do this with this kind of
term. In a later section we will learn of a technique that would allow us to differentiate this term
without combining exponents, however it will take significantly more work to do. Also don't
forget to move the term in the denominator of the third term up to the numerator. We can now
differentiate the function.
Make sure that you can deal with fractional exponents. You will see a lot of them in this class.
[Return to Problems]
(e) ( )
2
h x x x
π
= −
In all of the previous examples the exponents have been nice integers or fractions. That is usually
what we'll see in this class. However, the exponent only needs to be a number so don't get
excited about problems like this one. They work exactly the same.
( )
1 2 1
2 h x x x
π
π
− −
′ = −
The answer is a little messy and we won't reduce the exponents down to decimals. However, this
problem is not terribly difficult it just looks that way initially.
[Return to Problems]
Back when we first put down the properties we noted that we hadn't included a property for
products and quotients. That doesn't mean that we can't differentiate any product or quotient at
this point. There are some that we can do.
Example 2 Differentiate each of the following functions.
(a)
( )
3 2 2
2 y x x x = − [Solution]
(b) ( )
5 2
2
2 5 t t
h t
t
+ −
= [Solution]
Solution
(a)
( )
3 2 2
2 y x x x = −
In this function we can't just differentiate the first term, differentiate the second term and then
multiply the two back together. That just won't work. We will discuss this in detail in the next
section so if you're not sure you believe that hold on for a bit and we'll be looking at that soon as
well as showing you an example of why it won't work.
It is still possible to do this derivative however. All that we need to do is convert the radical to
fractional exponents (as we should anyway) and then multiply this through the parenthesis.
( )
2 5 8
2
3 3 3
2 2 y x x x x x = − = −
Now we can differentiate the function.
2 5
3 3
10 8
3 3
y x x ′ = −
[Return to Problems]
(b) ( )
5 2
2
2 5 t t
h t
t
+ −
=
As with the first part we can't just differentiate the numerator and the denominator and the put it
back together as a fraction. Again, if you're not sure you believe this hold on until the next
section and we'll take a more detailed look at this.
So, as we saw in this example there are a few products and quotients that we can differentiate. If
we can first do some simplification the functions will sometimes simplify into a form that can be
differentiated using the properties and formulas in this section.
Before moving on to the next section let's work a couple of examples to remind us once again of
some of the interpretations of the derivative.
Note that we rewrote the last term in the derivative back as a fraction. This is not something
we've done to this point and is only being done here to help with the evaluation in the next step.
It's often easier to do the evaluation with positive exponents.
Example 5 The position of an object at any time t (in hours) is given by,
( )
3 2
2 21 60 10 s t t t t = − + −
Determine when the object is moving to the right and when the object is moving to the left.
Solution
The only way that we'll know for sure which direction the object is moving is to have the velocity
in hand. Recall that if the velocity is positive the object is moving off to the right and if the
velocity is negative then the object is moving to the left.
Now, we need to determine where the derivative is positive and where the derivative is negative.
There are several ways to do this. The method that I tend to prefer is the following.
Since polynomials are continuous we know from the Intermediate Value Theorem that if the
polynomial ever changes sign then it must have first gone through zero. So, if we knew where
the derivative was zero we would know the only points where the derivative might change sign.
We can see from the factored form of the derivative that the derivative will be zero at 2 t = and
5 t = . Let's graph these points on a number line.
Now, we can see that these two points divide the number line into three distinct regions. In reach
of these regions we know that the derivative will be the same sign. Recall the derivative can only
change sign at the two points that are used to divide the number line up into the regions.
Here are the intervals in which the derivative is positive and negative.
positive : 2 & 5
negative : 2 5
t t
t
−∞ < < < < ∞
< <
We included negative t's here because we could even though they may not make much sense for
this problem. Once we know this we also can answer the question. The object is moving to the
right and left in the following intervals.
Make sure that you can do the kind of work that we just did in this example. You will be asked
numerous times over the course of the next two chapters to determine where functions are
positive and/or negative. If you need some review or want to practice these kinds of problems
you should check out the Solving Inequalities section of my Algebra/Trig Review.
Product and Quotient Rule
In the previous section we noted that we had to be careful when differentiating products or
quotients. It's now time to look at products and quotients and see why.
First let's take a look at why we have to be careful with products and quotients. Suppose that we
have the two functions ( )
3
f x x = and ( )
6
g x x = . Let's start by computing the derivative of
the product of these two functions. This is easy enough to do directly.
( ) ( ) ( )
3 6 9 8
9 f g x x x x
′ ′
′
= = =
Remember that on occasion we will drop the (x) part on the functions to simplify notation
somewhat. We've done that in the work above.
Solution
At this point there really aren't a lot of reasons to use the product rule. As we noted in the
previous section all we would need to do for either of these is to just multiply out the product and
then differentiate.
With that said we will use the product rule on these so we can see an example or two. As we add
more functions to our repertoire and as the functions become more complicated the product rule
will become more useful and in many cases required.
(a)
( )
3 2 2
2 y x x x = −
Note that we took the derivative of this function in the previous section and didn't use the product
rule at that point. We should however get the same result here as we did then.
Now let's do the problem here. There's not really a lot to do here other than use the product rule.
However, before doing that we should convert the radical to a fractional exponent as always.
( )
2
2
3
2 y x x x = −
Now let's take the derivative. So we take the derivative of the first function times the second
then add on to that the first function times the derivative of the second function.
Since it was easy to do we went ahead and simplified the results a little.
[Return to Problems]
Let's now work an example or two with the quotient rule. In this case, unlike the product rule
examples, a couple of these functions will require the quotient rule in order to get the derivative.
The last two however, we can avoid the quotient rule if we'd like to as we'll see.
(c) ( )
6
4
f x
x
=
It seems strange to have this one here rather than being the first part of this example given that it
definitely appears to be easier than any of the previous two. In fact, it is easier. There is a point
to doing it here rather than first. In this case there are two ways to do compute this derivative.
There is an easy way and a hard way and in this case the hard way is the quotient rule. That's the
point of this example.
Now, that was the "hard" way. So, what was so hard about it? Well actually it wasn't that hard,
there is just an easier way to do it that's all. However, having said that, a common mistake here
is to do the derivative of the numerator (a constant) incorrectly. For some reason many people
will give the derivative of the numerator in these kinds of problems as a 1 instead of 0! Also,
there is some simplification that needs to be done in these kinds of problems if you do the
quotient rule.
(d)
6
5
w
y =
This problem also seems a little out of place. However, it is here again to make a point. Do not
confuse this with a quotient rule problem. While you can do the quotient rule on this function
there is no reason to use the quotient rule on this. Simply rewrite the function as
Example 3 Suppose that the amount of air in a balloon at any time t is given by
( )
3
6
4 1
t
V t
t
=
+
Determine if the balloon is being filled with air or being drained of air at 8 t = .
Solution
If the balloon is being filled with air then the volume is increasing and if it's being drained of air
then the volume will be decreasing. In other words, we need to get the derivative so that we can
determine the rate of change of the volume at 8 t = .
With this section and the previous section we are now able to differentiate powers of x as well as
sums, differences, products and quotients of these kinds of functions. However, there are many
more functions out there in the world that are not in this form. The next few sections give many
of these functions as well as give their derivatives.
Derivatives of Trig Functions
With this section we're going to start looking at the derivatives of functions other than
polynomials or roots of polynomials. We'll start this process off by taking a look at the
derivatives of the six trig functions. Two of the derivatives will be derived. The remaining four
are left to the reader and will follow similar proofs for the two given here.
Before we actually get into the derivatives of the trig functions we need to give a couple of limits
that will show up in the derivation of two of the derivatives.
Fact
0 0
sin cos 1
lim 1 lim 0
θ θ
θ θ
θ θ
→ →
−
= =
See the Proof of Trig Limits section of the Extras chapter to see the proof of these two limits.
Before we start differentiating trig functions let's work a quick set of limit problems that this fact
now allows us to do.
There really isn't a whole lot to this limit. In fact, it's only here to contrast with the next example
so you can see the difference in how these work. In this case since there is only a 6 in the
denominator we'll just factor this out and then use the fact.
( )
0 0
sin 1 sin 1 1
lim lim 1
6 6 6 6
θ θ
θ θ
θ θ
→ →
= = =
[Return to Problems]
Now, in this case we can't factor the 6 out of the sine so we're stuck with it there and we'll need
to figure out a way to deal with it. To do this problem we need to notice that in the fact the
argument of the sine is the same as the denominator (i.e. both θ 's). So we need to get both of the
argument of the sine and the denominator to be the same. We can do this by multiplying the
numerator and the denominator by 6 as follows.
Note that we factored the 6 in the numerator out of the limit. At this point, while it may not look
like it, we can use the fact above to finish the limit.
To see that we can use the fact on this limit let's do a change of variables. A change of variables
is really just a renaming of portions of the problem to make something look more like something
we know how to deal with. They can't always be done, but sometimes, such as this case, they
can simplify the problem. The change of variables here is to let 6x θ = and then notice that as
0 x → we also have 0 θ → . When doing a change of variables in a limit we need to change all
the x's into θ 's and that includes the one in the limit.
And there we are. Note that we didn't really need to do a change of variables here. All we really
need to notice is that the argument of the sine is the same as the denominator and then we can use
the fact. A change of variables, in this case, is really only needed to make it clear that the fact
does work.
[Return to Problems]
With a little rewriting we can see that we do in fact end up needing to do a limit like the one we
did in the previous part. So, let's do the limit here and this time we won't bother with a change of
variable to help us out. All we need to do is multiply the numerator and denominator of the
fraction in the denominator by 7 to get things set up to use the fact. Here is the work for this
limit.
At this point we can see that this really is two limits that we've seen before. Here is the work for
each of these and notice on the second limit that we're going to work it a little differently than we
did in the previous part. This time we're going to notice that it doesn't really matter whether the
sine is in the numerator or the denominator as long as the argument of the sine is the same as
what's in the numerator the limit is still one.
This limit almost looks the same as that in the fact in the sense that the argument of the sine is the
same as what is in the denominator. However, notice that, in the limit, x is going to 4 and not 0 as
the fact requires. However, with a change of variables we can see that this limit is in fact set to
use the fact above regardless.
So, let x θ = − 4 and then notice that as 4 x → we have 0 θ → . Therefore, after doing the
change of variable the limit becomes,
The previous parts of this example all used the sine portion of the fact. However, we could just
have easily used the cosine portion so here is a quick example using the cosine portion to
illustrate this. We'll not put in much explanation here as this really does work in the same
manner as the sine portion.
All that is required to use the fact is that the argument of the cosine is the same as the
denominator.
[Return to Problems]
Okay, now that we've gotten this set of limit examples out of the way let's get back to the main
point of this section, differentiating trig functions.
We'll start with finding the derivative of the sine function. To do this we will need to use the
definition of the derivative. It's been a while since we've had to use this, but sometimes there
just isn't anything we can do about it. Here is the definition of the derivative for the sine
function.
( ) ( )
( ) ( )
0
sin sin
sin lim
h
x h x
d
x
dx h
→
+ −
=
Since we can't just plug in 0 h = to evaluate the limit we will need to use the following trig
formula on the first sine in the numerator.
( ) ( ) ( ) ( ) ( ) sin sin cos cos sin x h x h x h + = +
As you can see upon using the trig formula we can combine the first and third term and then
factor a sine out of that. We can then break up the fraction into two pieces, both of which can be
dealt with separately.
Now, both of the limits here are limits as h approaches zero. In the first limit we have a sin(x)
and in the second limit we have a cos(x). Both of these are only functions of x only and as h
moves in towards zero this has no affect on the value of x. Therefore, as far as the limits are
concerned, these two functions are constants and can be factored out of their respective limits.
Doing this gives,
( ) ( ) ( )
( )
( )
( )
0 0
cos 1 sin
sin sin lim cos lim
h h
h h
d
x x x
dx h h
→ →
−
= +
At this point all we need to do is use the limits in the fact above to finish out this problem.
( ) ( ) ( )( ) ( )( ) ( ) sin sin 0 cos 1 cos
d
x x x x
dx
= + =
Differentiating cosine is done in a similar fashion. It will require a different trig formula, but
other than that is an almost identical proof. The details will be left to you. When done with the
proof you should get,
( ) ( ) ( ) cos sin
d
x x
dx
= −
With these two out of the way the remaining four are fairly simple to get. All the remaining four
trig functions can be defined in terms of sine and cosine and these definitions, along with
appropriate derivative rules, can be used to get their derivatives.
The remaining three trig functions are also quotients involving sine and/or cosine and so can be
differentiated in a similar manner. We'll leave the details to you. Here are the derivatives of all
six of the trig functions.
we first looked at the product rule. When we first looked at the product rule the only functions
we knew how to differentiate were polynomials and in those cases all we really needed to do was
multiply them out and we could take the derivative without the product rule. We are now getting
into the point where we will be forced to do the product rule at times regardless of whether or not
we want to.
We will also need to be careful with the minus sign in front of the second term and make sure that
it gets dealt with properly. There are two ways to deal with this. One way it to make sure that
you use a set of parenthesis as follows,
Because the second term is being subtracted off of the first term then the whole derivative of the
second term must also be subtracted off of the derivative of the first term. The parenthesis make
this idea clear.
A potentially easier way to do this is to think of the minus sign as part of the first function in the
product. Or, in other words the two functions in the product, using this idea, are
2
w − and
( ) tan w . Doing this gives,
( ) ( ) ( )
5 2 2
12 2 tan sec h w w w w w w
−
′ = − − −
So, regardless of how you approach this problem you will get the same derivative.
[Return to Problems]
(c) ( ) ( ) ( ) 5sin cos 4csc y x x x = +
As with the previous part we'll need to use the product rule on the first term. We will also think
of the 5 as part of the first function in the product to make sure we deal with it correctly.
Alternatively, you could make use of a set of parenthesis to make sure the 5 gets dealt with
properly. Either way will work, but we'll stick with thinking of the 5 as part of the first term in
the product. Here's the derivative of this function.
Be careful with the signs when differentiating the denominator. The negative sign we get from
differentiating the cosine will cancel against the negative sign that is already there.
This appears to be done, but there is actually a fair amount of simplification that can yet be done.
To do this we need to factor out a "-2" from the last two terms in the numerator and the make use
of the fact that ( ) ( )
2 2
cos sin 1 θ θ + = .
As a final problem here let's not forget that we still have our standard interpretations to
derivatives.
Example 3 Suppose that the amount of money in a bank account is given by
( ) ( ) ( ) 500 100cos 150sin P t t t = + −
where t is in years. During the first 10 years in which the account is open when is the amount of
money in the account increasing?
Solution
To determine when the amount of money is increasing we need to determine when the rate of
change is positive. Since we know that the rate of change is given by the derivative that is the
first thing that we need to find.
( ) ( ) ( ) 100sin 150cos P t t t ′ = − −
Now, we need to determine where in the first 10 years this will be positive. This is equivalent to
asking where in the interval [0, 10] is the derivative positive. Recall that both sine and cosine are
continuous functions and so the derivative is also a continuous function. The Intermediate Value
Theorem then tells us that the derivative can only change sign if it first goes through zero.
If you don't recall how to solve trig equations go back and take a look at the sections on solving
trig equations in the Review chapter.
We are only interested in those solutions that fall in the range [0, 10]. Plugging in values of n
into the solutions above we see that the values we need are,
2.1588 2.1588 2 8.4420
5.3004
t t
t
π = = + =
=
So, much like solving polynomial inequalities all that we need to do is sketch in a number line
and add in these points. These points will divide the number line into regions in which the
derivative must always be the same sign. All that we need to do then is choose a test point from
each region to determine the sign of the derivative in that region.
Here is the number line with all the information on it.
So, it looks like the amount of money in the bank account will be increasing during the following
intervals.
2.1588 5.3004 8.4420 10 t t < < < <
Note that we can't say anything about what is happening after 10 t = since we haven't done any
work for t's after that point.
Also, it is important that we be able to solve trig equations as this is something that will arise off
and on in this course. It is also important that we can do the kinds of number lines that we used
in the last example to determine where a function is positive and where a function is negative.
This is something that we will be doing on occasion in both this chapter and the next.
Derivatives of Exponential and Logarithm Functions
The next set of functions that we want to take a look at are exponential and logarithm functions.
The most common exponential and logarithm functions in a calculus course are the natural
exponential function,
x
e , and the natural logarithm function, ( ) ln x . We will take a more
general approach however and look at the general exponential and logarithm function.
We want to differentiate this. The power rule that we looked at a couple of sections ago won't
work as that required the exponent to be a fixed number and the base to be a variable. That is
exactly the opposite from what we've got with this function. So, we're going to have to start with
the definition of the derivative.
Now, the
x
a is not affected by the limit since it doesn't have any h's in it and so is a constant as
far as the limit is concerned. We can therefore factor this out of the limit. This gives,
( )
0
1
lim
h
x
h
a
f x a
h
→
−
′ =
So, we are kind of stuck we need to know the derivative in order to get the derivative!
There is one value of a that we can deal with at this point. Back in the Exponential Functions
section of the Review chapter we stated that 2.71828182845905 = e … What we didn't do
however do actually define where e comes from. There are in fact a variety of ways to define e.
Here are a three of them.
So, how is this fact useful to us? Well recall that the natural exponential function and the natural
logarithm function are inverses of each other and we know what the derivative of the natural
exponential function is!
The last step just uses the fact that the two functions are inverses of each other.
Putting this all together gives,
( )
1
ln 0
d
x x
dx x
= >
Note that we need to require that 0 x > since this is required for the logarithm and so must also
be required for its derivative. In can also be shown that,
( )
1
ln 0
d
x x
dx x
= ≠
Using this all we need to avoid is 0 x = .
In this case, unlike the exponential function case, we can actually find the derivative of the
general logarithm function. All that we need is the derivative of the natural logarithm, which we
just found, and the change of base formula. Using the change of base formula we can write a
general logarithm as,
Solution
(a) This will be the only example that doesn't involve the natural exponential and natural
logarithm functions.
( )
5
4 ln 4
ln9
w
R w
w
′ = −
(b) Not much to this one. Just remember to use the product rule on the second term.
There's really not a lot to differentiating natural logarithms and natural exponential functions at
this point as long as you remember the formulas. In later sections as we get more formulas under
our belt they will become more complicated.
Next, we need to do our obligatory application/interpretation problem so we don't forget about
them.
Example 2 Suppose that the position of an object is given by
( )
t
s t t = e
Does the object ever stop moving?
at that point (provided there is one) the velocity will be zero and recall that the derivative of the
position function is the velocity of the object.
The derivative is,
( ) ( ) 1
t t t
s t t t ′ = + = + e e e
So, we need to determine if the derivative is ever zero. To do this we will need to solve,
( ) 1 0
t
t + = e
Now, we know that exponential functions are never zero and so this will only be zero at 1 t = − .
So, if we are going to allow negative values of t then the object will stop moving once at 1 t = − .
If we aren't going to allow negative values of t then the object will never stop moving.
Before moving on to the next section we need to go back over a couple of derivatives to make
sure that we don't confuse the two. The two derivatives are,
It is important to note that with the Power rule the exponent MUST be a constant and the base
MUST be a variable while we need exactly the opposite for the derivative of an exponential
function. For an exponential function the exponent MUST be a variable and the base MUST be a
constant.
It is easy to get locked into one of these formulas and just use it for both of these. We also
haven't even talked about what to do if both the exponent and the base involve variables. We'll
see this situation in a later section.
Derivatives of Inverse Trig Functions
In this section we are going to look at the derivatives of the inverse trig functions. In order to
derive the derivatives of inverse trig functions we'll need the formula from the last section
relating the derivatives of inverse functions. If f(x) and g(x) are inverse functions then,
( )
( ) ( )
1
g x
f g x
′ =
′
We'll go through inverse sine, inverse cosine and inverse tangent in detail here and leave the
other three to you to derive if you'd like to.
Inverse Sine
Let's start with inverse sine. Here is the definition of the inverse sine.
1
sin sin for
2 2
y x y x y
π π
−
= ⇔ = − ≤ ≤
So, evaluating an inverse trig function is the same as asking what angle (i.e. y) did we plug into
the sine function to get x. The restrictions on y given above are there to make sure that we get a
consistent answer out of the inverse sine. We know that there are in fact an infinite number of
angles that will work and we want a consistent value when we work with inverse sine. When
using the range of angles above gives all possible values of the sine function exactly once. If
you're not sure of that sketch out a unit circle and you'll see that that range of angles (the y's) will
cover all possible values of sine.
This is not a very useful formula. Let's see if we can get a better formula. Let's start by recalling
the definition of the inverse sine function.
( ) ( )
1
sin sin y x x y
−
= ⇒ =
Using the first part of this definition the denominator in the derivative becomes,
( ) ( )
1
cos sin cos x y
−
=
Now, recall that
2 2 2
cos sin 1 cos 1 sin y y y y + = ⇒ = −
Using this, the denominator is now,
( ) ( )
1 2
cos sin cos 1 sin x y y
−
= = −
Now, use the second part of the definition of the inverse sine function. The denominator is then,
( )
1 2 2
cos sin 1 sin 1 x y x
−
= − = −
Putting all of this together gives the following derivative.
( )
1
2
1
sin
1
d
x
dx
x
−
=
−
Inverse Cosine
Now let's take a look at the inverse cosine. Here is the definition for the inverse cosine.
1
cos cos for 0 y x y x y π
−
= ⇔ = ≤ ≤
As with the inverse since we've got a restriction on the angles, y, that we get out of the inverse
cosine function. Again, if you'd like to verify this a quick sketch of a unit circle should convince
you that this range will cover all possible values of cosine exactly once. Also, we also have
1 1 x − ≤ ≤ because ( ) 1 cos 1 y − ≤ ≤ .
Simplifying the denominator here is almost identical to the work we did for the inverse sine and
so isn't shown here. Upon simplifying we get the following derivative.
( )
1
2
1
cos
1
d
x
dx
x
−
= −
−
So, the derivative of the inverse cosine is nearly identical to the derivative of the inverse sine.
The only difference is the negative sign.
Inverse Tangent
Here is the definition of the inverse tangent.
1
tan tan for
2 2
y x y x y
π π
−
= ⇔ = − < <
Again, we have a restriction on y, but notice that we can't let y be either of the two endpoints in
the restriction above since tangent isn't even defined at those two points. To convince yourself
that this range will cover all possible values of tangent do a quick sketch of the tangent function
and we can see that in this range we do indeed cover all possible values of tangent. Also, in this
case there are no restrictions on x because tangent can take on all possible values.
Solution
Here we are asking,
tan 1 y =
where y satisfies the restrictions given above. From a unit circle we can see that
4
y
π
= .
Because there is no restriction on x we can ask for the limits of the inverse tangent function as x
goes to plus or minus infinity. To do this we'll need the graph of the inverse tangent function.
This is shown below.
Finally using the second portion of the definition of the inverse tangent function gives us,
( )
2 1 2 2
sec tan 1 tan 1 x y x
−
= + = +
The derivative of the inverse tangent is then,
( )
1
2
1
tan
1
d
x
dx x
−
=
+
There are three more inverse trig functions but the three shown here the most common ones.
Formulas for the remaining three could be derived by a similar process as we did those above.
Here are the derivatives of all six inverse trig functions.
Derivatives of Hyperbolic Functions
The last set of functions that we're going to be looking in this chapter at are the hyperbolic
functions. In many physical situations combinations of
x
e and
x −
e arise fairly often. Because of
this these combinations are given names. There are six hyperbolic functions and they are defined
as follows.
Because the hyperbolic functions are defined in terms of exponential functions finding their
derivatives is fairly simple provided you've already read through the next section. We haven't
however so we'll need the following formula that can be easily proved after we've covered the
next section.
( )
x x
d
dx
− −
= − e e
With this formula we'll do the derivative for hyperbolic sine and leave the rest to you as an
exercise.
Chain Rule
We've taken a lot of derivatives over the course of the last few sections. However, if you look
back they have all been functions similar to the following kinds of functions.
( ) ( ) ( ) ( ) ( )
50
tan ln
w
R z z f t t y x h w g x x = = = = = e
These are all fairly simple functions in that wherever the variable appears it is by itself. What
about functions like the following,
None of our rules will work on these functions and yet some of these functions are closer to the
derivatives that we're liable to run into than the functions in the first set.
Let's take the first one for example. Back in the section on the definition of the derivative we
actually used the definition to compute this derivative. In that section we found that,
( )
5
2 5 8
R z
z
′ =
−
If we were to just use the power rule on this we would get,
( )
1
2
1 1
5 8
2 2 5 8
z
z
−
− =
−
which is not the derivative that we computed using the definition. It is close, but it's not the
same. So, the power rule alone simply won't work to get the derivative here.
Each of these forms have their uses, however we will work mostly with the first form in this
class. To see the proof of the Chain Rule see the Proof of Various Derivative Formulas section of
the Extras chapter.
Now, let's go back and use the Chain Rule on the function that we used when we opened this
section.
Example 1 Use the Chain Rule to differentiate ( ) 5 8 R z z = − .
Solution
We've already identified the two functions that we needed for the composition, but let's write
them back down anyway and take their derivatives.
In general we don't really do all the composition stuff in using the Chain Rule. That can get a
little complicated and in fact obscures the fact that there is a quick and easy way of remembering
the chain rule that doesn't require us to think in terms of function composition.
Let's take the function from the previous example and rewrite it slightly.
( ) ( )
In general this is how we think of the chain rule. We identify the "inside function" and the
"outside function". We then we differentiate the outside function leaving the inside function
alone and multiply all of this by the derivative of the inside function. In its general form this is,
( ) ( ) ( ) ( )
derivative of
times derivative inside function
outside function
of inside function left alone
F x f g x g x ′ ′ ′ =
.
. .
We can always identify the "outside function" in the examples below by asking ourselves how we
would evaluate the function. For instance in the R(z) case if we were to ask ourselves what R(2)
is we would first evaluate the stuff under the radical and then finally take the square root of this
result. The square root is the last operation that we perform in the evaluation and this is also the
outside function. The outside function will always be the last operation you would perform if you
were going to evaluate the function.
(c) ( )
4 2
3 9 w w
h w
− +
= e
Identifying the outside function in the previous two was fairly simple since it really was the
"outside" function in some sense. In this case we need to be a little careful. Recall that the
outside function is the last operation that we would perform in an evaluation. In this case if we
were to evaluate this function the last operation would be the exponential. Therefore the outside
function is the exponential function and the inside function is its exponent.
Remember, we leave the inside function alone when we differentiate the outside function. So, the
derivative of the exponential function (with the inside left alone) is just the original function.
[Return to Problems]
Again remember to leave the inside function along when differentiating the outside function. So,
upon differentiating the logarithm we end up not with 1/x but instead with 1/(inside function).
[Return to Problems]
In this case the derivative of the outside function is ( ) ( ) sec tan x x . However, since we leave the
inside function alone we don't get x's in both. Instead we get 1 5x − in both.
[Return to Problems]
(f) ( ) ( ) ( )
4 4
cos cos P t t t = +
There are two points to this problem. First, there are two terms and each will require a different
application of the chain rule. That will often be the case so don't expect just a single chain rule
when doing these problems. Second, we need to be very careful in choosing the outside and
inside function for each term.
Recall that the first term can actually be written as,
( ) ( ) ( )
4
4
cos cos t t =
So, in the first term the outside function is the exponent of 4 and the inside function is the cosine.
In the second term it's exactly the opposite. In the second term the outside function is the cosine
and the inside function is
4
t . Here's the derivative for this function.
The formulas in this example are really just special cases of the Chain Rule but may be useful to
remember in order to quickly do some of these derivatives.
Now, let's also not forget the other rules that we've got for doing derivatives. For the most part
we'll not be explicitly identifying the inside and outside functions for the remainder of the
problems in this section. We will be assuming that you can see our choices based on the previous
examples and the work that we have shown.
When doing the chain rule with this we remember that we've got to leave the inside function
alone. That means that where we have the
2
x in the derivative of
1
tan x
−
we will need to have
( )
2
inside function .
[Return to Problems]
After factoring we were able to cancel some of the terms in the numerator against the
denominator. So even though the initial chain rule was fairly messy the final answer is
significantly simpler because of the factoring.
[Return to Problems]
The point of this last example is to not forget the other derivative rules that we've got. Most of
the examples in this section won't involve the product rule or the quotient rule to make the
problems a little shorter. However, in practice they will often be in the same problem.
Notice that we didn't actually do the derivative of the inside function yet. This is to allow us to
notice that when we do differentiate the second term we will require the chain rule again. Notice
as well that we will only need the chain rule on the exponential and not the first term. In many
functions we will be using the chain rule more than once so don't get excited about this when it
happens.
Be careful with the second application of the chain rule. Only the exponential gets multiplied by
the "-9" since that's the derivative of the inside function for that term only. One of the more
common mistakes in these kinds of problems is to multiply the whole thing by the "-9" and not
just the second term.
[Return to Problems]
(b) ( ) ( )
3
2
2 3 4 f y y y y = + +
We'll not put as many words into this example, but we're still going to be careful with this
derivative so make sure you can follow each of the steps here.
As with the first example the second term of the inside function required the chain rule to
differentiate it. Also note that again we need to be careful when multiplying by the derivative of
the inside function when doing the chain rule on the second term.
[Return to Problems]
This problem required a total of 4 chain rules to complete.
[Return to Problems]
Sometimes these can get quite unpleasant and require many applications of the chain rule.
Initially, in these cases it's usually best to be careful as we did in this previous set of examples
and write out a couple of extra steps rather than trying to do it all in one step in your head. Once
you get better at the chain rule you'll find that you can do these fairly quickly in your head.
Finally, before we move onto the next section there is one more issue that we need to address. In
the Derivatives of Exponential and Logarithm Functions section we claimed that,
( ) ( ) ( ) ln
x x
f x a f x a a ′ = ⇒ =
but at the time we didn't have the knowledge to do this. We now do. What we needed was the
Chain Rule.
This may seem kind of silly, but it is needed to compute the derivative. Now, using this we can
write the function as,
( )
( )
( )
( )
ln
ln
ln
x
x
x
a
a x
x a
f x a
a
=
=
=
=
=
e
e
e
Okay, now that we've gotten that taken care of all we need to remember is that a is a constant and
so ln a is also a constant. Now, differentiating the final version of this function is a (hopefully)
fairly simple Chain Rule problem.
( ) ( )
ln
ln
x a
f x a ′ = e
Now, all we need to do is rewrite the first term back as
x
a to get,
( ) ( ) ln
x
f x a a ′ =
So, not too bad if you can see the trick to rewrite a and with the Chain Rule.
Implicit Differentiation
To this point we've done quite a few derivatives, but they have all been derivatives of functions
of the form ( ) y f x = . Unfortunately not all the functions that we're going to look at will fall
into this form.
Let's take a look at an example of a function like this.
Example 1 Find y′ for 1 xy = .
Solution
There are actually two solution methods for this problem.
Solution 1 :
This is the simple way of doing the problem. Just solve for y to get the function in the form that
we're used to dealing with and then differentiate.
2
1 1
y y
x x
′ = ⇒ = −
So, that's easy enough to do. However, there are some functions for which this can't be done.
That's where the second solution technique comes into play.
Solution 2 :
In this case we're going to leave the function in the form that we were given and work with it in
that form. However, let's recall from the first part of this solution that if we could solve for y
then we will get y as a function of x. In other words, if we could solve for y (as we could in this
case, but won't always be able to do) we get ( ) y y x = . Let's rewrite the equation to note this.
( ) 1 xy x y x = =
Be careful here and note that when we write ( ) y x we don't mean y time x. What we are noting
here is that y is some (probably unknown) function of x. This is important to recall when doing
this solution technique.
The next step in this solution is to differentiate both sides with respect to x as follows,
( ) ( ) ( ) 1
d d
x y x
dx dx
=
Note that we dropped the ( ) x on the y as it was only there to remind us that the y was a function
of x and now that we've taken the derivative it's no longer really needed. We just wanted it in the
equation to recognize the product rule when we took the derivative.
So, let's now recall just what were we after. We were after the derivative, y′ , and notice that
there is now a y′ in the equation. So, to get the derivative all that we need to do is solve the
equation for y′ .
y
y
x
′ = −
There it is. Using the second solution technique this is our answer. This is not what we got from
the first solution however. Or at least it doesn't look like the same derivative that we got from the
first solution. Recall however, that we really do know what y is in terms of x and if we plug that
in we will get,
2
1
1
x
y
x x
′ = − = −
which is what we got from the first solution. Regardless of the solution technique used we should
get the same derivative.
The process that we used in the second solution to the previous example is called implicit
differentiation and that is the subject of this section. In the previous example we were able to
just solve for y and avoid implicit differentiation. However, in the remainder of the examples in
this section we either won't be able to solve for y or, as we'll see in one of the examples below,
the answer will not be in a form that we can deal with.
In the second solution above we replaced the y with ( ) y x and then did the derivative. Recall
that we did this to remind us that y is in fact a function of x. We'll be doing this quite a bit in
these problems, although we rarely actually write ( ) y x . So, before we actually work anymore
implicit differentiation problems let's do a quick set of "simple" derivatives that will hopefully
help us with doing derivatives of functions that also contain a ( ) y x .
Solution
These are written a little differently from what we're used to seeing here. This is because we
want to match up these problems with what we'll be doing in this section. Also, each of these
parts has several functions to differentiate starting with a specific function followed by a general
function. This again, is to help us with some specific parts of the implicit differentiation process
that we'll be doing.
and this is just the chain rule. We differentiated the outside function (the exponent of 5) and then
multiplied that by the derivative of the inside function (the stuff inside the parenthesis).
For the section function we're going to do basically the same thing. We're going to need to use
the chain rule. The outside function is still the exponent of 5 while the inside function this time is
simply ( ) f x . We don't have a specific function here, but that doesn't mean that we can't at
least write down the chain rule for this function. Here is the derivative for this function,
( ) ( ) ( )
5 4
5
d
f x f x f x
dx
′ = ⎡ ⎤ ⎡ ⎤
⎣ ⎦ ⎣ ⎦
We don't actually know what ( ) f x is so when we do the derivative of the inside function all we
can do is write down notation for the derivative, i.e. ( ) f x ′ .
With the final function here we simply replaced the f in the second function with a y since most
of our work in this section will involve y's instead of f's. Outside of that this function is identical
to the second. So, the derivative is,
( ) ( ) ( )
5 4
5
d
y x y x y x
dx
′ = ⎡ ⎤ ⎡ ⎤
⎣ ⎦ ⎣ ⎦
For the second function we didn't bother this time with using ( ) f x and just jumped straight to
( ) y x for the general version. This is still just a general version of what we did for the first
function. The outside function is still the sine and the inside is give by ( ) y x and while we don't
have a formula for ( ) y x and so we can't actually take its derivative we do have a notation for its
derivative. Here is the derivative for this function,
( ) ( ) ( ) ( ) ( )
sin cos
d
y x y x y x
dx
⎡ ⎤ ′ =
⎣ ⎦
[Return to Problems]
(c)
2
9 x x −
e ,
( ) y x
e
In this part we'll just give the answers for each and leave out the explanation that we had in the
first two parts.
So, in this set of examples we were just doing some chain rule problems where the inside function
was ( ) y x instead of a specific function. This kind of derivative shows up all the time in doing
implicit differentiation so we need to make sure that we can do them. Also note that we only did
this for three kinds of functions but there are many more kinds of functions that we could have
used here.
So, it's now time to do our first problem where implicit differentiation is required, unlike the first
example where we could actually avoid implicit differentiation by solving for y.
Example 3 Find y′ for the following function.
2 2
9 x y + =
Solution
How, this is just a circle and we can solve for y which would give,
2
9 y x = ± −
Prior to starting this problem we stated that we had to do implicit differentiation here because we
couldn't just solve for y and yet that's what we just did. So, why can't we use implicit
differentiation here? The problem is the " ± ". With this in the "solution" for y we see that y is in
fact two different functions. Which should we use? Should we use both? We only want a single
function for the derivative and at best we have two functions here.
So, in this example we really are going to need to do implicit differentiation so we can avoid this.
In this example ee'll do the same thing we did in the first example and remind ourselves that y is
really a function of x and write y as ( ) y x . Once we've done this all we need to do is
differentiate each term with respect to x.
( )
( )
( )
2
2
9
d d
x y x
dx dx
+ = ⎡ ⎤
⎣ ⎦
As with the first example the right side is easy. The left side is also pretty easy since all we need
to do is take the derivative of each term and note that the second term will be similar the part (a)
of the second example. All we need to do for the second term is use the chain rule.
At this point we can drop the ( ) x part as it was only in the problem to help with the
differentiation process. The final step is to simply solve the resulting equation for y′ .
2 2 0 x yy
x
y
y
′ + =
′ = −
Unlike the first example we can't just plug in for y since we wouldn't know which of the two
functions to use. Most answers from implicit differentiation will involve both x and y so don't get
excited about that when it happens.
As always, we can't forget our interpretations of derivatives.
Example 4 Find the equation of the tangent line to
2 2
9 x y + =
at the point
( )
2, 5 .
Solution
First note that unlike all the other tangent line problems we've done in previous sections we need
to be given both the x and the y values of the point. Notice as well that this point does lie on the
graph of the circle (you can check by plugging the points into the equation) and so it's okay to
talk about the tangent line at this point.
Now, let's work some more examples. In the remaining examples we will no longer write ( ) y x
for y. This is just something that we were doing to remind ourselves that y is really a function of
x to help with the derivatives. Seeing the ( ) y x reminded us that we needed to do the chain rule
on that portion of the problem. From this point on we'll leave the y's written as y's and in our
head we'll need to remember that they really are ( ) y x and that we'll need to do the chain rule.
There is an easy way to remember how to do the chain rule in these problems. The chain rule
really tells us to differentiate the function as we usually would, except we need to add on a
derivative of the inside function. In implicit differentiation this means that every time we are
differentiating a term with y in it the inside function is the y and we will need to add a y′ onto the
term since that will be the derivative of the inside function.
Solution
(a)
3 5 3
3 8 1 x y x y + = +
First differentiate both sides with respect to x and remember that each y is really ( ) y x we just
aren't going to write it that way anymore. This means that the first term on the left will be a
product rule.
We differentiated these kinds of functions involving y's to a power with the chain rule in the
Example 2 above. Also, recall the discussion prior to the start of this problem. When doing this
kind of chain rule problem all that we need to do is differentiate the y's as normal and then add on
a y′ , which is nothing more than the derivative of the "inside function".
Now all that we need to do is solve for the derivative, y′ . This is just basic solving algebra that
you are capable of doing. The main problem is that it's liable to be messier than what you're
used to doing. All we need to do is get all the terms with y′ in them on one side and all the terms
without y′ in them on the other. Then factor y′ out of all the terms containing it and divide both
side by the "coefficient" of the y′ . Here is the solving work for this one,
(c)
( )
2 3 2 3
ln
x y
x xy
+
= − e
We're going to need to be careful with this problem. We've got a couple chain rules that we're
going to need to deal with here that are a little different from those that we've dealt with prior to
this problem.
In both the exponential and the logarithm we've got a "standard" chain rule in that there is
something other than just an x or y inside the exponential and logarithm. So, this means we'll do
the chain rule as usual here and then when we do the derivative of the inside function for each
term we'll have to deal with differentiating y's.
In both of the chain rules note that the y′ didn't get tacked on until we actually differentiated the
y's in that term.
Now we need to solve for the derivative and this is liable to be somewhat messy. In order to get
the y′ on one side we'll need to multiply the exponential through the parenthesis and break up
the quotient.
Note that to make the derivative at least look a little nicer we converted all the fractions to
negative exponents.
[Return to Problems]
Okay, we've seen one application of implicit differentiation in the tangent line example above.
However, there is another application that we will be seeing in every problem in the next section.
In some cases we will have two (or more) functions all of which are functions of a third variable.
So, we might have ( ) x t and ( ) y t , for example and in these cases we will be differentiating
with respect to t. This is just implicit differentiation like we did in the previous examples, but
there is a difference however.
In the previous examples we have functions involving x's and y's and thinking of y as ( ) y x . In
these problems we differentiated with respect to x and so when faced with x's in the function we
differentiated as normal and when faced with y's we differentiated as normal except we then
added a y′ onto that term because we were really doing a chain rule.
In the new example we want to look at we're assume that ( ) x x t = and that ( ) y y t = and
differentiating with respect to t. This means that every time we are faced with an x or a y we'll be
doing the chain rule. This in turn means that when we differentiate an x we will need to add on
an x′ and whenever we differentiate a y we will add on a y′ .
These new types of problems are really the same kind of problem we've been doing in this
section. They are just expanded out a little to include more than one function that will require a
chain rule.
Let's take a look at an example of this kind of problem.
Example 6 Assume that ( ) x x t = and ( ) y y t = and differentiate the following equation with
respect to t.
( )
3 6 1 2
cos 5
x
x y y y
−
+ − = e
Solution
So, just differentiate as normal and add on an appropriate derivative at each step. Note as well
that the first term will be a product rule since both x and y are functions of t.
There really isn't all that much to this problem. Since there are two derivatives in the problem we
won't be bothering to solve for one of them. When we do this kind of problem in the next section
the problem will imply which one we need to solve for.
At this point there doesn't seem be any real reason for doing this kind of problem, but as we'll see
in the next section every problem that we'll be doing there will involve this kind of implicit
differentiation.
Related Rates
In this section we are going to look at an application of implicit differentiation. Most of the
applications of derivatives are in the next chapter however there are a couple of reasons for
placing it in this chapter as opposed to putting it into the next chapter with the other applications.
The first reason is that it's an application of implicit differentiation and so putting right after that
section means that we won't have forgotten how to do implicit differentiation. The other reason
is simply that after doing all these derivatives we need to be reminded that there really are actual
applications to derivatives. Sometimes it is easy to forget there really is a reason that we're
spending all this time on derivatives.
For these related rates problems it's usually best to just jump right into some problems and see
how they work.
Example 1 Air is being pumped into a spherical balloon at a rate of 5 cm
3
/min. Determine the
rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.
Solution
The first thing that we'll need to do here is to identify what information that we've been given
and what we want to find. Before we do that let's notice that both the volume of the balloon and
the radius of the balloon will vary with time and so are really functions of time, i.e. ( ) V t and
( ) r t .
We know that air is being pumped into the balloon at a rate of 5 cm
3
/min. This is the rate at
which the volume is increasing. Recall that rates of change are nothing more than derivatives and
so we know that,
( ) 5 V t ′ =
We want to determine the rate at which the radius is changing. Again, rates are derivatives and so
it looks like we want to determine,
( ) ( ) ? when 10cm
2
d
r t r t ′ = = =
Note that we needed to convert the diameter to a radius.
Now that we've identified what we have been given and what we want to find we need to relate
these two quantities to each other. In this case we can relate the volume and the radius with the
formula for the volume of a sphere.
( ) ( )
3 4
3
V t r t π = ⎡ ⎤
⎣ ⎦
Now we don't really want a relationship between the volume and the radius. What we really
want is a relationship between their derivatives. We can do this by differentiating both sides with
respect to t. In other words, we will need to do implicit differentiation on the above formula.
Doing this gives,
2
4 V r r π ′ ′ =
Note that at this point we went ahead and dropped the ( ) t from each of the terms. Now all that
we need to do is plug in what we know and solve for what we want to find.
( )
2
1
5 4 10 cm/min
80
r r π
π
′ ′ = ⇒ =
We can get the units of the derivative be recalling that,
dr
r
dt
′ =
The units of the derivative will be the units of the numerator (cm in the previous example)
divided by the units of the denominator (min in the previous example).
Let's work some more examples.
Example 2 A 15 foot ladder is resting against the wall. The bottom is initially 10 feet away
from the wall and is being pushed towards the wall at a rate of
1
4
ft/sec. How fast is the top of the
ladder moving up the wall 12 seconds after we start pushing?
Solution
The first thing to do in this case is to sketch picture that shows us what is going on.
We've defined the distance of the bottom of the latter from the wall to be x and the distance of the
top of the ladder from the floor to be y. Note as well that these are changing with time and so we
really should write ( ) x t and ( ) y t . However, as is often the case with related rates/implicit
differentiation problems we don't write the ( ) t part just try to remember this in our heads as we
proceed with the problem.
1
4
x′ = −
Note as well that the rate is negative since the distance from the wall, x, is decreasing. We always
need to be careful with signs with these problems.
We want to find the rate at which the top of the ladder is moving away from the floor. This is y′ .
Note as well that this quantity should be positive since y will be increasing.
As with the first example we first need a relationship between x and y. We can get this using
Pythagorean theorem.
( )
2
2 2
15 225 x y + = =
All that we need to do at this point is to differentiate both sides with respect to t, remembering
that x and y are really functions of t and so we'll need to do implicit differentiation. Doing this
gives an equation that shows the relationship between the derivatives.
2 2 0 xx yy ′ ′ + = (1)
Next, let's see which of the various parts of this equation that we know and what we need to find.
We know x′ and are being asked to determine y′ so it's okay that we don't know that.
However, we still need to determine x and y.
Determining x and y is actually fairly simple. We know that initially 10 x = and the end is being
pushed in towards the wall at a rate of
1
4
ft/sec and that we are interested in what has happened
after 12 seconds. We know that,
( )
distance rate time
1
12 3
4
= ×
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
So, the end of the ladder has been pushed in 3 feet and so after 12 seconds we must have 7 x = .
Note that we could have computed this in one step as follows,
( )
1
10 12 7
4
x = − =
To find y (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the
values of x that we just found above.
Notice that we got the correct sign for y′ . If we'd gotten a negative then we'd have known that
we had made a mistake and we could go back and look for it.
Example 3 Two people are 50 feet apart. One of them starts walking north at a rate so that the
angle shown in the diagram below is changing at a constant rate of 0.01 rad/min. At what rate is
distance between the two people changing when 0.5 θ = radians?
Solution
This example is not as tricky as it might at first appear. Let's call the distance between them at
any point in time x as noted above. We can then relate all the known quantities by one of two trig
formulas.
50
cos sec
50
x
x
θ θ = =
We want to find x′ and we could find x if we wanted to at the point in question using cosine since
we also know the angle at that point in time. However, if we use the second formula we won't
need to know x as you'll see. So, let's differentiate that formula.
sec tan
50
x
θ θ θ
′
′ =
As noted, there are no x's in this formula. We want to determine x′ and we know that 0.5 θ =
and 0.01 θ′ = (do you agree with it being positive?). So, just plug in and solve.
( )( ) ( ) ( ) 50 0.01 sec 0.5 tan 0.5 0.311254 ft / min x x ′ ′ = ⇒ =
So far we we've seen three related rates problems. While each one was worked in a very
different manner the process was essentially the same in each. In each problem we identified
what we were given and what we wanted to find. We next wrote down a relationship between all
the various quantities and used implicit differentiation to arrive at a relationship between the
various derivatives in the problem. Finally, we plugged into the equation to find the value we
were after.
quantities. This is often the hardest part of the problem. In many problems the best way to come
up with the relationship is to sketch a diagram that shows the situation. This often seems like a
silly step, but can make all the difference in whether we can find the relationship or not.
Let's work another problem that uses some different ideas and shows some of the different kinds
of things that can show up in related rates problems.
Example 4 A tank of water in the shape of a cone is leaking water at a constant rate of
3
2ft /hour . The base radius of the tank is 5 ft and the height of the tank is 14 ft.
(a) At what rate is the depth of the water in the tank changing when the depth of the
water is 6 ft?
(b) At what rate is the radius of the top of the water in the tank changing when the depth
of the water is 6 ft?
Solution
Okay, we should probably start off with a quick sketch (probably not to scale) of what is going on
here.
As we can see, the water in the tank actually forms a smaller cone with the same central angle as
the tank itself. The radius of the "water" cone at any time is given by r and the height of the
"water" cone at any time is given by h. The volume of water in the tank at any time t is given by,
2
1
3
V r h π =
and we've been given that 2 V′ = − .
(a) At what rate is the depth of the water in the tank changing when the depth of the water
is 6 ft?
formula that we've got that will relate the volume to the height also includes the radius and so if
we were to differentiate this with respect to t we would get,
2
2 1
3 3
V rr h r h π π ′ ′ ′ = +
So, in this equation we know V′ and h and want to find h′ , but we don't know r and r′ . As
we'll see finding r isn't too bad, but we just don't have enough information, at this point, that will
allow us to find r′ and h′ simultaneously.
To fix this we'll need to eliminate the r from the volume formula in some way. This is actually
easier than it might at first look. If we go back to our sketch above and look at just the right half
of the tank we see that we have two similar triangles and when we say similar we mean similar in
the geometric sense. Recall that two triangles are called similar if their angles are identical,
which in the case here. When we have two similar triangles then ratios of any two sides will be
equal. For our set this means that we have,
5 5
14 14
r
r h
h
= ⇒ =
If we take this and plug into our volume formula we have,
2
2 3
1 1 5 25
3 3 14 588
V r h h h h π π π
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
This gives us a volume formula that only involved the volume and the height of the water. Note
however that this volume formula is only valid for our cone, so don't be tempted to use it for
other cones! If we now differentiate this we have,
2
25
196
V h h π ′ ′ =
At this point all we need to do is plug in what we know and solve for h′ .
( )
2
25 98
2 6 0.1386
196 225
h h π
π
−
′ ′ − = ⇒ = = −
So, it looks like the height is decreasing at a rate of 0.1386 ft/hr.
(b) At what rate is the radius of the top of the water in the tank changing when the depth of
the water is 6 ft?
In this case we are asking for r′ and there is an easy way to do this part and a difficult (well,
more difficult than the easy way anyway….) way to do it. The "difficult" way is to redo the work
part (a) above only this time use,
That's not terribly difficult, but it is more work that we need to so. Recall from the first part that
we have,
5 5
14 14
r h r h ′ ′ = ⇒ =
So, as we can see if we take the relationship that relates r and h that we used in the first part and
differentiate it we get a relationship between r′ and h′ . At this point all we need to do here is
use the result from the first part to get,
5 98 7
0.04951
14 225 45
r
π π
− ⎛ ⎞
′ = = − = −
⎜ ⎟
⎝ ⎠
Much easier that redoing all of the first part. Note however, that we were only able to do this the
"easier" way because it was asking for r′ at exactly the same time that we asked for h′ in the
first part. If we hadn't been using the same time then we would have had no choice but to do this
the "difficult" way.
In the second part of the previous problem we saw an important idea in dealing with related rates.
In order to find the asked for rate all we need is an equations that relates the rate we're looking
for to a rate that we already know. Sometimes there are multiple equations that we can use and
sometimes one will be easier than another.
Also, this problem showed us that we will often have an equation that contains more variables
that we have information about and so, in these cases, we will need to eliminate one (or more) of
the variables. In this problem we eliminated the extra variable using the idea of similar triangles.
This will not always be how we do this, but many of these problems do use similar triangles so
make sure you can use that idea.
Let's work some more problems.
Example 5 A trough of water is 8 meters deep and its ends are in the shape of isosceles
triangles whose width is 5 meters and height is 2 meters. If water is being pumped in at a
constant rate of
3
6 m /sec . At what rate is the height of the water changing when the water has a
height of 120 cm?
Solution
Note that an isosceles triangle is just a triangle in which two of the sides are the same length. In
our case sides of the tank have the same length.
Now, in this problem we know that
3
6 m /sec V′ = and we want to determine h′ when
1.2m h = . Note that because V′ is in terms of meters we need to convert h into meters as well.
So, we need an equation that will relate these two quantities and the volume of the tank will do it.
The volume of this kind of tank is simple to compute. The volume is the area of the end times the
depth. For our case the volume of the water in the tank is,
As with the previous example we've got an extra quantity here, w, that is also changing with time
and so we need to get eliminate it from the problem. To do this we'll again make use of the idea
of similar triangles. If we look at the end of the tank we'll see that we again have two similar
triangles. One for the tank itself and on formed by the water in the tank. Again, remember that
with similar triangles that ratios of sides must be equal. In our case we'll use,
5
5 2 2
w h
w h = ⇒ =
Plugging this into the volume gives a formula for the volume (and only for this tank) that only
involved the height of the water.
Example 6 A light is on the top of a 12 ft tall pole and a 5ft 6in tall person is walking away
from the pole at a rate of 2 ft/sec.
(a) At what rate is the tip of the shadow moving away from the pole when the person is
25 ft from the pole?
(b) At what rate is the tip of the shadow moving away from the person when the person is
25 ft from the pole?
Solution
We'll definitely need a sketch of this situation to get us started here. The tip of the shadow is
defined by the rays of light just getting past the person and so we can form the following set of
similar triangles.
Here x is the distance of the tip of the shadow from the pole,
p
x is the distance of the person
from the pole and
s
x is the length of the shadow. Also note that we converted the persons height
over to 5.5 feet since all the other measurements are in feet.
(a) At what rate is the tip of the shadow moving away from the pole when the person is 25 ft
from the pole?
In this case we want to determine x′ when 25
p
x = given that 2
p
x′ = .
The tip of the shadow is then moving away from the pole at a rate of 3.6923 ft/sec. Notice as
well that we never actually had to use the fact that 25
p
x = for this problem. That will happen
on rare occasions.
(b) At what rate is the tip of the shadow moving away from the person when the person is
25 ft from the pole?
This part is actually quite simple if we have the answer from (a) in hand, which we do of course.
In this case we know that
s
x represents the length of the shadow, or the distance of the tip of the
shadow from the person so it looks like we want to determine
s
x′ when 25
p
x = .
Again, we can use
p s
x x x = + , however unlike the first part we now know that 2
p
x′ = and
3.6923 ft/sec x′ = so in this case all we need to do is differentiate the equation and plug in for
all the known quantities.
3.6923 2 1.6923 ft/sec
p s
s s
x x x
x x
′ ′ ′ = +
′ ′ = + =
The tip of the shadow is then moving away from the person at a rate of 1.6923 ft/sec.
Example 7 A spot light is on the ground 20 ft away from a wall and a 6 ft tall person is walking
towards the wall at a rate of 2.5 ft/sec. How fast the height of the shadow changing when the
person is 8 feet from the wall? Is the shadow increasing or decreasing in height at this time?
Solution
Let's start off with a sketch of this situation and the sketch here will be similar to that of the
previous problem. The top of the shadow will be defined by the light rays going over the head of
the person and so we again get yet another set of similar triangles.
In this case we want to determine y′ when the person is 8 ft from wall or 12 ft x = . Also, if the
person is moving towards the wall at 2.5 ft/sec then the person must be moving away from the
spotlight at 2.5 ft/sec and so we also know that 2.5 x′ = .
In all the previous problems that used similar triangles we used the similar triangles to eliminate
one of the variables from the equation we were working with. In this case however, we can get
the equation that relates x and y directly from the two similar triangles. In this case the equation
we're going to work with is,
20 120
6
y
y
x x
= ⇒ =
Example 8 Two people on bikes are separated by 350 meters. Person A starts riding north at a
rate of 5 m/sec and 7 minutes later Person B starts riding south at 3 m/sec. At what rate is the
distance separating the two people changing 25 minutes after Person A starts riding?
Solution
There is a lot to digest here with this problem. Let's start off with a sketch of the situation.
Now we are after z′ and we know that 5 x′ = and 3 y′ = . We want to know z′ after Person A
had been riding for 25 minutes and Person B has been riding for 25 7 18 − = minutes. After
converting these times to seconds (because our rates are all in m/sec) this means that at the time
we're interested in each of the bike riders has rode,
( ) ( ) 5 25 60 7500 m 3 18 60 3240 m x y = × = = × =
We've seen quite a few related rates problems in this section that cover a wide variety of possible
problems. There are still many more different kinds of related rates problems out there in the
world, but the ones that we've worked here should give you a pretty good idea on how to at least
start most of the problems that you're liable to run into.
By this point we should be able to differentiate this function without any problems. Doing this
we get,
( )
2
15 6 10 f x x x ′ = − +
Now, this is a function and so it can be differentiated. Here is the notation that we'll use for that,
as well as the derivative.
( ) ( ) ( )
30 6 f x f x x
′
′′ ′ = = −
This is called the second derivative and ( ) f x ′ is now called the first derivative.
Again, this is a function as so we can differentiate it again. This will be called the third
derivative. Here is that derivative as well as the notation for the third derivative.
( ) ( ) ( )
30 f x f x
′
′′′ ′′ = =
Continuing, we can differentiate again. This is called, oddly enough, the fourth derivative.
We're also going to be changing notation at this point. We can keep adding on primes, but that
will get cumbersome after awhile.
( ) ( ) ( )
(4)
0 f x f x
′
′′′ = =
This process can continue but notice that we will get zero for all derivatives after this point. This
set of derivatives leads us to the following fact about the differentiation of polynomials.
Fact
If p(x) is a polynomial of degree n (i.e. the largest exponent in the polynomial) then,
( )
( ) 0 for 1
k
p x k n = ≥ +
We will need to be careful with the "non-prime" notation for derivatives. Consider each of the
following.
( )
( ) ( )
( ) ( )
2
2
2
f x f x
f x f x
′′ =
= ⎡ ⎤
⎣ ⎦
The presence of parenthesis in the exponent denotes differentiation while the absence of
parenthesis denotes exponentiation.
Collectively the second, third, fourth, etc. derivatives are called higher order derivatives.
Notice that differentiating an exponential function is very simple. It doesn't change with each
differentiation.
[Return to Problems]
(b) cos y x =
Again, let's just do some derivatives.
( ) 4
cos
sin
cos
sin
cos
y x
y x
y x
y x
y x
=
′ = −
′′ = −
′′′ =
=
Note that cosine (and sine) will repeat every four derivatives. The other four trig functions will
not exhibit this behavior. You might want to take a few derivatives to convince yourself of this.
[Return to Problems]
Notice that each successive derivative will require a product and/or chain rule and that as noted
above this will not end up returning back to just a secant after four (or another other number for
that matter) derivatives as sine and cosine will.
[Return to Problems]
(b) ( )
3
1 2w
g w
−
= e
Again, let's start with the first derivative.
( )
3
2 1 2
6
w
g w w
−
′ = − e
As with the first example we will need the product rule for the second derivative.
As we saw in this last set of examples we will often need to use the product or quotient rule for
the higher order derivatives, even when the first derivative didn't require these rules.
Let's work one more example that will illustrate how to use implicit differentiation to find higher
order derivatives.
Example 3 Find y′′ for
2 4
10 x y + =
Solution
Okay, we know that in order to get the second derivative we need the first derivative and in order
to get that we'll need to do implicit differentiation. Here is the work for that.
3
3
2 4 0
2
x y y
x
y
y
′ + =
′ = −
Now, this is the first derivative. We get the second derivative by differentiating this, which will
require implicit differentiation again.
This is fine as far as it goes. However, we would like there to be no derivatives in the answer.
We don't, generally, mind having x's and/or y's in the answer when doing implicit differentiation,
but we really don't like derivatives in the answer. We can get rid of the derivative however by
acknowledging that we know what the first derivative is and substituting this into the second
derivative equation. Doing this gives,
Now that we've found some higher order derivatives we should probably talk about an
interpretation of the second derivative.
If the position of an object is given by s(t) we know that the velocity is the first derivative of the
position.
( ) ( ) v t s t ′ =
The acceleration of the object is the first derivative of the velocity, but since this is the first
derivative of the position function we can also think of the acceleration as the second derivative
of the position function.
( ) ( ) ( ) a t v t s t ′ ′′ = =
Alternate Notation
There is some alternate notation for higher order derivatives as well. Recall that there was a
fractional notation for the first derivative.
( )
df
f x
dx
′ =
Logarithmic Differentiation
There is one last topic to discuss in this section. Taking the derivatives of some complicated
functions can be simplified by using logarithms. This is called logarithmic differentiation.
It's easiest to see how this works in an example.
Example 1 Differentiate the function.
( )
5
2
1 10 2
x
y
x x
=
− +
Solution
Differentiating this function could be done with a product rule and a quotient rule. However, that
would be a fairly messy process. We can simplify things somewhat by taking logarithms of both
sides.
( )
5
2
ln ln
1 10 2
x
y
x x
⎛ ⎞
⎜ ⎟ =
⎜ ⎟
− +
⎝ ⎠
Of course, this isn't really simpler. What we need to do is use the properties of logarithms to
expand the right side as follows.
This doesn't look all the simple. However, the differentiation process will be simpler. What we
need to do at this point is differentiate both sides with respect to x. Note that this is really implicit
differentiation.
Depending upon the person doing this would probably be slightly easier than doing both the
product and quotient rule. The answer is almost definitely simpler that what we would have
gotten using the product and quotient rule.
So, as the first example has shown we can use logarithmic differentiation to avoid using the
product rule and/or quotient rule.
Neither of these two will work here since both require either the base or the exponent to be a
constant. In this case both the base and the exponent are variables and so we have no way to
differentiate this function using only known rules from previous sections.
With logarithmic differentiation we can do this however. First take the logarithm of both sides as
we did in the first example and use the logarithm properties to simplify things a little.
Solution
Now, this look much more complicated than the previous example, but is in fact only
slightly more complicated. The process is pretty much identical so we first take the log
of both sides and then simplify the right side.
( )
( )
( ) ( )
cos
ln ln 1 3 cos ln 1 3
x
y x x x
⎡ ⎤
= − = −
⎣ ⎦
It is sometimes easy to get these various functions confused and use the wrong rule for
differentiation. Always remember that each rule has very specific rules for where the variable
and constants must be. For example, the Power Rule requires that the base be a variable and the
exponent be a constant, while the exponential function requires exactly the opposite.
Introduction
In the previous chapter we focused almost exclusively on the computation of derivatives. In this
chapter will focus on applications of derivatives. It is important to always remember that we
didn't spend a whole chapter talking about computing derivatives just to be talking about them.
There are many very important applications to derivatives.
The two main applications that we'll be looking at in this chapter are using derivatives to
determine information about graphs of functions and optimization problems. These will not be
the only applications however. We will be revisiting limits and taking a look at an application of
derivatives that will allow us to compute limits that we haven't been able to compute previously.
We will also see how derivatives can be used to estimate solutions to equations.
Here is a listing of the topics in this section.
Rates of Change – The point of this section is to remind us of the application/interpretation of
derivatives that we were dealing with in the previous chapter. Namely, rates of change.
Critical Points – In this section we will define critical points. Critical points will show up in
many of the sections in this chapter so it will be important to understand them.
Minimum and Maximum Values – In this section we will take a look at some of the basic
definitions and facts involving minimum and maximum values of functions.
Finding Absolute Extrema – Here is the first application of derivatives that we'll look at in this
chapter. We will be determining the largest and smallest value of a function on an interval.
The Shape of a Graph, Part I – We will start looking at the information that the first derivatives
can tell us about the graph of a function. We will be looking at increasing/decreasing functions as
well as the First Derivative Test.
The Shape of a Graph, Part II – In this section we will look at the information about the graph
of a function that the second derivatives can tell us. We will look at inflection points, concavity,
and the Second Derivative Test.
The Mean Value Theorem – Here we will take a look that the Mean Value Theorem.
Optimization Problems – This is the second major application of derivatives in this chapter. In
this section we will look at optimizing a function, possible subject to some constraint.
More Optimization Problems – Here are even more optimization problems.
L'Hospital's Rule and Indeterminate Forms – This isn't the first time that we've looked at
indeterminate forms. In this section we will take a look at L'Hospital's Rule. This rule will
allow us to compute some limits that we couldn't do until this section.
Linear Approximations – Here we will use derivatives to compute a linear approximation to a
function. As we will see however, we've actually already done this.
Differentials – We will look at differentials in this section as well as an application for them.
Newton's Method – With this application of derivatives we'll see how to approximate solutions
to an equation.
Business Applications – Here we will take a quick look at some applications of derivatives to the
business field.
Rates of Change
The purpose of this section is to remind us of one of the more important applications of
derivatives. That is the fact that ( ) f x ′ represents the rate of change of ( ) f x . This is an
application that we repeatedly saw in the previous chapter. Almost every section in the previous
chapter contained at least one problem dealing with this application of derivatives. While this
application will arise occasionally in this chapter we are going to focus more on other
applications in this chapter.
So, to make sure that we don't forget about this application here is a brief set of examples
concentrating on the rate of change application of derivatives. Note that the point of these
examples is to remind you of material covered in the previous chapter and not to teach you how
to do these kinds of problems. If you don't recall how to do these kinds of examples you'll need
to go back and review the previous chapter.
Now, the function will not be changing if the rate of change is zero and so to answer this question
we need to determine where the derivative is zero. So, let's set this equal to zero and solve.
( ) ( ) 6 20sin 2 0 sin 2 0.3
20
x x
6
− + = ⇒ = =
So, the function is not changing at three values of t. Finally, to determine where the function is
increasing or decreasing we need to determine where the derivative is positive or negative.
Recall that if the derivative is positive then the function must be increasing and if the derivative is
negative then the function must be decreasing. The following number line gives this information.
So, from this number line we can see that we have the following increasing and decreasing
information.
Increasing : 1.159, 2.492 Decreasing : 1.159 2.492 t t t −∞ < < − < < ∞ − < <
If you don't remember how to solve polynomial and rational inequalities then you should check
out the appropriate sections in the Review Chapter.
Finally, we can't forget about Related Rates problems.
Example 3 Two cars start out 500 miles apart. Car A is to the west of Car B and starts driving
to the east (i.e. towards Car B) at 35 mph and at the same time Car B starts driving south at 50
mph. After 3 hours of driving at what rate is the distance between the two cars changing? Is it
increasing or decreasing?
Solution
The first thing to do here is to get sketch a figure showing the situation.
Now, to answer this question we will need to determine z′ given that 35 x′ = − and 50 y′ = . Do
you agree with the signs on the two given rates? Remember that a rate is negative if the quantity
is decreasing and positive if the quantity is increasing.
We can again use the Pythagorean theorem here. First, write it down and the remember that x, y,
and z are all changing with time and so differentiate the equation using Implicit Differentiation.
So, after three hours the distance between them is decreasing at a rate of 14.9696 mph.
So, in this section we covered three "standard" problems using the idea that the derivative of a
function gives its rate of change. As mentioned earlier, this chapter will e focusing more on other
applications than the idea of rate of change, however, we can't forget this application as it is a
very important one.
Critical Points
Critical points will show up throughout a majority of this chapter so we first need to define them
and work a few examples before getting into the sections that actually use them.
Definition
We say that x c = is a critical point of the function f(x) if ( ) f c exists and if either of the
following are true.
( ) ( ) 0 OR doesn't exist f c f c ′ ′ =
Note that we require that ( ) f c exists in order for x c = to actually be a critical point. This is an
important, and often overlooked, point.
The main point of this section is to work some examples finding critical points. So, let's work
some examples.
Example 1 Determine all the critical points for the function.
( )
5 4 3
6 33 30 100 f x x x x = + − +
Solution
We first need the derivative of the function in order to find the critical points and so let's get that
and notice that we'll factor it as much as possible to make our life easier when we go to find the
critical points.
Now, our derivative is a polynomial and so will exist everywhere. Therefore the only critical
points will be those values of x which make the derivative zero. So, we must solve.
( )( )
2
6 5 3 5 0 x x x − + =
Because this is the factored form of the derivative it's pretty easy to identify the three critical
points. They are,
3
5, 0,
5
x x x = − = =
Polynomials are usually fairly simple functions to find critical points for provided the degree
doesn't get so large that we have trouble finding the roots of the derivative.
We will need to be careful with this problem. When faced with a negative exponent it is often
best to eliminate the minus sign in the exponent as we did above. This isn't really required but it
can make our life easier on occasion if we do that.
Notice as well that eliminating the negative exponent in the second term allows us to correctly
identify why 0 t = is a critical point for this function. Once we move the second term to the
denominator we can clearly see that the derivative doesn't exist at 0 t = and so this will be a
critical point. If you don't get rid of the negative exponent in the second term many people will
incorrectly state that 0 t = is a critical point because the derivative is zero at 0 t = . While this
may seem like a silly point, after all in each case 0 t = is identified as a critical point, it is
sometimes important to know why a point is a critical point. In fact, in a couple of sections we'll
see a fact that only works for critical points in which the derivative is zero.
So, we've found one critical point (where the derivative doesn't exist), but we now need to
determine where the derivative is zero (provided it is of course…). To help with this it's usually
best to combine the two terms into a single rational expression. So, getting a common
denominator and combining gives us,
( )
1
3
10 2
3
t
g t
t
−
′ =
Notice that we factored a "-1" out of the numerator to help a little with finding the critical points.
This negative out in front will not affect the derivative whether or not the derivative is zero or not
exist but will make our work a little easier.
Now, we have two issues to deal with. First the derivative will not exist if there is division by
zero in the denominator. So we need to solve,
( )( )
2
6 3 2 0 w w w w − − = − + =
We didn't bother squaring this since if this is zero, then zero squared is still zero and if it isn't
zero then squaring it won't make it zero.
So, we can see from this that the derivative will not exist at 3 w = and 2 w = − . However, these
are NOT critical points since the function will also not exist at these points. Recall that in order
for a point to be a critical point the function must actually exist at that point.
At this point we need to be careful. The numerator doesn't factor, but that doesn't mean that
there aren't any critical points where the derivative is zero. We can use the quadratic formula on
the numerator to determine if the fraction as a whole is ever zero.
Again, remember that while the derivative doesn't exist at 3 w = and 2 w = − neither does the
function and so these two points are not critical points for this function.
So far all the examples have not had any trig functions, exponential functions, etc. in them. We
shouldn't expect that to always be the case. So, let's take a look at some examples that don't just
involve powers of x.
Now, this will exist everywhere and so there won't be any critical points for which the derivative
doesn't exist. The only critical points will come from points that make the derivative zero. We
will need to solve,
Don't forget the 2π n on these! There will be problems down the road in which we will miss
solutions without this! Also make sure that it gets put on at this stage! Now divide by 3 to get all
the critical points for this function.
This function will exist everywhere and so no critical points will come from that. Determining
where this is zero is easier than it looks. We know that exponentials are never zero and so the
only way the derivative will be zero is if,
2
2
2
1 2 0
1 2
1
2
t
t
t
− =
=
=
We will have two critical points for this function.
1
2
t = ±
Example 6 Determine all the critical points for the function.
( ) ( )
2
ln 3 6 f x x x = +
Solution
Before getting the derivative let's notice that since we can't take the log of a negative number or
zero we will only be able to look at 0 x > .
First note that, despite appearances, the derivative will not be zero for 0 x = . As noted above the
derivative doesn't exist at 0 x = because of the natural logarithm and so the derivative can't be
zero there!
This function will never be zero for any real value of x. The exponential is never zero of course
and the polynomial will only be zero if x is complex and recall that we only want real values of x
for critical points.
Therefore, this function will not have any critical points.
It is important to note that not all functions will have critical points! In this course most of the
functions that we will be looking at do have critical points. That is only because those problems
make for more interesting examples. Do not let this fact lead you to always expect that a function
will have critical points. Sometimes they don't as this final example has shown.
Minimum and Maximum Values
Many of our applications in this chapter will revolve around minimum and maximum values of a
function. While we can all visualize the minimum and maximum values of a function we want to
be a little more specific in our work here. In particular we want to differentiate between two
types of minimum or maximum values. The following definition gives the types of minimums
and/or maximums values that we'll be looking at.
Definition
1. We say that f(x) has an absolute (or global) maximum at x c = if ( ) ( ) f x f c ≤ for every x
in the domain we are working on.
2. We say that f(x) has a relative (or local) maximum at x c = if ( ) ( ) f x f c ≤ for every x in
some open interval around x c = .
3. We say that f(x) has an absolute (or global) minimum at x c = if ( ) ( ) f x f c ≥ for every x
in the domain we are working on.
4. We say that f(x) has a relative (or local) minimum at x c = if ( ) ( ) f x f c ≥ for every x in
some open interval around x c = .
Note that when we say an "open interval around x c = " we mean that we can find some interval
( ) , a b , not including the endpoints, such that a c b < < . Or, in other words, c will be contained
somewhere inside the interval and will not be either of the endpoints.
Also, we will collectively call the minimum and maximum points of a function the extrema of
the function. So, relative extrema will refer to the relative minimums and maximums while
absolute extrema refer to the absolute minimums and maximums.
Now, let's talk a little bit about the subtle difference between the absolute and relative in the
definition above.
We will have an absolute maximum (or minimum) at x c = provided f(c) is the largest (or
smallest) value that the function will ever take on the domain that we are working on. Also, when
we say the "domain we are working on" this simply means the range of x's that we have chosen
to work with for a given problem. There may be other values of x that we can actually plug into
the function but have excluded them for some reason.
A relative maximum or minimum is slightly different. All that's required for a point to be a
relative maximum or minimum is for that point to be a maximum or minimum in some interval of
x's around x c = . There may be larger or smaller values of the function at some other place, but
relative to x c = , or local to x c = , f(c) is larger or smaller than all the other function values that
are near it.
Note as well that in order for a point to be a relative extrema we must be able to look at function
values on both sides of x c = to see if it really is a maximum or minimum at that point. This
means that relative extrema do not occur at the end points of a domain. They can only occur
interior to the domain.
There is actually some debate on the preceding point. Some folks do feel that relative extrema
can occur on the end points of a domain. However, in this class we will be using the definition
that says that they can't occur at the end points of a domain.
It's usually easier to get a feel for the definitions by taking a quick look at a graph.
For the function shown in this graph we have relative maximums at x b = and x d = . Both of
these point are relative maximums since they are interior to the domain shown and are the largest
point on the graph in some interval around the point. We also have a relative minimum at x c =
since this point is interior to the domain and is the lowest point on the graph in an interval around
it. The far right end point, x e = , will not be a relative minimum since it is an end point.
The function will have an absolute maximum at x d = and an absolute minimum at x a = .
These two points are the largest and smallest that the function will ever be. We can also notice
that the absolute extrema for a function will occur at either the endpoints of the domain or at
relative extrema. We will use this idea in later sections so it's more important than it might seem
at the present time.
Let's take a quick look at some examples to make sure that we have the definitions of absolute
extrema and relative extrema straight.
Note that we used dots at the end of the graph to remind us that the graph ends at these points.
We can now identify the extrema from the graph. It looks like we've got a relative and absolute
minimum of zero at 0 x = and an absolute maximum of four at 2 x = . Note that 1 x = − is not a
relative maximum since it is at the end point of the interval.
This function doesn't have any relative maximums.
As we saw in the previous example functions do not have to have relative extrema. It is
completely possible for a function to not have a relative maximum and/or a relative minimum.
In this case we still have a relative and absolute minimum of zero at 0 x = . We also still have an
absolute maximum of four. However, unlike the first example this will occur at two points,
2 x = − and 2 x = .
Example 3 Identify the absolute extrema and relative extrema for the following function.
( )
2
f x x =
Solution
In this case we've given no domain and so the assumption is that we will take the largest possible
domain. For this function that means all the real numbers. Here is the graph.
In this case the graph doesn't stop increasing at either end and so there are no maximums of any
kind for this function. No matter which point we pick on the graph there will be points both
larger and smaller than it on either side so we can't have any maximums (or any kind, relative or
absolute) in a graph.
We still have a relative and absolute minimum value of zero at 0 x = .
So, some graphs can have minimums but not maximums. Likewise, a graph could have
maximums but not minimums.
As this example has shown a graph can in fact have extrema occurring at a large number (infinite
in this case) of points.
We've now worked quite a few examples and we can use these examples to see a nice fact about
absolute extrema. First let's notice that all the functions above were continuous functions. Next
notice that every time we restricted the domain to a closed interval (i.e. the interval contains its
end points) we got absolute maximums and absolute minimums. Finally, in only one of the three
examples in which we did not restrict the domain did we get both an absolute maximum and an
absolute minimum.
These observations lead us the following theorem.
Extreme Value Theorem
Suppose that ( ) f x is continuous on the interval [a,b] then there are two numbers , a c d b ≤ ≤ so
that ( ) f c is an absolute maximum for the function and ( ) f d is an absolute minimum for the
function.
So, if we have a continuous function on an interval [a,b] then we are guaranteed to have both an
absolute maximum and an absolute minimum for the function somewhere in the interval. The
theorem doesn't tell us where they will occur or if they will occur more than once, but at least it
tells us that they do exist somewhere. Sometimes, all that we need to know is that they do exist.
This theorem doesn't say anything about absolute extrema if we aren't working on an interval.
We saw examples of functions above that had both absolute extrema, one absolute extrema, and
no absolute extrema when we didn't restrict ourselves down to an interval.
The requirement that a function be continuous is also required in order for us to use the theorem.
Consider the case of
( )
2
1
on [ 1,1] f x
x
= −
Here's the graph.
This function is not continuous at 0 x = as we move in towards zero the function approaching
infinity. So, the function does not have an absolute maximum. Note that it does have an absolute
minimum however. In fact the absolute minimum occurs twice at both 1 x = − and 1 x = .
the function would now have both absolute extrema. We may only run into problems if the
interval contains the point of discontinuity. If it doesn't then the theorem will hold.
We should also point out that just because a function is not continuous at a point that doesn't
mean that it won't have both absolute extrema in an interval that contains that point. Below is the
graph of a function that is not continuous at a point in the given interval and yet has both absolute
extrema.
This graph is not continuous at x c = , yet it does have both an absolute maximum ( x b = ) and an
absolute minimum ( x c = ). Also note that, in this case one of the absolute extrema occurred at
the point of discontinuity, but it doesn't need to. The absolute minimum could just have easily
been at the other end point or at some other point interior to the region. The point here is that this
graph is not continuous and yet does have both absolute extrema
The point of all this is that we need to be careful to only use the Extreme Value Theorem when
the conditions of the theorem are met and not misinterpret the results if the conditions aren't met.
In order to use the Extreme Value Theorem we must have an interval and the function must be
continuous on that interval. If we don't have an interval and/or the function isn't continuous on
the interval then the function may or may not have absolute extrema.
We need to discuss one final topic in this section before moving on to the first major application
of the derivative that we're going to be looking at in this chapter.
Fermat's Theorem
If ( ) f x has a relative extrema at x c = and ( ) f c ′ exists then x c = is a critical point of
( ) f x . In fact, it will be a critical point such that ( ) 0 f c ′ = .
Note that we can say that ( ) 0 f c ′ = because we are also assuming that ( ) f c ′ exists.
This theorem tells us that there is a nice relationship between relative extrema and critical points.
In fact it will allow us to get a list of all possible relative extrema. Since a relative extrema must
be a critical point the list of all critical points will give us a list of all possible relative extrema.
Consider the case of ( )
2
f x x = . We saw that this function had a relative minimum at 0 x = in
several earlier examples. So according to Fermat's theorem 0 x = should be a critical point. The
derivative of the function is,
( ) 2 f x x ′ =
Sure enough 0 x = is a critical point.
Be careful not to misuse this theorem. It doesn't say that a critical point will be a relative
extrema. To see this, consider the following case.
( ) ( )
3 2
3 f x x f x x ′ = =
Clearly 0 x = is a critical point. However we saw in an earlier example this function has no
relative extrema of any kind. So, critical points do not have to be relative extrema.
Also note that this theorem says nothing about absolute extrema. An absolute extrema may or
may not be a critical point.
To see the proof of this theorem see the Proofs From Derivative Applications section of the
Extras chapter.
Finding Absolute Extrema
It's now time to see our first major application of derivatives in this chapter. Given a continuous
function, f(x), on an interval [a,b] we want to determine the absolute extrema of the function. To
do this we will need many of the ideas that we looked at in the previous section.
First, since we have an interval and we are assuming that the function is continuous the Extreme
Value Theorem tells us that we can in fact do this. This is a good thing of course. We don't want
to be trying to find something that may not exist.
Next, we saw in the previous section that absolute extrema can occur at endpoints or at relative
extrema. Also, from Fermat's Theorem we know that the list of critical points is also a list of all
possible relative extrema. So the endpoints along with the list of all critical points will in fact be
a list of all possible absolute extrema.
Now we just need to recall that the absolute extrema are nothing more than the largest and
smallest values that a function will take so all that we really need to do is get a list of possible
absolute extrema, plug these points into our function and then identify the largest and smallest
values.
Here is the procedure for finding absolute extrema.
Finding Absolute Extrema of f(x) on [a,b].
0. Verify that the function is continuous on the interval [a,b].
1. Find all critical points of f(x) that are in the interval [a,b]. This makes sense if you think
about it. Since we are only interested in what the function is doing in this interval we
don't care about critical points that fall outside the interval.
2. Evaluate the function at the critical points found in step 1 and the end points.
3. Identify the absolute extrema.
There really isn't a whole lot to this procedure. We called the first step in the process step 0,
mostly because all of the functions that we're going to look at here are going to be continuous,
but it is something that we do need to be careful with. This process will only work if we have a
function that is continuous on the given interval. The most labor intensive step of this process is
the second step (step 1) where we find the critical points. It is also important to note that all we
want are the critical points that are in the interval.
It looks like we'll have two critical points, 2 t = − and 1 t = . Note that we actually want
something more than just the critical points. We only want the critical points of the function that
lie in the interval in question. Both of these do fall in the interval as so we will use both of them.
That may seem like a silly thing to mention at this point, but it is often forgotten, usually when it
becomes important, and so we will mention it at every opportunity to make it's not forgotten.
Now we evaluate the function at the critical points and the end points of the interval.
( ) ( )
( ) ( )
2 24 1 3
4 28 2 8
g g
g g
− = = −
− = − =
Absolute extrema are the largest and smallest the function will ever be and these four points
represent the only places in the interval where the absolute extrema can occur. So, from this list
we see that the absolute maximum of g(t) is 24 and it occurs at 2 t = − (a critical point) and the
absolute minimum of g(t) is -28 which occurs at 4 t = − (an endpoint).
In this example we saw that absolute extrema can and will occur at both endpoints and critical
points. One of the biggest mistakes that students make with these problems is to forget to check
the endpoints of the interval.
Example 2 Determine the absolute extrema for the following function and interval.
( ) [ ]
3 2
2 3 12 4 on 0, 2 g t t t t = + − +
Solution
Note that this problem is almost identical to the first problem. The only difference is the interval
that we're working on. This small change will completely change our answer however. With this
change we have excluded both of the answers from the first example.
The first step is to again find the critical points. From the first example we know these are
2 t = − and 1 t = .. At this point it's important to recall that we only want the critical points that
actually fall in the interval in question. This means that we only want 1 t = since 2 t = − falls
outside the interval.
Now evaluate the function at the single critical point in the interval and the two endpoints.
( ) ( ) ( ) 1 3 0 4 2 8 g g g = − = =
From this list of values we see that the absolute maximum is 8 and will occur at 2 t = and the
absolute minimum is -3 which occurs at 1 t = .
As we saw in this example a simple change in the interval can completely change the answer. It
also has shown us that we do need to be careful to exclude critical points that aren't in the
interval. Had we forgotten this and included 2 t = − we would have gotten the wrong absolute
maximum!
This is the other big mistakes that students make in these problems. All too often they forget to
exclude critical points that aren't in the interval. If your instructor is anything like me this will
mean that you will get the wrong answer. It's not to hard to make sure that a critical point outside
of the interval is larger or smaller than any of the points in the interval.
Example 3 Suppose that the population (in thousands) of a certain kind of insect after t months
is given by the following formula.
( ) ( ) 3 sin 4 100 P t t t = + +
Determine the minimum and maximum population in the first 4 months.
Solution
The question that we're really asking is to find the absolute extrema of P(t) on the interval [0,4].
Since this function is continuous everywhere we know we can do this.
Let's start with the derivative.
( ) ( ) 3 4cos 4 P t t ′ = +
We need the critical points of the function. The derivative exists everywhere so there are no
critical points from that. So, all we need to do is determine where the derivative is zero.
From these evaluations it appears that the minimum population is 100,000 (remember that P is in
thousands…) which occurs at 0 t = and the maximum population is 111,900 which occurs at
3.7463 t = .
Make sure that you can correctly solve trig equations. If we had forgotten the 2 n π we would
have missed the last three critical points in the interval and hence gotten the wrong answer since
the maximum population was at the final critical point.
Also, note that we do really need to be very careful with rounding answers here. If we'd rounded
to the nearest integer, for instance, it would appear that the maximum population would have
occurred at two different locations instead of only one.
Example 4 Suppose that the amount of money in a bank account after t years is given by,
( )
2
5
8
2000 10
t
A t t
−
= − e
Determine the minimum and maximum amount of money in the account during the first 10 years
that it is open.
Solution
Here we are really asking for the absolute extrema of A(t) on the interval [0,10]. As with the
previous examples this function is continuous everywhere and so we know that this can be done.
The derivative exists everywhere and the exponential is never zero. Therefore the derivative will
only be zero where,
2
2
1 0 4 2
4
t
t t − + = ⇒ = ⇒ = ±
We've got two critical points, however only 2 t = is actually in the interval so that is only critical
point that we'll use.
Let's now evaluate the function at the lone critical point and the end points of the interval. Here
are those function evaluations.
( ) ( ) ( ) 0 2000 2 199.66 10 1999.94 A A A = = =
So, the maximum amount in the account will be $2000 which occurs at 0 t = and the minimum
amount in the account will be $199.66 which occurs at the 2 year mark.
In this example there are two important things to note. First, if we had included the second
critical point we would have gotten an incorrect answer for the maximum amount so it's
important to be careful with which critical points to include and which to exclude.
All of the problems that we've worked to this point had derivatives that existed everywhere and
so the only critical points that we looked at where those for which the derivative is zero. Do not
get too locked into this always happening. Most of the problems that we run into will be like this,
but they won't all be like this.
Let's work another example to make this point.
Example 5 Determine the absolute extrema for the following function and interval.
( ) ( ) [ ]
2
3
3 4 on 5, 1 Q y y y = + − −
Solution
Again, as with all the other examples here, this function is continuous on the given interval and so
we know that this can be done.
The function has an absolute maximum of zero at 4 y = − and the function will have an absolute
minimum of -15 at 5 y = − .
So, if we had ignored or forgotten about the critical point where the derivative doesn't exist
( 4 y = − ) we would not have gotten the correct answer.
In this section we've seen how we can use a derivative to identify the absolute extrema of a
function. This is an important application of derivatives that will arise from time to time so don't
forget about it.
The Shape of a Graph, Part I
In the previous section we saw how to use the derivative to determine the absolute minimum and
maximum values of a function. However, there is a lot more information about a graph that can
be determined from the first derivative of a function. We will start looking at that information in
this section. The main idea we'll be looking at in this section we will be identifying all the
relative extrema of a function.
Let's start this section off by revisiting a familiar topic from the previous chapter. Let's suppose
that we have a function, ( ) f x . We know from our work in the previous chapter that the first
derivative, ( ) f x ′ , is the rate of change of the function. We used this idea to identify where a
function was increasing, decreasing or not changing.
Before reviewing this idea let's first write down the mathematical definition of increasing and
decreasing. We all know what the graph of an increasing/decreasing function looks like but
sometimes it is nice to have a mathematical definition as well. Here it is.
This definition will actually be used in the proof of the next fact in this section.
Now, recall that in the previous chapter we constantly used the idea that if the derivative of a
function was positive at a point then the function was increasing at that point and if the derivative
was negative at a point then the function was decreasing at that point. We also used the fact that
if the derivative of a function was zero at a point then the function was not changing at that point.
We used these ideas to identify the intervals in which a function is increasing and decreasing.
The following fact summarizes up what we were doing in the previous chapter.
Fact
1. If ( ) 0 f x ′ > for every x on some interval I, then ( ) f x is increasing on the interval.
2. If ( ) 0 f x ′ < for every x on some interval I, then ( ) f x is decreasing on the interval.
3. If ( ) 0 f x ′ = for every x on some interval I, then ( ) f x is constant on the interval.
The proof of this fact is in the Proofs From Derivative Applications section of the Extras chapter.
Let's take a look at an example. This example has two purposes. First, it will remind us of the
increasing/decreasing type of problems that we were doing in the previous chapter. Secondly,
and maybe more importantly, it will now incorporate critical points into the solution. We didn't
know about critical points in the previous chapter, but if you go back and look at those examples,
the first step in almost every increasing/decreasing problem is to find the critical points of the
function.
Note that when we factored the derivative we first factored a "-1" out to make the rest of the
factoring a little easier.
From the factored form of the derivative we see that we have three critical points : 2 x = − ,
0 x = , and 4 x = . We'll need these in a bit.
We now need to determine where the derivative is positive and where it's negative. We've done
this several times now in both the Review chapter and the previous chapter. Since the derivative
is a polynomial it is continuous and so we know that the only way for it to change signs is to first
go through zero.
In other words, the only place that the derivative may change signs is at the critical points of the
function. We've now got another use for critical points. So, we'll build a number line, graph the
critical points and pick test points from each region to see if the derivative is positive or negative
in each region.
Make sure that you test your points in the derivative. One of the more common mistakes here is
to test the points in the function instead! Recall that we know that the derivative will be the same
sign in each region. The only place that the derivative can change signs is at the critical points
and we've marked the only critical points on the number line.
So, it looks we've got the following intervals of increase and decrease.
Note that often the fact that only a single point separates the two intervals of increase will be
ignored and the interval will be written 2 4 x − < < .
In this example we used the fact that the only place that a derivative can change sign is at the
critical points. Also, the critical points for this function were those for which the derivative was
zero. However, the same thing can be said for critical points where the derivative doesn't exist.
This is nice to know. A function can change signs where it is zero or doesn't exist. In the
previous chapter all our examples of this type had only critical points where the derivative was
zero. Now, that we know more about critical points we'll also see an example or two later on
with critical points where the derivative doesn't exist.
How that we have the previous "reminder" example out of the way let's move into some new
material. Once we have the intervals of increasing and decreasing for a function we can use this
information to get a sketch of the graph. Note that the sketch, at this point, may not be super
accurate when it comes to the curvature of the graph, but it will at least have the basic shape
correct. To get the curvature of the graph correct we'll need the information from the next
section.
Let's attempt to get a sketch of the graph of the function we used in the previous example.
Note that we are only after a sketch of the graph. As noted before we started this example we
won't be able to accurately predict the curvature of the graph at this point. However, even
without this information we will still be able to get a basic idea of what the graph should look
like.
To get this sketch we start at the very left of the graph and knowing that the graph must be
decreasing and will continue to decrease until we get to 2 x = − . At this point the function will
continue to increase until it gets to 4 x = . However, note that during the increasing phase it does
need to go through the point at 0 x = and at this point we also know that the derivative is zero
here and so the graph goes through 0 x = horizontally. Finally, once we hit 4 x = the graph
starts, and continues, to decrease. Also, note that just like at 0 x = the graph will need to be
horizontal when it goes through the other two critical points as well.
Here is the graph of the function. We, of course, used a graphical program to generate this graph,
however, outside of some potential curvature issues if you followed the increasing/decreasing
information and had all the critical points plotted first you should have something similar to this.
Let's use the sketch from this example to give us a very nice test for classifying critical points as
relative maximums, relative minimums or neither minimums or maximums.
Recall Fermat's Theorem from the Minimum and Maximum Values section. This theorem told
us that all relative extrema (provided the derivative exists at that point of course) of a function
will be critical points. The graph in the previous example has two relative extrema and both
occur at critical points as the Fermat's Theorem predicted. Note as well that we've got a critical
point that isn't a relative extrema ( 0 x = ). This is okay since Fermat's theorem doesn't say that
all critical points will be relative extrema. It only states that relative extrema will be critical
points.
In the sketch of the graph from the previous example we can see that to the left of 2 x = − the
graph is decreasing and to the right of 2 x = − the graph is increasing and 2 x = − is a relative
minimum. In other words, the graph is behaving around the minimum exactly as it would have to
be in order for 2 x = − to be a minimum. The same thing can be said for the relative maximum at
4 x = . The graph is increasing of the left and decreasing on the right exactly as it must be in
order for 4 x = to be a maximum. Finally, the graph is increasing on both sides of 0 x = and so
this critical point can't be a minimum or a maximum.
These ideas can be generalized to arrive at a nice way to test if a critical point is a relative
minimum, relative maximum or neither. If x c = is a critical point and the function is decreasing
to the left of x c = and is increasing to the right then x c = must be a relative minimum of the
function. Likewise, if the function is increasing to the left of x c = and decreasing to the right
then x c = must be a relative maximum of the function. Finally, if the function is increasing on
both sides of x c = or decreasing on both sides of x c = then x c = can be neither a relative
minimum nor a relative maximum.
These idea can be summarized up in the following test.
First Derivative Test
Suppose that x c = is a critical point of ( ) f x then,
1. If ( ) 0 f x ′ > to the left of x c = and ( ) 0 f x ′ < to the right of x c = then x c = is a relative
maximum.
2. If ( ) 0 f x ′ < to the left of x c = and ( ) 0 f x ′ > to the right of x c = then x c = is a relative
minimum.
3. If ( ) f x ′ is the same sign on both sides of x c = then x c = is neither a relative maximum
nor a relative minimum.
It is important to note here that the first derivative test will only classify critical points as relative
extrema and not as absolute extrema. As we recall from the Finding Absolute Extrema section
absolute extrema are largest and smallest function value and may not even exist or be critical
points if they do exist.
The first derivative test is exactly that, a test using the first derivative. It doesn't ever use the
value of the function and so no conclusions can be drawn from the test about the relative "size" of
the function at the critical points (which would be needed to identify absolute extrema) and can't
even begin to address the fact that absolute extrema may not occur at critical points.
Example 3 Find and classify all the critical points of the following function. Give the intervals
where the function is increasing and decreasing.
( )
3 2
4 g t t t = −
Solution
First we'll need the derivative so we can get our hands on the critical points. Note as well that
we'll do some simplification on the derivative to help us find the critical points.
Finding the intervals of increasing and decreasing will also give the classification of the critical
points so let's get those first. Here is a number line with the critical points graphed and test
points.
So, it looks like we've got the following intervals of increasing and decreasing.
From this it looks like 2 t = − and 2 t = are neither relative minimum or relative maximums
since the function is increasing on both side of them. On the other hand,
12
5
t = − is a relative
maximum and
12
5
t = is a relative minimum.
For completeness sake here is the graph of the function.
In the previous example the two critical points where the derivative didn't exist ended up not
being relative extrema. Do not read anything into this. They often will be relative extrema.
Check out this example in the Absolute Extrema to see an example of one such critical point.
Example 5 The population of rabbits (in hundreds) after t years in a certain area is given by the
following function,
( ) ( )
2
ln 3 6 P t t t = +
Determine if the population ever decreases in the first two years.
Solution
So, again we are really after the intervals and increasing and decreasing in the interval [0,2].
We found the only critical point to this function back in the Critical Points section to be,
1
0.202
3
x = =
e
Here is a number line for the intervals of increasing and decreasing.
So, it looks like the population will decrease for a short period and then continue to increase
forever.
Also, while the problem didn't ask for it we can see that the single critical point is a relative
minimum.
In this section we've seen how we can use the first derivative of a function to give us some
information about the shape of a graph and how we can use this information in some applications.
Using the first derivative to give us information about a whether a function is increasing or
decreasing is a very important application of derivatives and arises on a fairly regular basis in
many areas.
The Shape of a Graph, Part II
In the previous section we saw how we could use the first derivative of a function to get some
information about the graph of a function. In this section we are going to look at the information
that the second derivative of a function can give us a about the graph of a function.
Before we do this we will need a couple of definitions out of the way. The main concept that
we'll be discussing in this section is concavity. Concavity is easiest to see with a graph (we'll
give the mathematical definition in a bit).
So a function is concave up if it "opens" up and the function is concave down if it "opens"
down. Notice as well that concavity has nothing to do with increasing or decreasing. A function
can be concave up and either increasing or decreasing. Similarly, a function can be concave
down and either increasing or decreasing.
It's probably not the best way to define concavity by saying which way it "opens" since this is a
somewhat nebulous definition. Here is the mathematical definition of concavity.
Definition 1
Given the function ( ) f x then
1. ( ) f x is concave up on an interval I if all of the tangents to the curve on I are below the
graph of ( ) f x .
2. ( ) f x is concave down on an interval I if all of the tangents to the curve on I are above
the graph of ( ) f x .
So, as you can see, in the two upper graphs all of the tangent lines sketched in are all below the
graph of the function and these are concave up. In the lower two graphs all the tangent lines are
above the graph of the function and these are concave down.
Again, notice that concavity and the increasing/decreasing aspect of the function is completely
separate and do not have anything to do with the other. This is important to note because students
often mix these two up and use information about one to get information about the other.
There's one more definition that we need to get out of the way.
Definition 2
A point x c = is called an inflection point if the function is continuous at the point and the
concavity of the graph changes at that point.
Now that we have all the concavity definitions out of the way we need to bring the second
derivative into the mix. We did after all start off this section saying we were going to be using
the second derivative to get information about the graph. The following fact relates the second
derivative of a function to its concavity. The proof of this fact is in the Proofs From Derivative
Applications section of the Extras chapter.
Fact
Given the function ( ) f x then,
1. If ( ) 0 f x ′′ > for all x in some interval I then ( ) f x is concave up on I.
Notice that this fact tells us that a list of possible inflection points will be those points where the
second derivative is zero or doesn't exist. Be careful however to not make the assumption that
just because the second derivative is zero or doesn't exist that the point will be an inflection point.
We will only know that it is an inflection point once we determine the concavity on both sides of
it. It will only be an inflection point if the concavity is different on both sides of the point.
Now that we know about concavity we can use this information as well as the
increasing/decreasing information from the previous section to get a pretty good idea of what a
graph should look like. Let's take a look at an example of that.
Example 1 For the following function identify the intervals where the function is increasing and
decreasing and the intervals where the function is concave up and concave down. Use this
information to sketch the graph.
( )
5 3
3 5 3 h x x x = − +
Solution
Okay, we are going to need the first two derivatives so let's get those first.
and decreasing. This only difference is that we will be using the second derivative instead of the
first derivative.
The first thing that we need to do is identify the possible inflection points. These will be where
the second derivative is zero or doesn't exist. The second derivative in this case is a polynomial
and so will exist everywhere. It will be zero at the following points.
1
0, 0.7071
2
x x = = ± = ±
As with the increasing and decreasing part we can draw a number line up and use these points to
divide the number line into regions. In these regions we know that the second derivative will
always have the same value since these three points are the only places where the function may
change sign. Therefore, all that we need to do is pick a point from each region and plug it into
the second derivative. The second derivative will then have that sign in the whole region from
which the point came from
Using all this information to sketch the graph gives the following graph.
We can use the previous example to get illustrate another way to classify some of the critical
points of a function as relative maximums or relative minimums.
Notice that 1 x = − is a relative maximum and that the function is concave down at this point.
This means that ( ) 1 f ′′ − must be negative. Likewise, 1 x = is a relative minimum and the
function is concave up at this point. This means that ( ) 1 f ′′ must be positive.
As we'll see in a bit we will need to be very careful with 0 x = . In this case the second
derivative is zero, but that will not actually mean that 0 x = is not a relative minimum or
maximum. We'll see some examples of this in a bit, but we need to get some other information
taken care of first.
It is also important to note here that all of the critical points in this example were critical points in
which the first derivative were zero and this is required for this to work. We will not be able to
use this test on critical points where the derivative doesn't exist.
Here is the test that can be used to classify some of the critical points of a function. The proof of
this test is in the Proofs From Derivative Applications section of the Extras chapter.
The third part of the second derivative test is important to notice. If the second derivative is zero
then the critical point can be anything. Below are the graphs of three functions all of which have
a critical point at 0 x = , the second derivative of all of the functions is zero at 0 x = and yet all
three possibilities are exhibited.
The first is the graph of ( )
4
f x x = . This graph has a relative minimum at 0 x = .
Next is the graph of ( )
4
f x x = − which has a relative maximum at 0 x = .
So, we can see that we have to be careful if we fall into the third case. For those times when we
do fall into this case we will have to resort to other methods of classifying the critical point. This
is usually done with the first derivative test.
Let's go back and relook at the critical points from the first example and use the Second
Derivative Test on them, if possible.
The three critical points ( 1 x = − , 0 x = , and 1 x = ) of this function are all critical points where
the first derivative is zero so we know that we at least have a chance that the Second Derivative
Test will work. The value of the second derivative for each of these are,
( ) ( ) ( ) 1 30 0 0 1 30 h h h ′′ ′′ ′′ − = − = =
The second derivative at 1 x = − is negative so by the Second Derivative Test this critical point
this is a relative maximum as we saw in the first example. The second derivative at 1 x = is
positive and so we have a relative minimum here by the Second Derivative Test as we also saw in
the first example.
In the case of 0 x = the second derivative is zero and so we can't use the Second Derivative Test
to classify this critical point. Note however, that we do know from the First Derivative Test we
used in the first example that in this case the critical point is not a relative extrema.
Example 3 For the following function find the inflection points and use the second derivative
test, if possible, to classify the critical points. Also, determine the intervals of increase/decrease
and the intervals of concave up/concave down and sketch the graph of the function.
( ) ( )
2
3
6 f t t t = −
18
3.6 6
5
t t = = =
Notice as well that we won't be able to use the second derivative test on 6 t = to classify this
critical point since the derivative doesn't exist at this point. To classify this we'll need the
increasing/decreasing information that we'll get to sketch the graph.
We can however, use the Second Derivative Test to classify the other critical point so let's do that
before we proceed with the sketching work. Here is the value of the second derivative at 3.6 t = .
( ) 3.6 1.245 0 f ′′ = − <
So, according to the second derivative test 3.6 t = is a relative maximum.
Now let's proceed with the work to get the sketch of the graph and notice that once we have the
increasing/decreasing information we'll be able to classify 6 t = .
Here is the number line for the first derivative.
So, according to the first derivative test we can verify that 3.6 t = is in fact a relative maximum.
We can also see that 6 t = is a relative minimum.
we have the test that just because we can't use the Second Derivative Test or the Test doesn't tell
us anything about a critical point doesn't mean that the critical point will not be a relative
extrema. This is a common mistake that many students make so be careful when using the
Second Derivative Test.
Okay, let's finish the problem out. We will need the list of possible inflection points. These are,
72
6 7.2
10
t t = = =
Here is the number line for the second derivative. Note that we will need this to see if the two
points above are in fact inflection points.
So, the concavity only changes at 7.2 t = and so this is the only inflection point for this function.
Here is the sketch of the graph.
The change of concavity at 7.2 t = is hard to see, but it is there it's just a very subtle change in
concavity.
The Mean Value Theorem
In this section we want to take a look at the Mean Value Theorem. In most traditional textbooks
this section comes before the sections containing the First and Second Derivative Tests because
the many of the proofs in those sections need the Mean Value Theorem. However, we feel that
from a logical point of view it's better to put the Shape of a Graph sections right after the absolute
extrema section. So, if you've been following the proofs from the previous two sections you've
probably already read through this section.
Before we get to the Mean Value Theorem we need to cover the following theorem.
Solution
From basic Algebra principles we know that since ( ) f x is a 5
th
degree polynomial there it will
have five roots. What we're being asked to prove here is that only one of those 5 is a real number
and the other 4 must be complex roots.
First, we should show that it does have at least one real root. To do this note that ( ) 0 2 f = − and
that ( ) 1 10 f = and so we can see that ( ) ( ) 0 0 1 f f < < . Now, because ( ) f x is a polynomial
we know that it is continuous everywhere and so by the Intermediate Value Theorem there is a
number c such that 0 1 c < < and ( ) 0 f c = . In other words ( ) f x has at least one real root.
is called contradiction proof. What we'll do is assume that ( ) f x has at least two real roots.
This means that we can find real numbers a and b (there might be more, but all we need for this
particular argument is two) such that ( ) ( ) 0 f a f b = = . But if we do this then we know from
Rolle's Theorem that there must then be another number c such that ( ) 0 f c ′ = .
This is a problem however. The derivative of this function is,
( )
4 2
20 3 7 f x x x ′ = + +
Because the exponents on the first two terms are even we know that the first two terms will
always be greater than or equal to zero and we are then going to add a positive number onto that
and so we can see that the smallest the derivative will ever be is 7 and this contradicts the
statement above that says we MUST have a number c such that ( ) 0 f c ′ = .
We reached these contradictory statements by assuming that ( ) f x has at least two roots. Since
this assumption leads to a contradiction the assumption must be false and so we can only have a
single real root.
The reason for covering Rolle's Theorem is that it is needed in the proof of the Mean Value
Theorem. To see the proof see the Proofs From Derivative Applications section of the Extras
chapter. Here is the theorem.
Mean Value Theorem
Suppose ( ) f x is a function that satisfies both of the following.
1. ( ) f x is continuous on the closed interval [a,b].
2. ( ) f x is differentiable on the open interval (a,b).
Then there is a number c such that a < c < b and
( )
( ) ( ) f b f a
f c
b a
−
′ =
−
Or,
( ) ( ) ( )( ) f b f a f c b a ′ − = −
Note that the Mean Value Theorem doesn't tell us what c is. It only tells us that there is at least
one number c that will satisfy the conclusion of the theorem.
Also note that if it weren't for the fact that we needed Rolle's Theorem to prove this we could
think of Rolle's Theorem as a special case of the Mean Value Theorem. To see that just assume
that ( ) ( ) f a f b = and then the result of the Mean Value Theorem gives the result of Rolle's
Theorem.
Before we take a look at a couple of examples let's think about a geometric interpretation of the
Mean Value Theorem. First define ( ) ( )
, A a f a = and ( ) ( )
, B b f b = and then we know from
the Mean Value theorem that there is a c such that a c b < < and that
( )
( ) ( ) f b f a
f c
b a
−
′ =
−
Now, if we draw in the secant line connecting A and B then we can know that the slope of the
secant line is,
( ) ( ) f b f a
b a
−
−
Likewise, if we draw in the tangent line to ( ) f x at x c = we know that its slope is ( ) f c ′ .
What the Mean Value Theorem tells us is that these two slopes must be equal or in other words
the secant line connecting A and B and the tangent line at x c = must be parallel. We can see this
in the following sketch.
Let's now take a look at a couple of examples using the Mean Value Theorem.
Example 2 Determine all the numbers c which satisfy the conclusions of the Mean Value
Theorem for the following function.
( ) [ ]
3 2
2 on 1, 2 f x x x x = + − −
Solution
There isn't really a whole lot to this problem other than to notice that since ( ) f x is a polynomial
it is both continuous and differentiable (i.e. the derivative exists) on the interval given.
Notice that only one of these is actually in the interval given in the problem. That means that we
will exclude the second one (since it isn't in the interval). The number that we're after in this
problem is,
0.7863 c =
Be careful to not assume that only one of the numbers will work. It is possible for both of them
to work.
Example 3 Suppose that we know that ( ) f x is continuous and differentiable on [6, 15]. Let's
also suppose that we know that ( ) 6 2 f = − and that we know that ( ) 10 f x ′ ≤ . What is the
largest possible value for ( ) 15 f ?
Now we know that ( ) 10 f x ′ ≤ so in particular we know that ( ) 10 f c ′ ≤ . This gives us the
following,
( ) ( )
( )
15 2 9
2 9 10
88
f f c ′ = − +
≤ − +
=
All we did was replace ( ) f c ′ with its largest possible value.
This means that the largest possible value for ( ) 15 f is 88.
Example 4 Suppose that we know that ( ) f x is continuous and differentiable everywhere.
Let's also suppose that we know that ( ) f x has two roots. Show that ( ) f x ′ must have at least
one root.
Solution
It is important to note here that all we can say is that ( ) f x ′ will have at least one root. We can't
say that it will have exactly one root. So don't confuse this problem with the first one we
worked.
This is actually a fairly simple thing to prove. Since we know that ( ) f x has two roots let's
suppose that they are a and b. Now, by assumption we know that ( ) f x is continuous and
differentiable everywhere and so in particular it is continuous on [a,b] and differentiable on (a,b).
Therefore, by the Mean Value Theorem there is a number c that is between a and b (this isn't
needed for this problem, but it's true so it should be pointed out) and that,
( )
( ) ( ) f b f a
f c
b a
−
′ =
−
But we now need to recall that a and b are roots of ( ) f x and so this is,
( )
0 0
0 f c
b a
−
′ = =
−
It is completely possible to generalize the previous example significantly. For instance if we
know that ( ) f x is continuous and differentiable everywhere and has three roots we can then
show that not only will ( ) f x ′ have at least two roots but that ( ) f x ′′ will have at least one root.
We'll leave it to you to verify this, but the ideas involved are identical to those in the previous
example.
We'll close this section out with a couple of nice facts that can be proved using the Mean Value
Theorem. Note that in both of these facts we are assuming the functions are continuous and
differentiable on the interval [a,b].
This fact is very easy to prove so let's do that here. Take any two x's in the interval ( ) , a b , say
1
x and
2
x . Then since ( ) f x is continuous and differential on [a,b] it must also be continuous
and differentiable on
[ ]
1 2
, x x . This means that we can apply the Mean Value Theorem for these
two values of x. Doing this gives,
( ) ( ) ( )( )
2 1 2 1
f x f x f c x x ′ − = −
where
1 2
x c x < < . But by assumption ( ) 0 f x ′ = for all x in an interval ( ) , a b and so in
particular we must have,
( ) 0 f c ′ =
Now, since
1
x and
2
x where any two values of x in the interval ( ) , a b we can see that we must
have ( ) ( )
2 1
f x f x = for all
1
x and
2
x in the interval and this is exactly what it means for a
function to be constant on the interval and so we've proven the fact.
However, by assumption ( ) ( ) f x g x ′ ′ = for all x in an interval ( ) , a b and so we must have that
( ) 0 h x ′ = for all x in an interval ( ) , a b . So, by Fact 1 ( ) h x must be constant on the interval.
Optimization
In this section we are going to look at optimization problems. In optimization problems we are
looking for the largest value or the smallest value that a function can take. We saw how to one
kind of optimization problem in the Absolute Extrema section where we found the largest and
smallest value that a function would take on an interval.
In this section we are going to look at another type of optimization problem. Here we will be
looking for the largest or smallest value of a function subject to some kind of constraint. The
constraint will be some condition (that can usually be described by some equation) that must
absolutely, positively be true no matter what our solution is. On occasion, the constraint will not
be easily described by an equation, but in these problems it will be easy to deal with as we'll see.
This section is generally one of the more difficult for students taking a Calculus course. One of
the main reasons for this is that a subtle change of wording can completely change the problem.
There is also the problem of identifying the quantity that we'll be optimizing and the quantity that
is the constraint and writing down equations for each.
The first step in all of these problems should be to very carefully read the problem. Once you've
done that the next step is to identify the quantity to be optimized and the constraint.
In identifying the constraint remember that the constraint is something that must true regardless
of the solution. In almost every one of the problems we'll be looking at here one quantity will be
clearly indicated as having a fixed value and so must be the constraint. Once you've got that
identified the quantity to be optimized should be fairly simple to get. It is however easy to
confuse the two if you just skim the problem so make sure you carefully read the problem first!
Let's start the section off with a simple problem to illustrate the kinds of issues will be dealing
with here.
Example 1 We need to enclose a field with a fence. We have 500 feet of fencing material and a
building is on one side of the field and so won't need any fencing. Determine the dimensions of
the field that will enclose the largest area.
Solution
In all of these problems we will have two functions. The first is the function that we are actually
trying to optimize and the second will be the constraint. Sketching the situation will often help
us to arrive at these equations so let's do that.
fencing material. So, the area will be the function we are trying to optimize and the amount of
fencing is the constraint. The two equations for these are,
Maximize :
Contraint : 500 2
A xy
x y
=
= +
Okay, we know how to find the largest or smallest value of a function provided it's only got a
single variable. The area function (as well as the constraint) has two variables in it and so what
we know about finding absolute extrema won't work. However, if we solve the constraint for one
of the two variables we can substitute this into the area and we will then have a function of a
single variable.
So, let's solve the constraint for x. Note that we could have just as easily solved for y but that
would have led to fractions and so, in this case, solving for x will probably be best.
500 2 x y = −
Now we want to find the largest value this will have on the interval [0,250]. Note that the
interval corresponds to taking 0 y = (i.e. no sides to the fence) and 250 y = (i.e. only two sides
and no width, also if there are two sides each must be 250 ft to use the whole 500ft…).
Note that the endpoints of the interval won't make any sense from a physical standpoint if we
actually want to enclose some area because they would both give zero area. They do, however,
give us a set of limits on y and so the Extreme Value Theorem tells us that we will have a
maximum value of the area somewhere between the two endpoints. Having these limits will also
mean that we can use the process we discussed in the Finding Absolute Extrema section earlier
in the chapter to find the maximum value of the area.
So, recall that the maximum value of a continuous function (which we've got here) on a closed
interval (which we also have here) will occur at critical points and/or end points. As we've
already pointed out the end points in this case will give zero area and so don't make any sense.
That means our only option will be the critical points.
So let's get the derivative and find the critical points.
( ) 500 4 A y y ′ = −
Setting this equal to zero and solving gives a lone critical point of 125 y = . Plugging this into
the area gives an area of 31250 ft
2
. So according to the method from Absolute Extrema section
this must be the largest possible area, since the area at either endpoint is zero.
Finally, let's not forget to get the value of x and then we'll have the dimensions since this is what
the problem statement asked for. We can get the x by plugging in our y into the constraint.
( ) 500 2 125 250 x = − =
The dimensions of the field that will give the largest area, subject to the fact that we used exactly
500 ft of fencing material, are 250 x 125.
Don't forget to actually read the problem and give the answer that was asked for. These types of
problems can take a fair amount of time/effort to solve and it's not hard to sometimes forget what
the problem was actually asking for.
In the previous problem we used the method from the Finding Absolute Extrema section to find
the maximum value of the function we wanted to optimize. However, as we'll see in later
examples we won't always have easy to find endpoints and/or dealing with the endpoints may not
be easy to deal with. Not only that, but this method requires that the function we're optimizing be
continuous on the interval we're looking at, including the endpoints, and that may not always be
the case.
So, before proceeding with the anymore examples let's spend a little time discussing some
methods for determining if our solution is in fact the absolute minimum/maximum value that
we're looking for. In some examples all of these will work while in others one or more won't be
all that useful. However, we will always need to use some method for making sure that our
answer is in fact that optimal value that we're after.
Method 1 : Use the method used in Finding Absolute Extrema.
This is the method used in the first example above. Recall that in order to use this method the
range of possible optimal values, let's call it I, must have finite endpoints. Also, the function
we're optimizing (once it's down to a single variable) must be continuous on I, including the
endpoints. If these conditions are met then we know that the optimal value, either the maximum
or minimum depending on the problem, will occur at either the endpoints of the range or at a
critical point that is inside the range of possible solutions.
There are two main issues that will often prevent this method from being used however. First,
not every problem will actually have a range of possible solutions that have finite endpoints at
both ends. We'll see at least one example of this as we work through the remaining examples.
Also, many of the functions we'll be optimizing will not be continuous once we reduce them
down to a single variable and this will prevent us from using this method.
continuous at the end points if the endpoint is finite and the function actually exists at the
endpoint. We'll see several problems where the function we're optimizing doesn't actually exist
at one of the endpoints. This will not prevent this method from being used.
Let's suppose that x c = is a critical point of the function we're trying to optimize, ( ) f x . We
already know from the First Derivative Test that if ( ) 0 f x ′ > immediately to the left of x c =
(i.e. the function is increasing immediately to the left) and if ( ) 0 f x ′ < immediately to the right
of x c = (i.e. the function is decreasing immediately to the right) then x c = will be a relative
maximum for ( ) f x .
Now, this does not mean that the absolute maximum of ( ) f x will occur at x c = . However,
suppose that we knew a little bit more information. Suppose that in fact we knew that ( ) 0 f x ′ >
for all x in I such that x c < . Likewise, suppose that we knew that ( ) 0 f x ′ < for all x in I such
that x c > . In this case we know that to the left of x c = , provided we stay in I of course, the
function is always increasing and to the right of x c = , again staying in I, we are always
decreasing. In this case we can say that the absolute maximum of ( ) f x in I will occur at x c = .
Similarly, if we know that to the left of x c = the function is always decreasing and to the right
of x c = the function is always increasing then the absolute minimum of ( ) f x in I will occur at
x c = .
Before we give a summary of this method let's discuss the continuity requirement a little.
Nowhere in the above discussion did the continuity requirement apparently come into play. We
require that that the function we're optimizing to be continuous in I to prevent the following
situation.
In this case, a relative maximum of the function clearly occurs at x c = . Also, the function is
always decreasing to the right and is always increasing to the left. However, because of the
discontinuity at x d = , we can clearly see that ( ) ( ) f d f c > and so the absolute maximum of
the function does not occur at x c = . Had the discontinuity at x d = not been there this would
not have happened and the absolute maximum would have occurred at x c = .
Here is a summary of this method.
First Derivative Test for Absolute Extrema
Let I be the interval of all possible optimal values of ( ) f x and further suppose that ( ) f x is
continuous on I , except possibly at the endpoints. Finally suppose that x c = is a critical point
of ( ) f x and that c is in the interval I. If we restrict x to values from I (i.e. we only consider
possible optimal values of the function) then,
There are actually two ways to use the second derivative to help us identify the optimal value of a
function and both use the Second Derivative Test to one extent or another.
The first way to use the second derivative doesn't actually help us to identify the optimal value.
What it does do is allow us to potentially exclude values and knowing this can simplify our work
somewhat and so is not a bad thing to do.
Suppose that we are looking for the absolute maximum of a function and after finding the critical
points we find that we have multiple critical points. Let's also suppose that we run all of them
through the second derivative test and determine that some of them are in fact relative minimums
of the function. Since we are after the absolute maximum we know that a maximum (of any
kind) can't occur at relative minimums and so we immediately know that we can exclude these
points from further consideration. We could do a similar check if we were looking for the
absolute minimum. Doing this may not seem like all that great of a thing to do, but it can, on
occasion, lead to a nice reduction in the amount of work that we need to do in later steps.
values, I and the endpoint(s) may or may not be finite. We'll also need to require that the
function, ( ) f x be continuous everywhere in I except possibly at the endpoints as above.
Now, suppose that x c = is a critical point and that ( ) 0 f c ′′ > . The second derivative test tells
us that x c = must be a relative minimum of the function. Suppose however that we also knew
that ( ) 0 f x ′′ > for all x in I. In this case we would know that the function was concave up in all
of I and that would in turn mean that the absolute minimum of ( ) f x in I would in fact have to
be at x c = .
Likewise if x c = is a critical point and ( ) 0 f x ′′ < for all x in I then we would know that the
function was concave down in I and that the absolute maximum of ( ) f x in I would have to be
at x c = .
Here is a summary of this method.
Second Derivative Test for Absolute Extrema
Let I be the range of all possible optimal values of ( ) f x and further suppose that ( ) f x is
continuous on I , except possibly at the endpoints. Finally suppose that x c = is a critical point
of ( ) f x and that c is in the interval I. Then,
1. If ( ) 0 f x ′′ > for all x in I then ( ) f c will be the absolute minimum value of ( ) f x on
the interval I.
2. If ( ) 0 f x ′′ < for all x in I then ( ) f c will be the absolute maximum value of ( ) f x on
the interval I.
Before proceeding with some more examples we need to once again acknowledge that not every
method discussed above will work for every problem and that, in some problems, more than one
method will work. There are also problems were we may need to use a combination of these
methods to identify the optimal value. Each problem will be different and we'll need to see what
we've got once we get the critical points before we decide which method might be best to use.
Okay, let's work some more examples.
Example 2 We want to construct a box whose base length is 3 times the base width. The
material used to build the top and bottom cost $10/ft
2
and the material used to build the sides cost
$6/ft
2
. If the box must have a volume of 50ft
3
determine the dimensions that will minimize the
cost to build the box.
We want to minimize the cost of the materials subject to the constraint that the volume must be
50ft
3
. Note as well that the cost for each side is just the area of that side times the appropriate
cost.
So, it looks like we've got two critical points here. The first is obvious, 0 w = , and it's also just
as obvious that this will not be the correct value. We are building a box here and w is the box's
width and so since it makes no sense to talk about a box with zero width we will ignore this
critical point. This does not mean however that you should just get into the habit of ignoring zero
when it occurs. There are other types of problems where it will be a valid point that we will need
to consider.
The next critical point will come from determining where the numerator is zero.
So, once we throw out 0 w = , we've got a single critical point and we now have to verify that
this is in fact the value that will give the absolute minimum cost.
In this case we can't use Method 1 from above. First, the function is not continuous at one of the
endpoints, 0 w = , of our interval of possible values. Secondly, there is no theoretical upper limit
to the width that will give a box with volume of 50 ft
3
. If w is very large then we would just need
to make h very small.
The second method listed above would work here, but that's going to involve some calculations,
not difficult calculations, but more work nonetheless.
The third method however, will work quickly and simply here. First, we know that whatever the
value of w that we get it will have to be positive and we can see second derivative above that
provided 0 w > we will have ( ) 0 C w ′′ > and so in the interval of possible optimal values the
cost function will always be concave up and so 1.8821 w = must give the absolute minimum
cost.
Also, even though it was not asked for, the minimum cost is : ( ) 1.8821 $637.60 C = .
Example 3 We want to construct a box with a square base and we only have 10 m
2
of material
to use in construction of the box. Assuming that all the material is used in the construction
process determine the maximum volume that the box can have.
Solution
This example is in many ways the exact opposite of the previous example. In this case we want
to optimize the volume and the constraint this time is the amount of material used. We don't
have a cost here, but if you think about it the cost is nothing more than the amount of material
used times a cost and so the amount of material and cost are pretty much tied together. If you can
do one you can do the other as well. Note as well that the amount of material used is really just
the surface area of the box.
Note as well here that provided 0 w > , which we know from a physical standpoint will be true,
then the volume function will be concave down and so if we get a single critical point then we
know that it will have to be the value that gives the absolute maximum.
Setting the first derivative equal to zero and solving gives us the two critical points,
5
1.2910
3
w = ± = ±
In this case we can exclude the negative critical point since we are dealing with a length of a box
and we know that these must be positive. Do not however get into the habit of just excluding any
negative critical point. There are problems where negative critical points are perfectly valid
possible solutions.
Now, as noted above we got a single critical point, 1.2910, and so this must be the value that
gives the maximum volume and since the maximum volume is all that was asked for in the
problem statement the answer is then : ( )
3
1.2910 2.1517m V = .
1.2910 w > then ( ) 0 V w ′ < and so if we are to the left of the critical point the volume is always
increasing and if we are to the right of the critical point the volume is always decreasing and so
by the Method 2 above we can also see that the single critical point must give the absolute
maximum of the volume.
In the last two examples we've seen that many of these optimization problems can in both
directions so to speak. In both examples we have essentially the same two equations: volume and
surface area. However, in Example 2 the volume was the constraint and the cost (which is
directly related to the surface area) was the function we were trying to optimize. In Example 3,
on the other hand, we were trying to optimize the volume and the surface area was the constraint.
It is important to not get so locked into one way of doing these problems that we can't do it in the
opposite direction as needed as well. This is one of the more common mistakes that students
make with these kinds of problems. They see one problem and then try to make every other
problem that seems to be the same conform to that one solution even if the problem needs to be
worked differently. Keep an open mind with these problems and make sure that you understand
what is being optimized and what the constraint is before you jump into the solution.
Also, as seen in the last example we used two different methods of verifying that we did get the
optimal value. Do not get too locked into one method of doing this verification that you forget
about the other methods.
Let's work some another example that this time doesn't involve a rectangle or box.
Example 4 A manufacturer needs to make a cylindrical can that will hold 1.5 liters of liquid.
Determine the dimensions of the can that will minimize the amount of material used in its
construction.
We'll need the surface area of this can and that will be the surface area of the walls of the can
(which is really just a cylinder) and the area of the top and bottom caps (which are just disks, and
don't forget that there are two of them).
Note that if you think of a cylinder of height h and radius r as just a bunch of disks/circles of
radius r stacked on top of each other the equations for the surface area and volume are pretty
simple to remember. The volume is just the area of each of the disks times the height. Similarly,
the surface area is just the circumference of the each circle times the. The equations for the
volume and surface area of a cylinder are then,
( )( ) ( )( )
2 2
2 2 V r h r h A r h rh π π π π = = = =
Next, we're also going to need the required volume in a better set of units than liters. Since we
want length measurements for the radius and height we'll need to use the fact that 1 Liter = 1000
cm
3
to convert the 1.5 liters into 1500 cm
3
. This will in turn give a radius and height in terms of
centimeters.
Here are the equations that we'll need for this problem and don't forget that there two caps and so
we'll need the area from each.
From this we can see that we have two critical points : 0 r = and
750
3
6.2035 r
π
= = . The first
critical point doesn't make sense from a physical standpoint and so we can ignore that one.
So we only have a single critical point to deal with here and notice that 6.2035 is the only value
for which the derivative will be zero and hence the only place (with 0 r > of course) that the
derivative may change sign. It's not difficult to check that if 6.2035 r < then ( ) 0 A r ′ < and
likewise if 6.2035 r > then ( ) 0 A r ′ > . The variant of the First Derivative Test above then tells
us that the absolute minimum value of the area (for 0 r > ) must occur at 6.2035 r = .
All we need to do this is determine height of the can and we'll be done.
( )
2
1500
12.4070
6.2035
h
π
= =
Therefore if the manufacturer makes the can with a radius of 6.2035 cm and a height of 12.4070
cm the least amount of material will be used to make the can.
As an interesting side problem and extension to the above example you might want to show that
for a given volume, L, the minimum material will be used if 2 h r = regardless of the volume of
the can.
In the examples to this point we've put in quite a bit of discussion in the solution. In the
remaining problems we won't be putting in quite as much discussion and leave it to you to fill in
any missing details.
Example 5 We have a piece of cardboard that is 14 inches by 10 inches and we're going to cut
out the corners as shown below and fold up the sides to form a box, also shown below.
Determine the height of the box that will give a maximum volume.
Setting the first derivative equal to zero and solving gives the following two critical points,
12 39
1.9183, 6.0817
3
h
±
= =
We now have an apparent problem. We have two critical points and we'll need to determine
which one is the value we need. In this case, this is easier than it looks. Go back to the figure in
the problem statement and notice that we can quite easily find limits on h. The smallest h can be
is 0 h = even though this doesn't make much sense as we won't get a box in this case. Also
from the 10 inch side we can see that the largest h can be is 5 h = although again, this doesn't
make much sense physically.
So, knowing that whatever h is it must be in the range 0 5 h ≤ ≤ we can see that the second
critical point is outside this range and so the only critical point that we need to worry about is
1.9183.
Finally, since the volume is defined and continuous on 0 5 h ≤ ≤ all we need to do is plug in the
critical points and endpoints into the volume to determine which gives the largest volume. Here
are those function evaluations.
( ) ( ) ( ) 0 0 1.9183 120.1644 5 0 V V V = = =
So, if we take 1.9183 h = we get a maximum volume.
Example 6 A printer need to make a poster that will have a total area of 200 in
2
and will have 1
inch margins on the sides, a 2 inch margin on the top and a 1.5 inch margin on the bottom. What
dimensions will give the largest printed area?
Solution
This problem is a little different from the previous problems. Both the constraint and the function
we are going to optimize are areas. The constraint is that the overall area of the poster must be
200 in
2
while we want to optimize the printed area (i.e. the area of the poster with the margins
taken out).
Here is a sketch of the poster and we can see that once we've taken the margins into account the
width of the printed area is 2 w− and the height of the printer area is 3.5 h − .
The first and second derivatives are,
( ) ( )
2
2 2 3
400 400 3.5 800
3.5
w
A w A w
w w w
−
′ ′′ = − + = = −
From the first derivative we have the following three critical points.
400
3.5
0 10.6904 w w = = ± = ±
However, since we're dealing with the dimensions of a piece of paper we know that we must
have 0 w > and so only 10.6904 will make sense.
Also notice that provided 0 w > the second derivative will always be negative and so in the
range of possible optimal values of the width the area function is always concave down and so we
know that the maximum printed area will be at 10.6904 inches w = .
The height of the paper that gives the maximum printed area is then,
200
18.7084 inches
10.6904
h = =
We've worked quite a few examples to this point and we have quite a few more to work.
However this section has gotten quite lengthy so let's continue our examples in the next section.
This is being done mostly because these notes are also being presented on the web and this will
help to keep the load times on the pages down somewhat.
More Optimization Problems
Because these notes are also being presented on the web we've broken the optimization examples
up into several sections to keep the load times to a minimum. Do not forget the various methods
for verifying that we have the optimal value that we looked at in the previous section. In this
section we'll just use them without acknowledging so make sure you understand them and can
use them. So let's get going on some more examples.
Example 1 A window is being built and the bottom is a rectangle and the top is a semicircle. If
there is 12 meters of framing materials what must the dimensions of the window be to let in the
most light?
Solution
Okay, let's ask this question again is slightly easier to understand terms. We want a window in
the shape described above to have a maximum area (and hence let in the most light) and have a
perimeter of 12 m (because we have 12 m of framing material). Little bit easier to understand in
those terms.
Here's a sketch of the window. The height of the rectangular portion is h and because the
semicircle is on top we can think of the width of the rectangular portion at 2r.
The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half
the circumference of a circle of radius r. The area (what we want to maximize) is the area of the
rectangle plus half the area of a circle of radius r. Here are the equations we'll be working with
in this example.
We can also see that the second derivative is always negative (in fact it's a constant) and so we
can see that the maximum area must occur at this point. So, for the maximum area the semicircle
on top must have a radius of 1.6803 and the rectangle must have the dimensions 3.3606 x 1.6803
(h x 2r).
Example 2 Determine the area of the largest rectangle that can be inscribed in a circle of radius
4.
Solution
Huh? This problem is best described with a sketch. Here is what we're looking for.
We want the area of the largest rectangle that we can fit inside a circle and have all of its corners
touching the circle.
To do this problem it's easiest to assume that the circle (and hence the rectangle) is centered at
the origin. Doing this we know that the equation of the circle will be
2 2
16 x y + =
and that the right upper corner of the rectangle will have the coordinates ( ) , x y . This means that
the width of the rectangle will be 2x and the height of the rectangle will be 2y. The area of the
rectangle will then be,
( )( ) 2 2 4 A x y xy = =
So, we've got the function we want to maximize (the area), but what is the constraint? Well since
the coordinates of the upper right corner must be on the circle we know that x and y must satisfy
the equation of the circle. In other words, the equation of the circle is the constraint.
The first thing to do then is to solve the constraint for one of the variables.
2
16 y x = ± −
Since the point that we're looking at is in the first quadrant we know that y must be positive and
so we can take the "+" part of this. Plugging this into the area and computing the first derivative
gives,
Before getting the critical points let's notice that we can limit x to the range 0 4 x ≤ ≤ since we
are assuming that x is in the first quadrant and must stay inside the circle. Now the four critical
points we get (two from the numerator and two from the denominator) are,
2
2
16 0 4
64 8 0 2 2
x x
x x
− = ⇒ = ±
− = ⇒ = ±
We only want critical points that are in the range of possible optimal values so that means that we
have two critical points to deal with : 2 2 x = and 4 x = . Notice however that the second
critical point is also one of the endpoints of our interval.
Now, area function is continuous and we have an interval of possible solution with finite
endpoints so,
( )
( )
( ) 0 0 2 2 32 4 0 A A A = = =
So, we can see that we'll get the maximum area if 2 2 x = and the corresponding value of y is,
( )
2
16 2 2 8 2 2 y = − = =
It looks like the maximum area will be found if the inscribed rectangle is in fact a square.
So, we're looking for the shortest length of the dashed line. Notice as well that if the shortest
distance isn't at 0 x = there will be two points on the graph, as we've shown above, that will give
the shortest distance. This is because the parabola is symmetric to the y-axis and the point in
question is on the y-axis. This won't always be the case of course so don't always expect two
points in these kinds of problems.
In this case we need to minimize the distance between the point (0,2) and any point that is one the
graph (x,y). Or,
( ) ( ) ( )
2 2 2
2
0 2 2 d x y x y = − + − = + −
If you think about the situation here it makes sense that the point that minimizes the distance will
also minimize the square of the distance and so since it will be easier to work with we will use the
square of the distance and minimize that. If you aren't convinced of this we'll take a closer look
at this after this problem. So, the function that we're going to minimize is,
( )
2
2 2
2 D d x y = = + −
The constraint in this case is the function itself since the point must lie on the graph of the
function.
Solution 1
In this case we will use the constraint in probably the most obvious way. We already have the
constraint solved for y so let's plug that into the square of the distance and get the derivatives.
So, it looks like there are three critical points for the square of the distance and notice that this
time, unlike pretty much every previous example we've worked, we can't exclude zero or
negative numbers. They are perfectly valid possible optimal values this time.
So, 0 x = is a relative maximum and so can't possibly be the minimum distance. That means
that we've got two critical points. The question is how do we verify that these give the minimum
distance and yes we did mean to say that both will give the minimum distance. Recall from our
sketch above that if x gives the minimum distance then so will –x and so if gives the minimum
distance then the other should as well.
None of the methods we discussed in the previous section will really work here. We don't have
an interval of possible solutions with finite endpoints and both the first and second derivative
change sign. In this case however, we can still verify that they are the points that give the
minimum distance.
First, notice that if we are working on the interval
1 1
2 2
,
⎡ ⎤
−
⎣ ⎦
then the endpoints of this interval
(which are also the critical points) are in fact where the absolute minimum of the function occurs
in this interval.
Next we can see that if
1
2
x < − then ( ) 0 D x ′ < . Or in other words, if
1
2
x < − the function is
decreasing until it hits
1
2
x = − and so must always be larger than the function at
1
2
x = − .
So, putting all of this together tells us that we do in fact have an absolute minimum at
1
2
x = ± .
All that we need to do is to find the value of y for these points.
1 3
:
2 2
1 3
:
2 2
x y
x y
= =
= − =
So, the points on the graph that are closest to (0,2) are,
1 3 1 3
, ,
2 2 2 2
⎛ ⎞ ⎛ ⎞
−
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Solution 2
The first solution that we worked was actually the long solution. There is a much shorter solution
to this problem. Instead of plugging y into the square of the distance let's plug in x. From the
constraint we get,
2
1 x y = −
and notice that the only place x show up in the square of the distance it shows up as x
2
and let's
just plug this into the square of the distance. Doing this gives,
There is now a single critical point,
3
2
y = , and since the second derivative is always positive we
know that this point must give the absolute minimum. So all that we need to do at this point is
find the value(s) of x that go with this value of y.
examples. Again, be careful to not get into the habit of always excluding them as we do many of
the examples we'll work.
Next, some of these problems will have multiple solution methods and sometimes one will be
significantly easier than the other. The method you use is up to you and often the difficulty of
any particular method is dependent upon the person doing the problem. One person may find one
way easier and other person may find a different method easier.
Finally, as we saw in the first solution method sometimes we'll need to use a combination of the
optimal value verification methods we discussed in the previous section.
Now, before we move onto the next example let's take a look at the claim above that we could
find the location of the point that minimizes the distance by finding the point that minimizes the
square of the distance. We'll generalize things a little bit,
Fact
Suppose that we have a positive function, ( ) 0 f x > , that exists everywhere then ( ) f x and
( ) ( ) g x f x = will have the same critical points and the relative extrema will occur at the same
points.
By assumption we know that ( ) f c exists and ( ) 0 f c > and therefore the denominator of this
will always exist and will never be zero. We'll need this in several places so we can't forget this.
If ( ) 0 f c ′ = then because we know that the denominator will not be zero here we must also have
( ) 0 g c ′ = . Likewise, if ( ) 0 g c ′ = then we must have ( ) 0 f c ′ = . So, ( ) f x and ( ) g x will
have the same critical points in which the derivatives will be zero.
not exist. Therefore, ( ) f x and ( ) g x will have the same critical points in which the derivatives
does not exist.
So, the upshot of all this is that ( ) f x and ( ) g x will have the same critical points.
Next, let's notice that because we know that ( ) 2 0 f x > then ( ) f x ′ and ( ) g x ′ will have the
same sign and so if we apply the first derivative test (and recalling that they have the same critical
points) to each of these functions we can see that the results will be the same and so the relative
extrema will occur at the same points.
Note that we could also use the second derivative test to verify that the critical points will have
the same classification if we wanted to. The second derivative is (and you should see if you can
use the quotient rule to verify this),
( )
( ) ( ) ( ) ( )
( )
1
2
2
2
4
f x f x f x f x
g x
f x
−
′′ ′ − ⎡ ⎤ ⎡ ⎤
⎣ ⎦ ⎣ ⎦
′′ =
Then if x c = is a critical point such that ( ) 0 f c ′ = (and so we can use the second derivative
test) we get,
Now, because we know that ( ) 2 0 f c > and by assumption ( ) 0 f c > we can see that ( ) f c ′′
and ( ) g c ′′ will have the same sign and so will have the same conclusion from the second
derivative test.
So, now that we have that out of the way let's work some more examples.
Example 4 A 2 feet piece of wire is cut into two pieces and once piece is bent into a square and
the other is bent into an equilateral triangle. Where should the wire cut so that the total area
enclosed by both is minimum and maximum?
Solution
Before starting the solution recall that an equilateral triangle is a triangle with three equal sides
and each of the interior angles are
3
π
(or 60° ).
Now, this is another problem where the constraint isn't really going to be given by an equation, it
is simply that there is 2 ft of wire to work with and this will be taken into account in our work.
So, let's cut the wire into two pieces. The first piece will have length x which we'll bend into a
square and each side will have length
4
x
. The second piece will then have length 2 x − (we just
used the constraint here…) and we'll bend this into an equilateral triangle and each side will have
length ( )
1
3
2 x − . Here is a sketch of all this.
As noted in the sketch above we also will need the height of the triangle. This is easy to get if
you realize that the dashed line divides the equilateral triangle into two other triangles. Let's look
at the right one. The hypotenuse is ( )
1
3
2 x − while the lower right angle is
3
π
. Finally the
height is then the opposite side to the lower right angle so using basic right triangle trig we arrive
at the height of the triangle as follows.
Now, let's notice that the problem statement asked for both the minimum and maximum enclosed
area and we got a single critical point. This clearly can't be the answer to both, but this is not the
problem that it might seem to be.
Let's notice that x must be in the range 0 2 x ≤ ≤ and since the area function is continuous we
use the basic process for finding absolute extrema of a function.
( ) ( ) ( ) 0 0.1925 0.8699 0.1087 2 0.25 A A A = = =
So, it looks like the minimum area will arise if we take 0.8699 x = while the maximum area will
arise if we take the whole piece of wire and bend it into a square.
As the previous problem illustrated we can't get too locked into the answers always occurring at
the critical points as they have to this point. That will often happen, but one of the extrema in the
previous problem was at an endpoint and that will happen on occasion.
Example 5 A piece of pipe is being carried down a hallway that is 10 feet wide. At the end of
the hallway the there is a right-angled turn and the hallway narrows down to 8 feet wide. What is
the longest pipe that can be carried (always keeping it horizontal) around the turn in the hallway?
Solution
Let's start off with a sketch of the situation so we can get a grip on what's going on and how
we're going to have to go about solving this.
The largest pipe that can go around the turn will do so in the position shown above. One end will
be touching the outer wall of the hall way at A and C and the pipe will touch the inner corner at B.
Let's assume that the length of the pipe in the small hallway is
1
L while
2
L is the length of the
pipe in the large hallway. The pipe then has a length of
1 2
L L L = + .
Now, if 0 θ = then the pipe is completely in the wider hallway and we can see that as 0 θ →
then L → ∞ . Likewise, if
2
π
θ = the pipe is completely in the narrow hallway and as
2
π
θ →
we also have L → ∞ . So, somewhere in the interval
2
0
π
θ < < is an angle that will minimize L
and oddly enough that is the length that we're after. The largest pipe that will fit around the turn
will in fact be the minimum value of L.
So, if 0.8226 θ = radians then the pipe will have a minimum length and will just fit around the
turn. Anything larger will not fit around the turn and so the largest pipe that can be carried
around the turn is,
( ) ( ) 8sec 0.8226 10csc 0.8226 25.4033 feet L = + =
Example 6 minimum amount of wire is used?
The total length of the wire is
1 2
L L L = + and we need to determine the value of x that will
minimize this. The constraint in this problem is that the poles must be 20 meters apart and that x
must be in the range 0 20 x ≤ ≤ . The first thing that we'll need to do here is to get the length of
wire in terms of x, which is fairly simple to do using the Pythagorean Theorem.
Not the nicest function we've had to work with but there it is. Note however, that it is a
continuous function and we've got an interval with finite endpoints and so finding the absolute
minimum won't require much more work than just getting the critical points of this function. So,
let's do that. Here's the derivative.
It's probably been quite a while since you've been asked to solve something like this. To solve
this we'll need to square both sides to get rid of the roots, but this will cause problems as well
soon see. Let's first just square both sides and solve that equation.
Note that if you can't do that factoring done worry, you can always just use the quadratic formula
and you'll get the same answers.
Okay two issues that we need to discuss briefly here. The first solution above (note that I didn't
call it a critical point…) doesn't make any sense because it is negative and outside of the range of
possible solutions and so we can ignore it.
time that this would cause problems. We'll we've hit those problems. In squaring both sides
we've inadvertently introduced a new solution to the equation. When you do something like this
you should ALWAYS go back and verify that the solutions that you are in fact solutions to the
original equation. In this case we were lucky and the "bad" solution also happened to be outside
the interval of solutions we were interested in but that won't always be the case.
So, if we go back and do a quick verification we can in fact see that the only critical point is
40
7
5.7143 x = = and this is nicely in our range of acceptable solutions.
Now all that we need to do is plug this critical point and the endpoints of the wire into the length
formula and identify the one that gives the minimum value.
( ) ( ) ( )
40
7
0 31 29 20 35.8806 L L L = = =
So, we will get the minimum length of wire if we stake it to the ground
40
7
feet from the smaller
pole.
Let's do a modification of the above problem that asks a completely different question.
Example 7 angle formed by the two pieces of wire at
the stake is a maximum?
Solution
Here's a sketch for this example.
The equation that we're going to need to work with here is not obvious. Let's start with the
following fact.
180 δ θ ϕ π + + = =
( ) ( )
1 1
6 15
20
tan tan
x x
θ π
− −
−
= − −
Note that this is the reason for the π in our equation. The inverse tangents give angles that are in
radians and so can't use the 180 that we're used to in this kind of equation.
Next we'll need the derivative so hopefully you'll recall how to differentiate inverse tangents.
Setting this equal to zero and solving give the following two critical points.
40 4810
3
36.4514, 9.7847 x
− ±
= = −
The first critical point is not in the interval of possible solutions and so we can exclude it.
It's not difficult to show that if 0 9.7847 x ≤ ≤ that 0 θ′ > and if 9.7847 20 x ≤ ≤ that 0 θ′ <
and so when 9.7847 x = we will get the maximum value of θ .
Example 8 A trough for holding water is be formed by taking a piece of sheet metal 60 cm wide
and folding the 20 cm on either end up as shown below. Determine the angle θ that will
maximize the amount of water that the trough can hold.
Solution
Now, in this case we are being asked to maximize the volume that a trough can hold, but if you
think about it the volume of a trough in this shape is nothing more than the cross-sectional area
times the length of the trough. So for a given length in order to maximize the volume all you
really need to do is maximize the cross-sectional area.
We can think of the cross-sectional area as a rectangle in the middle with width 20 and height h
and two identical triangles on either end with height h, base b and hypotenuse 20. Also note that
basic geometry tells us that the angle between the hypotenuse and the base must also be the same
angle θ that we had in our original sketch.
Also, basic right triangle trig tells us that the base and height can be written as,
20cos 20sin b h θ θ = =
However, we can see that θ must be in the interval
2
0
π
θ ≤ ≤ or we won't get a trough in the
proper shape. Therefore, the second critical point makes no sense and also note that we don't
need to add on the standard " 2 n π + " for the same reason.
Finally, since the equation for the area is continuous all we need to do is plug in the critical point
and the end points to find the one that gives the maximum area.
( ) ( ) ( )
3 2
0 0 519.6152 400 A A A
π π
= = =
So, we will get a maximum cross-sectional area, and hence a maximum volume, when
3
π
θ = .
Indeterminate Forms and L'Hospital's Rule
Back in the chapter on Limits we saw methods for dealing with the following limits.
2 2
2
4
16 4 5
lim lim
4 1 3
x x
x x x
x x
→ →∞
− −
− −
In the first limit if we plugged in 4 x = we would get 0/0 and in the second limit if we "plugged"
in infinity we would get ∞ −∞ (recall that as x goes to infinity a polynomial will behave in the
same fashion that it's largest power behaves). Both of these are called indeterminate forms. In
both of these cases there are competing interests or rules and it's not clear which will win out.
In the case of 0/0 we typically think of a fraction that has a numerator of zero as being zero.
However, we also tend to think of fractions in which the denominator is going to zero as infinity
or might not exist at all. Likewise, we tend to think of a fraction in which the numerator and
denominator are the same as one. So, which will win out? Or will neither win out and they all
"cancel out" and the limit will reach some other value?
In the case of ∞ −∞ we have a similar set of problems. If the numerator of a fraction is going to
infinity we tend to think of the whole fraction going to infinity. Also if the denominator is going
to infinity we tend to think of the fraction as going to zero. We also have the case of a fraction in
which the numerator and denominator are the same (ignoring the minus sign) and so we might get
-1. Again, it's not clear which of these will win out, if any of them will win out.
With the second limit there is the further problem that infinity isn't really a number and so we
really shouldn't even treat it like a number. Much of the time it simply won't behave as we
would expect it to if it was a number. To look a little more into this check out the Types of
Infinity section in the Extras chapter at the end of this document.
This is the problem with indeterminate forms. It's just not clear what is happening in the limit.
There are other types of indeterminate forms as well. Some other types are,
( )( )
0 0
0 1 0
∞
±∞ ∞ ∞−∞
These all have competing interests or rules that tell us what should happen and it's just not clear
which, if any, of the interests or rules will win out. The topic of this section is how to deal with
these kinds of limits.
As already pointed out we do know how to deal with some kinds of indeterminate forms already.
For the two limits above we work them as follows.
( )
2
4 4
16
lim lim 4 8
4
x x
x
x
x
→ →
−
= + =
−
In the first case we simply factored, canceled and took the limit and in the second case we
factored out an
2
x from both the numerator and the denominator and took the limit. Notice as
well that none of the competing interests or rules in these cases won out! That is often the case.
So we can deal with some of these. However what about the following two limits.
2
0
sin
lim lim
x
x x
x
x x
→ →∞
e
This first is a 0/0 indeterminate form, but we can't factor this one. The second is an ∞ ∞
indeterminate form, but we can't just factor an
2
x out of the numerator. So, nothing that we've
got in our bag of tricks will work with these two limits.
This is where the subject of this section comes into play.
L'Hospital's Rule
Suppose that we have one of the following cases,
( )
( )
( )
( )
0
lim OR lim
0
x a x a
f x f x
g x g x
→ →
±∞
= =
±∞
where a can be any real number, infinity or negative infinity. In these cases we have,
( )
( )
( )
( )
lim lim
x a x a
f x f x
g x g x
→ →
′
=
′
So, L'Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞ ∞all we need to do
is differentiate the numerator and differentiate the denominator and then take the limit.
Before proceeding with examples let me address the spelling of "L'Hospital". The more correct
spelling is "L'Hôpital". However, when I first learned Calculus I my teacher used the spelling
that I use in these notes and the first text book that I taught Calculus out of also used the spelling
that I use here. So, I'm used to spelling it that way and that is the way that I've spelled it here.
So, we have already established that this is a 0/0 indeterminate form so let's just apply
L'Hospital's Rule.
0 0
sin cos 1
lim lim 1
1 1
x x
x x
x
→ →
= = =
[Return to Problems]
(b)
4 2
3
1
5 4 1
lim
10 9
t
t t
t t
→
− −
− −
In this case we also have a 0/0 indeterminate form and if we were really good at factoring we
could factor the numerator and denominator, simplify and take the limit. However, that's going
to be more work than just using L'Hospital's Rule.
L'Hospital's Rule works great on the two indeterminate forms 0/0 and ±∞ ±∞. However, there
are many more indeterminate forms out there as we saw earlier. Let's take a look at some of
those and see how we deal with those kinds of indeterminate forms.
We'll start with the indeterminate form ( ) ( ) 0 ±∞ .
Example 2 Evaluate the following limit.
0
lim ln
x
x x
+
→
Solution
Note that we really do need to do the right-hand limit here. We know that the natural logarithm
only defined for positive x and so this is the only limit that makes any sense.
Now, in the limit, we get the indeterminate form ( )( ) 0 −∞ . L'Hospital's Rule won't work on
products, it only works on quotients. However, we can turn this into a fraction if we rewrite
things a little.
0 0
ln
lim ln lim
1 x x
x
x x
x
+ +
→ →
=
The function is the same, just rewritten, and the limit is now in the form −∞ ∞ and we can now
use L'Hospital's Rule.
Example 3 Evaluate the following limit.
lim
x
x
x
→−∞
e
Solution
So, it's in the form ( )( ) 0 ∞ . This means that we'll need to write it as a quotient. Moving the x
to the denominator worked in the previous example so let's try that with this problem as well.
lim lim
1
x
x
x x
x
x
→−∞ →−∞
=
e
e
Writing the product in this way gives us a product that has the form 0/0 in the limit. So,
let's use L'Hospital's Rule on the quotient.
Hummmm…. This doesn't seem to be getting us anywhere. With each application of
L'Hospital's Rule we just end up with another 0/0 indeterminate form and in fact the derivatives
seem to be getting worse and worse. Also note that if we simplified the quotient back into a
product we would just end up with either ( )( ) 0 ∞ or ( )( ) 0 −∞ and so that won't do us any
good.
to simplify the quotient up a little. This will help us when it comes time to take some derivatives.
The quotient is now an indeterminate form of −∞ ∞ and use L'Hospital's Rule gives,
1
lim lim lim 0
x
x x
x x x
x
x
− −
→−∞ →−∞ →−∞
= = =
−
e
e e
So, when faced with a product ( )( ) 0 ±∞ we can turn it into a quotient that will allow us to use
L'Hospital's Rule. However, as we saw in the last example we need to be careful with how we
do that on occasion. Sometimes we can use either quotient and in other cases only one will work.
Linear Approximations
In this section we're going to take a look at an application not of derivatives but of the tangent
line to a function. Of course, to get the tangent line we do need to take derivatives, so in some
way this is an application of derivatives as well.
Take a look at the following graph of a function and its tangent line.
From this graph we can see that near x a = the tangent line and the function have nearly the
same graph. On occasion we will use the tangent line, ( ) L x , as an approximation to the
function, ( ) f x , near x a = . In these cases we call the tangent line the linear approximation to
the function at x a = .
So, at 8.05 x = this linear approximation does a very good job of approximating the actual value.
However, at 25 x = it doesn't do such a good job.
This shouldn't be too surprising if you think about. Near 8 x = both the function and the linear
approximation have nearly the same slope and since they both pass through the point ( ) 8, 2 they
should have nearly the same value as long as we stay close to 8 x = . However, as we move away
from 8 x = the linear approximation is a line and so will always have the same slope while the
functions slope will change as x changes and so the function will, in all likelihood, move away
from the linear approximation.
Here's a quick sketch of the function and it's linear approximation at 8 x = .
As noted above, the farther from 8 x = we get the more distance separates the function itself and
its linear approximation.
we're using. Also, there will often be no easy way of prediction how far away from x a = we
can get and still have a "good" approximtation.
Let's take a look at another example that is actually used fairly heavily in some places.
Example 2 Determine the linear approximation for sinθ at 0 θ = .
Solution
Again, there really isn't a whole lot to this example. All that we need to do is compute the
tangent line to sinθ at 0 θ = .
( ) ( )
( ) ( )
sin cos
0 0 0 1
f f
f f
θ θ θ θ ′ = =
′ = =
The linear approximation is,
( ) ( ) ( )( )
( )( )
0 0
0 1 0
L f f a θ θ
θ
θ
′ = + −
= + −
=
So, as long as θ stays small we can say that sinθ θ ≈ .
This is actually a somewhat important linear approximation. In optics this linear approximation
is often used to simplify formulas. This linear approximation is also used to help describe the
motion of a pendulum and vibrations in a string.
Solution
Before working any of these we should first discuss just what we're being asked to find here. We
defined two differentials earlier and here we're being asked to compute a differential.
So, which differential are we being asked to compute? In this kind of problem we're being asked
to compute the differential of the function. In other words, dy for the first problem, dw for the
second problem and df for the third problem.
Here are the solutions. Not much to do here other than take a derivative and don't forget to add
on the second differential to the derivative.
There is a nice application to differentials. If we think of x Δ as the change in x then
( ) ( ) y f x x f x Δ = + Δ − is the change in y corresponding to the change in x. Now, if x Δ is
small we can assume that y dy Δ ≈ . Let's see an illustration of this idea.
Next, the change in x from 2 x = to 2.03 x = is 0.03 x Δ = and so we then assume that
0.03 dx x ≈ Δ = . This gives an approximate change in y of,
( ) ( ) ( )
( )
2
2 2 sin 2 1 1 0.03 0.085070913 dy = − + − =
We can see that in fact we do have that y dy Δ ≈ provided we keep x Δ small.
We can use the fact that y dy Δ ≈ in the following way.
Example 3 A sphere was measured and its radius was found to be 45 inches with a possible
error of no more that 0.01 inches. What is the maximum possible error in the volume if we use
this value of the radius?
Be careful to not assume this is a large error. On the surface it looks large, however if we
compute the actual volume for 45 r = we get
3
381, 703.51 in V = . So, in comparison the error
in the volume is,
Newton's Method
The next application that we'll take a look at in this chapter is an important application that is
used in many areas. If you've been following along in the chapter to this point it's quite possible
that you've gotten the impression that many of the applications that we've looked at are just made
up by us to make you work. This is unfortunate because all of the applications that we've looked
at to this point are real applications that really are used in real situations. The problem is often
that in order to work more meaningful examples of the applications we would need more
knowledge than we generally have about the science and/or physics behind the problem. Without
that knowledge we're stuck doing some fairly simplistic examples that often don't seem very
realistic at all and that makes it hard to understand that the application we're looking at is a real
application.
That is going to change in this section. This is an application that we can all understand and we
can all understand needs to be done on occasion even if we don't understand the physics/science
behind an actual application.
In this section we are going to look at a method for approximating solutions to equations. We all
know that equations need to be solved on occasion and in fact we've solved quite a few equations
ourselves to this point. In all the examples we've looked at to this point we were able to actually
find the solutions, but it's not always possible to do that exactly and/or do the work by hand.
That is where this application comes into play. So, let's see what this application is all about.
Let's suppose that we want to approximate the solution to ( ) 0 f x = and let's also suppose that
we have somehow found an initial approximation to this solution say, x
0
. This initial
approximation is probably not all that good and so we'd like to find a better approximation. This
is easy enough to do. First we will get the tangent line to ( ) f x at x
0
.
( ) ( )( )
0 0 0
y f x f x x x ′ = + −
The blue line (if you're reading this in color anyway…) is the tangent line at x
0
. We can see that
this line will cross the x-axis much closer to the actual solution to the equation than x
0
does. Let's
call this point where the tangent at x
0
crosses the x-axis x
1
and we'll use this point as our new
approximation to the solution.
So, how do we find this point? Well we know it's coordinates, ( )
1
, 0 x , and we know that it's on
the tangent line so plug this point into the tangent line and solve for x
1
as follows,
So, we can find the new approximation provided the derivative isn't zero at the original
approximation.
Now we repeat the whole process to find an even better approximation. We form up the tangent
line to ( ) f x at x
1
and use its root, which we'll call x
2
, as a new approximation to the actual
solution. If we do this we will arrive at the following formula.
( )
( )
1
2 1
1
f x
x x
f x
= −
′
This point is also shown on the graph above and we can see from this graph that if we continue
following this process will get a sequence of numbers that are getting very close the actual
solution. This process is called Newton's Method.
Here is the general Newton's Method
Newton's Method
If x
n
is an approximation a solution of ( ) 0 f x = and if ( ) 0
n
f x ′ ≠ the next approximation is
given by,
( )
( )
1
n
n n
n
f x
x x
f x
+
= −
′
This should lead to the question of when do we stop? How many times do we go through this
process? One of the more common stopping points in the process is to continue until two
successive approximations agree to a given number of decimal places.
Before working any examples we should address two issues. First, we really do need to be
solving ( ) 0 f x = in order Newton's Method to be applied. This isn't really all that much of an
issue but we do need to make sure that the equation is in this form prior to using the method.
Secondly, we do need to somehow get our hands on an initial approximation to the solution (i.e.
we need
0
x somehow). One of the more common ways of getting our hands on
0
x is to sketch
the graph of the function and use that to get an estimate of the solution which we then use as
0
x .
Another common method is if we know that there is a solution to a function in an interval then we
can use the midpoint of the interval as
0
x .
Let's work an example of Newton's Method.
Example 1 Use Newton's Method to determine an approximation to the solution to cos x x =
that lies in the interval [0,2]. Find the approximation to six decimal places.
Solution
First note that we weren't given an initial guess. We were however, given an interval in which to
look. We will use this to get our initial guess. As noted above the general rule of thumb in these
cases is to take the initial approximation to be the midpoint of the interval. So, we'll use
0
1 x =
as our initial guess.
Next, recall that we must have the function in the form ( ) 0 f x = . Therefore, we first rewrite
the equation as,
cos 0 x x − =
We can now write down the general formula for Newton's Method. Doing this will often
simplify up the work a little so it's generally not a bad idea to do this.
1
cos
sin 1
n n
x x
x x
x
+
−
= −
− −
Let's now get the first approximation.
( )
( )
1
cos 1 1
1 0.7503638679
sin 1 1
x
−
= − =
− −
At this point we should point out that the phrase "six decimal places" does not mean just get x
1
to
six decimal places and then stop. Instead it means that we continue until two successive
approximations agree to six decimal places.
Alright, we're making progress. We've got the approximation to 1 decimal place. Let's do
another one, leaving the details of the computation to you.
3
0.7390851334 x =
We've got it to three decimal places. We'll need another one.
4
0.7390851332 x =
And now we've got two approximations that agree to 9 decimal places and so we can stop. We
will assume that the solution is approximately
4
0.7390851332 x = .
In this last example we saw that we didn't have to do too many computations in order for
Newton's Method to give us an approximation in the desired range of accuracy. This will not
always be the case. Sometimes it will take many iterations through the process to get to the
desired accuracy and on occasion it can fail completely.
The following example is a little silly but it makes the point about the method failing.
Example 2 Use
0
1 x = to find the approximation to the solution to
3
0 x = .
Solution
Yes, it's a silly example. Clearly the solution is 0 x = , but it does make a very important point.
Let's get the general formula for Newton's method.
1
3
1 2
3
3 2
1
3
n
n n n n n
n
x
x x x x x
x
+
−
= − = − = −
In fact we don't really need to do any computations here. These computations get farther and
farther away from the solution, 0 x = ,with each iteration. Here are a couple of computations to
make the point.
So, we need to be a little careful with Newton's method. It will usually quickly find an
approximation to an equation. However, there are times when it will take a lot of work or when it
won't work at all.
Business Applications
In the final section of this chapter let's take a look at some applications of derivatives in the
business world. For the most part these are really applications that we've already looked at, but
they are now going to be approached with an eye towards the business world.
Let's start things out with a couple of optimization problems. We've already looked at more than
a few of these in previous sections so there really isn't anything all that new here except for the
fact that they are coming out of the business world.
How many apartments should they rent in order to maximize their profit?
Solution
All that we're really being asked to do here is to maximize the profit subject to the constraint that
x must be in the range 0 250 x ≤ ≤ .
First, we'll need the derivative and the critical point(s) that fall in the range 0 250 x ≤ ≤ .
( )
3200
16
16 3200 3200 16 0 200 P x x x x ′ = − + ⇒ − = ⇒ = =
Since the profit function is continuous and we have an interval with finite bounds we can find the
maximum value by simply plugging in the only critical point that we have (which nicely enough
in the range of acceptable answers) and the end points of the range.
( ) ( ) ( ) 0 80, 000 200 240, 000 250 220, 000 P P P = − = =
So, it looks like they will generate the most profit if they only rent out 200 of the apartments
instead of all 250 of them.
Note that with these problems you shouldn't just assume that renting all the apartments will
generate the most profit. Do not forget that there are all sorts of maintenance costs and that the
more tenants renting apartments the more the maintenance costs will be. With this analysis we
can see that, for this complex at least, something probably needs to be done to get the maximum
profit more towards full capacity. This kind of analysis can help them determine just what they
need to do to move towards goal that whether it be raising rent or find a way to reduce
maintenance costs.
slightly under full capacity to take this into account. Again, another reason to not just assume
that maximum profit will always be at the upper limit of the range.
Let's take a quick look at another problem along these lines.
Example 4 A production facility is capable of producing 60,000 widgets in a day and the total
daily cost of producing x widgets in a day is given by,
( )
200, 000, 000
250, 000 0.08 C x x
x
= + +
How many widgets per day should they produce in order to minimize production costs?
Solution
Here we need to minimize the cost subject to the constraint that x must be in the range
0 60, 000 x ≤ ≤ . Note that in this case the cost function is not continuous at the left endpoint and
so we won't be able to just plug critical points and endpoints into the cost function to find the
minimum value.
Now, clearly the negative value doesn't make any sense in this setting and so we have a single
critical point in the range of possible solutions : 50,000.
Now, as long as 0 x > the second derivative is positive and so, in the range of possible solutions
the function is always concave up and so producing 50,000 widgets will yield the absolute
minimum production cost.
Now, we shouldn't walk out of the previous two examples with the idea that the only applications
to business are just applications we've already looked at but with a business "twist" to them.
(a) What is the cost to produce the 301
st
widget?
(b) What is the rate of change of the cost at 300 x = ?
Solution
(a) We can't just compute ( ) 301 C as that is the cost of producing 301 widgets while we are
looking for the actual cost of producing the 301
st
widget. In other words, what we're looking for
here is,
( ) ( ) 301 300 97, 695.91 97, 400.00 295.91 C C − = − =
So, the cost of producing the 301
st
widget is $295.91.
(b) In the part all we need to do is get the derivative and then compute ( ) 300 C′ .
( ) ( ) 350 0.18 300 296.00 C x x C ′ ′ = − ⇒ =
Okay, so just what did we learn in this example? The cost to produce an additional item is called
the marginal cost and as we've seen in the above example the marginal cost is approximated by
the rate of change of the cost function, ( ) C x . So, we define the marginal cost function to be
the derivative of the cost function or, ( ) C x ′ . Let's work a quick example of this.
Note that it is important to note that ( ) C n ′ is the approximate cost of production the ( )
st
1 n +
item and NOT the n
th
item as it may seem to imply!
Let's now turn our attention to the average cost function. If ( ) C x is the cost function for some
item then the average cost function is,
( )
( ) C x
C x
x
=
Here is the sketch of the average cost function from Example 4 above.
We can see from this that the average cost function has an absolute minimum. We can also see
that this absolute minimum will occur at a critical point with ( ) 0 C x ′ = since it clearly will have
a horizontal tangent there.
Now, we could get the average cost function, differentiate that and then find the critical point.
However, this average cost function is fairly typical for average cost functions so let's instead
differentiate the general formula above using the quotient rule and see what we have.
( )
( ) ( )
2
xC x C x
C x
x
′ −
′ =
We should note however that not all average cost functions will look like this and so you
shouldn't assume that this will always be the case.
Let's now move onto the revenue and profit functions. First, let's suppose that the price that
some item can be sold at if there is a demand for x units is given by ( ) p x . This function is
typically called either the demand function or the price function.
The revenue function is then how much money is made by selling x items and is,
( ) ( ) R x x p x =
So, upon producing and selling the 2501
st
widget it will cost the company approximately $25 to
produce the widget and they will see and added $175 in revenue and $150 in profit.
On the other hand when they produce and sell the 7501
st
widget it will cost an additional $325
and they will receive an extra $125 in revenue, but lose $200 in profit.
We'll close this section out with a brief discussion on maximizing the profit. If we assume that
the maximum profit will occur at a critical point such that ( ) 0 P x ′ = we can then say the
following,
In this section we took a brief look at some of the ideas in the business world that involve
calculus. Again, it needs to be stressed however that there is a lot more going on here and to
really see how these applications are done you should really take some business courses. The
point of this section was to just give a few idea on how calculus is used in a field other than the
sciences.
Introduction
In this chapter we will be looking at integrals. Integrals are the third and final major topic that
will be covered in this class. As with derivatives this chapter will be devoted almost exclusively
to finding and computing integrals. Applications will be given in the following chapter. There
are really two types of integrals that we'll be looking at in this chapter : Indefinite Integrals and
Definite Integrals. The first half of this chapter is devoted to indefinite integrals and the last half
is devoted to definite integrals. As we will see in the last half of the chapter if we don't know
indefinite integrals we will not be able to do definite integrals.
Here is a quick listing of the material that is in this chapter.
Indefinite Integrals – In this section we will start with the definition of indefinite integral. This
section will be devoted mostly to the definition and properties of indefinite integrals.
Computing Indefinite Integrals – In this section we will compute some indefinite integrals and
take a look at a quick application of indefinite integrals.
Substitution Rule for Indefinite Integrals – Here we will look at the Substitution Rule as it
applies to indefinite integrals. Many of the integrals that we'll be doing later on in the course and
in later courses will require use of the substitution rule.
More Substitution Rule – Even more substitution rule problems.
Area Problem – In this section we start off with the motivation for definite integrals and give
one of the interpretations of definite integrals.
Definition of the Definite Integral – We will formally define the definite integral in this section
and give many of its properties. We will also take a look at the first part of the Fundamental
Theorem of Calculus.
Computing Definite Integrals – We will take a look at the second part of the Fundamental
Theorem of Calculus in this section and start to compute definite integrals.
Substitution Rule for Definite Integrals – In this section we will revisit the substitution rule as
it applies to definite integrals.
Indefinite Integrals
In the past two chapters we've been given a function, ( ) f x , and asking what the derivative of
this function was. Starting with this section we are not going to turn things around. We now
want to ask what function we differentiated to get the function ( ) f x .
Let's take a quick look at an example to get us started.
Example 1 What function did we differentiate to get the following function.
( )
4
3 9 f x x x = + −
Solution
Let's actually start by getting the derivative of this function to help us see how we're going to
have to approach this problem. The derivative of this function is,
( )
3
4 3 f x x ′ = +
The point of this was to remind us of how differentiation works. When differentiating powers of
x we multiply the term by the original exponent and then drop the exponent by one.
Now, let's go back and work the problem. In fact let's just start with the first term. We got x
4
by
differentiating a function and since we drop the exponent by one it looks like we must have
differentiated x
5
. However, if we had differentiated x
5
we would have 5x
4
and we don't have a 5
in front our first term, so the 5 needs to cancel out after we've differentiated. It looks then like
we would have to differentiate
5
1
5
x in order to get x
4
.
Likewise for the second term, in order to get 3x after differentiating we would have to
differentiate
2
3
2
x . Again, the fraction is there to cancel out the 2 we pick up in the
differentiation.
The third term is just a constant and we know that if we differentiate x we get 1. So, it looks like
we had to differentiate -9x to get the last term.
There were two points to this last example. The first point was to get you thinking about how to
do these problems. It is important initially to remember that we are really just asking what we
differentiated to get the given function.
The other point is to recognize that there are actually an infinite number of functions that we
could use and they will all differ by a constant.
Now that we've worked an example let's get some of the definitions and terminology out of the
way.
Note that often we will just say integral instead of indefinite integral (or definite integral for that
matter when we get to those). It will be clear from the context of the problem that we are talking
about an indefinite integral (or definite integral).
The process of finding the indefinite integral is called integration or integrating f(x). If we
need to be specific about the integration variable we will say that we are integrating f(x) with
respect to x.
Let's rework the first problem in light of the new terminology.
Example 2 Evaluate the following indefinite integral.
4
3 9 x x dx + −
∫
Solution
Since this is really asking for the most general anti-derivative we just need to reuse the final
answer from the first example.
The indefinite integral is,
4 5 2
1 3
3 9 9
5 2
x x dx x x x c + − = + − +
∫
A couple of warnings are now in order. One of the more common mistakes that students make
with integrals (both indefinite and definite) is to drop the dx at the end of the integral. This is
required! Think of the integral sign and the dx as a set of parenthesis. You already know and are
probably quite comfortable with the idea that every time you open a parenthesis you must close it.
With integrals, think of the integral sign as an "open parenthesis" and the dx as a "close
parenthesis".
If you drop the dx it won't be clear where the integrand ends. Consider the following variations
of the above example.
You only integrate what is between the integral sign and the dx. Each of the above integrals end
in a different place and so we get different answers because we integrate a different number of
terms each time. In the second integral the "-9" is outside the integral and so is left alone and not
integrated. Likewise, in the third integral the "3 9 x − " is outside the integral and so is left alone.
Knowing which terms to integrate is not the only reason for writing the dx down. In the
Substitution Rule section we will actually be working with the dx in the problem and if we aren't
in the habit of writing it down it will be easy to forget about it and then we will get the wrong
answer at that stage.
The moral of this is to make sure and put in the dx! At this stage it may seem like a silly thing to
do, but it just needs to be there, if for no other reason than knowing where the integral stops.
On a side note, the dx notation should seem a little familiar to you. We saw things like this a
couple of sections ago. We called the dx a differential in that section and yes that is exactly what
it is. The dx that ends the integral is nothing more than a differential.
The next topic that we should discuss here is the integration variable used in the integral.
Actually there isn't really a lot to discuss here other than to note that the integration variable
doesn't really matter. For instance,
Changing the integration variable in the integral simply changes the variable in the answer. It is
important to notice however that when we change the integration variable in the integral we also
changed the differential (dx, dt, or dw) to match the new variable. This is more important that we
might realize at this point.
Another use of the differential at the end of integral is to tell us what variable we are integrating
with respect to. At this stage that may seem unimportant since most of the integrals that we're
going to be working with here will only involve a single variable. However, if you are on a
degree track that will take you into multi-variable calculus this will be very important at that stage
since there will be more than one variable in the problem. You need to get into the habit of
writing the correct differential at the end of the integral so when it becomes important in those
classes you will already be in the habit of writing it down.
To see why this is important take a look at the following two integrals.
2 2 x dx t dx
∫ ∫
The second integral is also fairly simple, but we need to be careful. The dx tells us that we are
integrating x's. That means that we only integrate x's that are in the integrand and all other
variables in the integrand are considered to be constants. The second integral is then,
2 2 t dx tx c = +
∫
So, it may seem silly to always put in the dx, but it is a vital bit of notation that can cause us to get
the incorrect answer if we neglect to put it in.
Now, there are some important properties of integrals that we should take a look at.
See the Proof of Various Integral Formulas section of the Extras chapter to see the proof of
this property.
2. ( ) ( ) f x dx f x dx − = −
∫ ∫
. This is really the first property with 1 k = − and so no proof of
this property will be given.
3. ( ) ( ) ( ) ( ) f x g x dx f x dx g x dx ± = ±
∫ ∫ ∫
. In other words, the integral of a sum or
difference of functions is the sum or difference of the individual integrals. This rule can be
extended to as many functions as we need.
See the Proof of Various Integral Formulas section of the Extras chapter to see the proof of
this property.
Notice that when we worked the first example above we used the first and third property in the
discussion. We integrated each term individually, put any constants back in and then put
everything back together with the appropriate sign.
Not listed in the properties above were integrals of products and quotients. The reason for this is
simple. Just like with derivatives each of the following will NOT work.
With derivatives we had a product rule and a quotient rule to deal with these cases. However,
with integrals there are no such rules. When faced with a product and quotient in an integral we
will have a variety of ways of dealing with it depending on just what the integrand is.
There is one final topic to be discussed briefly in this section. On occasion we will be given
( ) f x ′ and will ask what ( ) f x was. We can now answer this question easily with an indefinite
integral.
( ) ( ) f x f x dx ′ =
∫
In this section we kept evaluating the same indefinite integral in all of our examples. The point of
this section was not to do indefinite integrals, but instead to get us familiar with the notation and
some of the basic ideas and properties of indefinite integrals. The next couple of sections are
devoted to actually evaluating indefinite integrals.
Computing Indefinite Integrals
In the previous section we started looking at indefinite integrals and in that section we
concentrated almost exclusively on notation, concepts and properties of the indefinite integral. In
this section we need to start thinking about how we actually compute indefinite integrals. We'll
start off with some of the basic indefinite integrals.
The first integral that we'll look at is the integral of a power of x.
1
, 1
1
n
n
x
x dx c n
n
+
= + ≠ −
+
∫
The general rule when integrating a power of x we add one onto the exponent and then divide by
the new exponent. It is clear (hopefully) that we will need to avoid 1 n = − in this formula. If we
allow 1 n = − in this formula we will end up with division by zero. We will take care of this case
in a bit.
Next is one of the easier integrals but always seems to cause problems for people.
, and are constants k dx kx c c k = +
∫
If you remember that all we're asking is what did we differentiate to get the integrand this is
pretty simple, but it does seem to cause problems on occasion.
Notice that we only integrated two of the six trig functions here. The remaining four integrals are
really integrals that give the remaining four trig functions. Also, be careful with signs here. It is
easy to get the signs for derivatives and integrals mixed up. Again, remember that we're asking
what function we differentiated to get the integrand.
We will be able to integrate the remaining four trig functions in a couple of sections, but they all
require the Substitution Rule.
Integrating logarithms requires a topic that is usually taught in Calculus II and so we won't be
integrating a logarithm in this class. Also note the third integrand can be written in a couple of
ways and don't forget the absolute value bars in the x in the answer to the third integral.
Finally, let's take care of the inverse trig and hyperbolic functions.
As with logarithms integrating inverse trig functions requires a topic usually taught in Calculus II
and so we won't be integrating them in this class. There is also a different answer for the second
integral above. Recalling that since all we are asking here is what function did we differentiate to
get the integrand the second integral could also be,
1
2
1
cos
1
dx x c
x
−
= − +
−
⌠
⎮
⌡
Traditionally we use the first form of this integral.
Okay, now that we've got most of the basic integrals out of the way let's do some indefinite
integrals. In all of these problems remember that we can always check our answer by
differentiating and making sure that we get the integrand.
with the appropriate signs. Next, we can ignore any coefficients until we are done with
integrating that particular term and then put the coefficient back in. Also, do not forget the "+c"
at the end it is important and must be there.
So, let's evaluate some integrals.
(a)
3 6
5 10 4 t t dt
−
− +
∫
There's not really a whole lot to do here other than use the first two formulas from the beginning
of this section. Remember that when integrating powers (that aren't -1 of course) we just add one
onto the exponent and then divide by the new exponent.
Be careful when integrating negative exponents. Remember to add one onto the exponent. One
of the more common mistakes that students make when integrating negative exponents is to "add
one" and end up with an exponent of "-7" instead of the correct exponent of "-5".
[Return to Problems]
(b)
8 8
x x dx
−
+
∫
This is here just to make sure we get the point about integrating negative exponents.
8 8 9 7
1 1
9 7
x x dx x x c
− −
+ = − +
∫
[Return to Problems]
(c)
3 4
5
7 1
3
6
x dx
x x
+ +
⌠
⎮
⌡
In this case there isn't a formula for explicitly dealing with radicals or rational expressions.
However, just like with derivatives we can write all these terms so they are in the numerator and
they all have an exponent. This should always be your first step when faced with this kind of
integral just as it was when differentiating.
this is equivalent to multiplying by the reciprocal of the new exponent and so that is what we will
usually do.
[Return to Problems]
(d) dy
∫
Don't make this one harder that it is…
1 dy dy y c = = +
∫ ∫
In this case we are really just integrating a one!
[Return to Problems]
(e)
( )( )
2 3
4 w w w dw + −
∫
We've got a product here and as we noted in the previous section there is no rule for dealing with
products. However, in this case we don't need a rule. All that we need to do is multiply things
out (taking care of the radicals at the same time of course) and then we will be able to integrate.
As with the previous part it's not really a problem that we don't have a rule for quotients for this
integral. In this case all we need to do is break up the quotient and then integrate the individual
terms.
Be careful to not think of the third term as x to a power for the purposes of integration. Using that
rule on the third term will NOT work. The third term is simply a logarithm. Also, don't get
excited about the 15. The 15 is just a constant and so it can be factored out of the integral. In
other words, here is what we did to integrate the third term.
Always remember that you can't integrate products and quotients in the same way that we
integrate sums and differences. At this point the only way to integrate products and quotients is
to multiply the product out or break up the quotient. Eventually we'll see some other products
and quotients that can be dealt with in other ways. However, there will never be a single rule that
will work for all products and there will never be a single rule that will work for all quotients.
Every product and quotient is different and will need to be worked on a case by case basis.
The first set of examples focused almost exclusively on powers of x (or whatever variable we
used in each example). It's time to do some examples that involve other functions.
Solution
Most of the problems in this example will simply use the formulas from the beginning of this
section. More complicated problems involving most of these functions will need to wait until we
reach the Substitution Rule.
(a)
2
3 5cos 10sec
x
x x dx + −
∫
e
There isn't anything to this one other than using the formulas.
Rewriting the second integrand will help a little with the integration at this early stage. We can
think of the 6 in the denominator as a 1/6 out in front of the term and then since this is a constant
it can be factored out of the integral. The answer is then,
1 1
2sec tan 2sec ln
6 6
w w dw w w c
w
+ = + +
⌠
⎮
⌡
Note that we didn't factor the 2 out of the first integral as we factored the 1/6 out of the second.
In fact, we will generally not factor the 1/6 out either in later problems. It was only done here to
make sure that you could follow what we were doing.
[Return to Problems]
(c)
2
23 9
6csc cot
1
y y dy
y y
+ +
+
⌠
⎮
⌡
In this one we'll just use the formulas from above and don't get excited about the coefficients.
They are just multiplicative constants and so can be ignored while we integrate each term and
then once we're done integrating a given term we simply put the coefficients back in.
As shown in the last part of this example we can do some fairly complicated looking quotients at
this point if we remember to do simplifications when we see them. In fact, this is something that
you should always keep in mind. In almost any problem that we're doing here don't forget to
simplify where possible. In almost every case this can only help the problem and will rarely
complicate the problem.
In the next problem we're going to take a look at a product and this time we're not going to be
able to just multiply the product out. However, if we recall the comment about simplifying a
little this problem becomes fairly simple.
Solution
There are several ways to do this integral and most of them require the next section. However,
there is a way to do this integral using only the material from this section. All that is required is
to remember the trig formula that we can use to simplify the integrand up a little. Recall the
following double angle formula.
( ) sin 2 2sin cos t t t =
As noted earlier there is another method for doing this integral. In fact there are two alternate
methods. To see all three check out the section on Constant of Integration in the Extras chapter
but be aware that the other two do require the material covered in the next section.
The formula/simplification in the previous problem is a nice "trick" to remember. It can be used
on occasion to greatly simplify some problems.
(b) ( ) ( ) ( )
3
5
15 5 6, 1 , 4 404
4
f x x x f f ′′ = + + = − =
This one is a little different form the first one. In order to get the function we will need the first
derivative and we have the second derivative. We can however, use an integral to get the first
derivative from the second derivative, just as we used an integral to get the function from the first
derivative.
Do not get excited about integrating the c. It's just a constant and we know how to integrate
constants. Also, there will be no reason to think the constants of integration from the integration
in each step will be the same and so we'll need to call each constant of integration something
different.
There are many new formulas in this section that we'll now have to know. However, if you think
about it, they aren't really new formulas. They are really nothing more than derivative formulas
that we should already know written in terms of integrals. If you remember that you should find
it easier to remember the formulas in this section.
Always remember that integration is asking nothing more than what function did we differentiate
to get the integrand. If you can remember that many of the basic integrals that we saw in this
section and many of the integrals in the coming sections aren't too bad.
All of these look considerably more difficult than the first set. However, they aren't too bad once
you see how to do them. Let's start with the first one.
2 3 4
18 6 5 x x dx +
∫
In this case let's notice that if we let
3
6 5 u x = +
and we compute the differential (you remember how to compute these right?) for this we get,
2
18 du x dx =
Now, let's go back to our integral and notice that we can eliminate every x that exists in the
integral and write the integral completely in terms of u using both the definition of u and its
differential.
A natural question at this stage is how to identify the correct substitution. Unfortunately, the
answer is it depends on the integral. However, there is a general rule of thumb that will work for
many of the integrals that we're going to be running across.
When faced with an integral we'll ask ourselves what we know how to integrate. With the
integral above we can quickly recognize that we know how to integrate
4
x dx
∫
However, we didn't have just the root we also had stuff in front of the root and (more importantly
in this case) stuff under the root. Since we can only integrate roots if there is just an x under the
root a good first guess for the substitution is then to make u be the stuff under the root.
Another way to think of this is to ask yourself what portion of the integrand has an inside function
and can you do the integral with that inside function present. If you can't then there is a pretty
good chance that the inside function will be the substitution.
We will have to be careful however. There are times when using this general rule can get us in
trouble or overly complicate the problem. We'll eventually see problems where there are more
than one "inside function" and/or integrals that will look very similar and yet use completely
different substitutions. The reality is that the only way to really learn how to do substitutions is to
just work lots of problems and eventually you'll start to get a feel for how these work and you'll
find it easier and easier to identify the proper substitutions.
Now, with that out of the way we should ask the following question. How, do we know if we got
the correct substitution? Well, upon computing the differential and actually performing the
substitution every x in the integral (including the x in the dx) must disappear in the substitution
process and the only letters left should be u's (including a du). If there are x's left over then there
is a pretty good chance that we chose the wrong substitution. Unfortunately, however there is at
least one case (we'll be seeing an example of this in the next section) where the correct
substitution will actually leave some x's and we'll need to do a little more work to get it to work.
Again, it cannot be stressed enough at this point that the only way to really learn how to do
substitutions is to just work lots of problems. There are lots of different kinds of problems and
after working many problems you'll start to get a real feel for these problems and after you work
enough of them you'll reach the point where you'll be able to do simple substitutions in your
head without having to actually write anything down.
3 2
3 10 30 u x du x dx = − = −
Okay, now we have a small problem. We've got an x
2
out in front of the parenthesis but we don't
have a "-30". This is not really the problem it might appear to be at first. We will simply rewrite
the differential as follows.
2
1
30
x dx du = −
With this we can now substitute the "x
2
dx" away. In the process we will pick up a constant, but
that isn't a problem since it can always be factored out of the integral.
In the previous set of examples the substitution was generally pretty clear. There was exactly one
term that had an "inside function" that we also couldn't integrate. Let's take a look at some more
complicated problems to make sure we don't come to expect all substitutions are like those in the
previous set of examples.
In this problem there are two "inside functions". There is the 1 x − that is inside the two trig
functions and there is also the term that is raised to the 4
th
power.
There are two ways to proceed with this problem. The first idea that many students have is
substitute the 1 x − away. There is nothing wrong with doing this but it doesn't eliminate the
problem of the term to the 4
th
power. That's still there and if we used this idea we would then
need to do a second substitution to deal with that.
The second (and much easier) way of doing this problem is to just deal with the stuff raised to the
4
th
power and see what we get. The substitution in this case would be,
As seen in this example sometimes there will seem to be two substitutions that will need to be
done however, if one of them is buried inside of another substitution then we'll only really need
to do one. Recognizing this can save a lot of time in working some of these problems.
[Return to Problems]
(b) ( ) ( )
10
cos 3 sin 3 z z dz
∫
This one is a little tricky at first. We can see the correct substitution by recalling that,
( ) ( ) ( )
10
10
sin 3 sin 3 z z =
Note that the one third in front of the integral came about from the substitution on the differential
and we just factored it out to the front of the integral. This is what we will usually do with these
constants.
[Return to Problems]
The most important thing to remember in substitution problems is that after the substitution all
the original variables need to disappear from the integral. After the substitution the only variables
that should be present in the integral should be the new variable from the substitution (usually u).
Note as well that this includes the variables in the differential!
This next set of examples, while not particular difficult, can cause trouble if we aren't paying
attention to what we're doing.
We haven't seen a problem quite like this one yet. Let's notice that if we take the denominator
and differentiate it we get just a constant and the only thing that we have in the numerator is also
a constant. This is a pretty good indication that we can use the denominator for our substitution
so,
Remember that we can just factor the 3 in the numerator out of the integral and that makes the
integral a little clearer in this case.
[Return to Problems]
(b)
2
3
5 4
y
dy
y +
⌠
⎮
⌡
The integral is very similar to the previous one with a couple of minor differences but notice that
again if we differentiate the denominator we get something that is different from the numerator
by only a multiplicative constant. Therefore we'll again take the denominator as our substitution.
Now, this one is almost identical to the previous part except we added a power onto the
denominator. Notice however that if we ignore the power and differentiate what's left we get the
same thing as the previous example so we'll use the same substitution here.
Be careful in this case to not turn this into a logarithm. After working problems like the first two
in this set a common error is to turn every rational expression into a logarithm. If there is a power
on the whole denominator then there is a good chance that it isn't a logarithm.
The idea that we used in the last three parts to determine the substitution is not a bad idea to
remember. If we've got a rational expression try differentiating the denominator (ignoring any
powers that are on the whole denominator) and if the result is the numerator or only differs from
the numerator by a multiplicative constant then we can usually use that as our substitution.
[Return to Problems]
(d)
2
3
5 4
dy
y +
⌠
⎮
⌡
Now, this part is completely different from the first three and yet seems similar to them as well.
In this case if we differentiate the denominator we get a y that is not in the numerator and so we
can't use the denominator as our substitution.
In fact, because we have y
2
in the denominator and no y in the numerator is an indication of how
to work this problem. This integral is going to be an inverse tangent when we are done. To key to
seeing this is to recall the following formula,
1
2
1
tan
1
du u c
u
−
= +
+
⌠
⎮
⌡
We clearly don't have exactly this but we do have something that is similar. The denominator
has a squared term plus a constant and the numerator is just a constant. So, with a little work and
the proper substitution we should be able to get our integral into a form that will allow us to use
this formula.
Notice that in the last step we rewrote things a little in the denominator. This will help us to see
what the substitution needs to be. In order to get this integral into the formula above we need to
end up with a u
2
in the denominator. Our substitution will then need to be something that upon
squaring gives us
2
5
4
y
. With the rewrite we can see what that we'll need to use the following
substitution.
5 5 2
2 2 5
y
u du dy dy du = = ⇒ =
Don't get excited about the root in the substitution, these will show up on occasion. Upon
plugging our substitution in we get,
This is a fairly common occurrence and so you will need to be able to deal with these kinds of
issues. There are many integrals that on the surface look very similar and yet will use a
completely different substitution or will yield a completely different answer when using the same
substitution.
Let's take a look at another set of examples to give us more practice in recognizing these kinds of
issues. Note however that we won't be putting as much detail into these as we did with the
previous examples.
The only difference between this problem and the previous one is the denominator. In the
previous problem the whole denominator is cubed and in this problem the denominator has no
power on it. The same substitution will work in this problem but because we no longer have the
power the problem will be different.
The integral in this problem is nearly this. The only difference is the presence of the coefficient
of 4 on the x
2
. With the correct substitution this can be dealt with however. To see what this
substitution should be let's rewrite the integral a little. We need to figure out what we squared to
get 4x
2
and that will be our substitution.
( )
2 2
1 1
1 4
1 2
dx dx
x
x
=
−
−
⌠
⌠
⎮ ⎮
⌡
⌡
With this rewrite it looks like we can use the following substitution.
Since this document is also being presented on the web we're going to put the rest of the
substitution rule examples in the next section. With all the examples in one section the section
was becoming too large for web presentation.
More Substitution Rule
In order to allow these pages to be displayed on the web we've broken the substitution rule
examples into two sections. The previous section contains the introduction to the substitution
rule and some fairly basic examples. The examples in this section tend towards the slightly more
difficult side. Also, we'll not be putting quite as much explanation into the solutions here as we
did in the previous section.
In the first couple of sets of problems in this section the difficulty is not with the actual
integration itself, but with the set up for the integration. Most of the integrals are fairly simple
and most of the substitutions are fairly simple. The problems arise in getting the integral set up
properly for the substitution(s) to be done. Once you see how these are done it's easy to see what
you have to do, but the first time through these can cause problems if you aren't on the lookout
for potential problems.
Example 1 Evaluate each of the following integrals.
(a) ( ) ( )
2
sec 2 tan 2
t
t t dt +
∫
e [Solution]
(b) ( ) ( ) ( ) ( )
3 2
sin 4cos 6cos 8 t t t dt + −
∫
[Solution]
(c)
( )
2
2
cos 1
1
x
x x dx
x
+ +
+
⌠
⎮
⌡
[Solution]
Solution
(a) ( ) ( )
2
sec 2 tan 2
t
t t dt +
∫
e
This first integral has two terms in it and both will require the same substitution. This means that
we won't have to do anything special to the integral. One of the more common "mistakes" here
is to break the integral up and do a separate substitution on each part. This isn't really mistake
but will definitely increase the amount of work we'll need to do. So, since both terms in the
integral use the same substitution we'll just do everything as a single integral using the following
substitution.
Often a substitution can be used multiple times in an integral so don't get excited about that if it
happens. Also note that since there was a
1
2
in front of the whole integral there must also be a
1
2
This integral is similar to the previous one, but it might not look like it at first glance. Here is the
substitution for this problem,
( ) ( ) ( ) cos sin sin u t du t dt t dt du = = − ⇒ = −
We'll plug the substitution into the problem twice (since there are two cosines) and will only
work because there is a sine multiplying everything. Without that sine in front we would not be
able to use this substitution.
Since each term had an x in it and we'll need that for the differential we factored that out of both
terms to get it into the front. This integral is now very similar to the previous one. Here's the
substitution.
Be careful with this kind of integral. One of the more common mistakes here is do the following
"shortcut".
2 1 2 x u
x dx x du
−
+ = − +
∫ ∫
e e
In other words, some students will try do the substitution just the second term without breaking
up the integral. There are two issues with this. First, there is a "-" in front of the whole integral
that shouldn't be there. It should only be on the second term because that is the term getting the
substitution. Secondly, and probably more importantly, there are x's in the integral and we have a
du for the differential. We can't mix variables like this. When we do integrals all the variables in
the integrand must match the variable in the differential.
[Return to Problems]
In this set of examples we saw that sometimes one (or potentially more than one) term in the
integrand will not require a substitution. In these cases we'll need to break up the integral into
two integrals, one involving the terms that don't need a substitution and another with the term(s)
that do need a substitution.
We've now seen a set of integrals in which we need to do more than one substitution. In these
cases we will need to break up the integral into separate integrals and do separate substitutions for
each.
We now need to move onto a different set of examples that can be a little tricky. Once you've
seen how to do these they aren't too bad, but doing them for the first time can be difficult if you
aren't ready for them.
The first question about this problem is probably why is it here? Substitution rule problems
generally require more than a single function. The key to this problem is to realize that there
really are two functions here. All we need to do is remember the definition of tangent and we can
write the integral as,
sin
tan
cos
x
x dx dx
x
=
⌠
⎮
⌡
∫
Written in this way we can see that the following substitution will work for us,
cos sin sin u x du x dx x dx du = = − ⇒ = −
The integral is then,
1
tan
ln
ln cos
x dx du
u
u c
x c
= −
= − +
= − +
⌠
⎮
⌡
∫
Now, while this is a perfectly serviceable answer that minus sign in front is liable to cause
problems if we aren't careful. So, let's rewrite things a little. Recalling a property of logarithms
we can move the minus sign in front to an exponent on the cosine and then do a little
simplification.
This is the formula that is typically given for the integral of tangent.
Note that we could integrate cotangent in a similar manner.
[Return to Problems]
(b) sec y dy
∫
This problem also at first appears to not belong in the substitution rule problems. This is even
more of a problem upon noticing that we can't just use the definition of the secant function to
write this in a form that will allow the use of the substitution rule.
This problem is going to require a technique that isn't used terribly often at this level, but is a
useful technique to be aware of. Sometimes we can make an integral doable by multiplying the
top and bottom by a common term. This will not always work and even when it does it is not
always clear what we should multiply by but when it works it is very useful.
Here is how we'll use this idea for this problem.
( )
( )
sec tan
sec
sec
1 sec tan
y y
y
y dy dy
y y
+
=
+
⌠
⎮
⌡
∫
First, we will think of the secant as a fraction and then multiply the top and bottom of the fraction
by the same term. It is probably not clear why one would want to do this here but doing this will
actually allow us to use the substitution rule. To see how this will work let's simplify the
integrand somewhat.
2
sec tan sec
sec
sec tan
y y y
y dy dy
y y
+
=
+
⌠
⎮
⌡
∫
We can now use the following substitution.
( )
2
sec tan sec tan sec u y y du y y y dy = + = +
The integral is then,
1
sec
ln
ln sec tan
y dy du
u
u c
y y c
=
= +
= + +
⌠
⎮
⌡
∫
Sometimes multiplying the top and bottom of a fraction by a carefully chosen term will allow us
to work a problem. It does however take some thought sometimes to determine just what the
term should be.
Some substitutions can be really tricky to see and it's not unusual that you'll need to do some
simplification and/or rewriting to get a substitution to work.
[Return to Problems]
(e)
3 2
2 1 x x dx +
∫
This last problem in this set is different from all the other substitution problems that we've
worked to this point. Given the fact that we've got more than an x under the root it makes sense
that the substitution pretty much has to be,
2
1 2 u x du x dx = + =
However, if we use this substitution we will get the following,
( )
3 2 2 2
1
2
2
2 1 1 2 x x dx x x x dx
x u du
+ = +
=
∫ ∫
∫
This is a real problem. Our integrals can't have two variables in them. Normally this would
mean that we chose our substitution incorrectly. However, in this case we can write the
substitution as follows,
2
1 x u = −
and now, we can eliminate the remaining x's from our integral. Doing this gives,
This kind of problem doesn't arise all that often and when it does there will sometimes be
alternate methods of doing the integral. However, it will often work out that the easiest method
of doing the integral is to do what we just did here.
[Return to Problems]
This final set of examples isn't too bad once you see the substitutions and that is the point with
this set of problems. These all involve substitutions that we've not seen prior to this and so we
need to see some of these kinds of problems.
In this case we know that we can't integrate a logarithm by itself and so it makes some sense
(hopefully) that the logarithm will need to be in the substitution. Here is the substitution for this
problem.
1
ln u x du dx
x
= =
So the x in the denominator of the integrand will get substituted away with the differential. Here
is the integral for this problem.
1 1
ln
ln
ln ln
dx du
x x u
u c
x c
=
= +
= +
⌠
⌠
⎮
⎮
⌡
⌡
[Return to Problems]
(b)
2
2
1
t
t
dt
+
⌠
⎮
⌡
e
e
Again, the substitution here may seem a little tricky. In this case the substitution is,
In this case we can't use the same type of substitution that we used in the previous problem. In
order to use the substitution in the previous example the exponential in the numerator and the
denominator need to be the same and in this case they aren't.
This integral is similar to the first problem in this set. Since we don't know how to integrate
inverse sine functions it seems likely that this will be our substitution. If we use this as our
substitution we get,
( )
1
2
1
sin
1
u x du dx
x
−
= =
−
Over the last couple of sections we've seen a lot of substitution rule examples. There are a
couple of general rules that we will need to remember when doing these problems. First, when
doing a substitution remember that when the substitution is done all the x's in the integral (or
whatever variable is being used for that particular integral) should all be substituted away. This
includes the x in the dx. After the substitution only u's should be left in the integral. Also,
sometimes the correct substitution is a little tricky to find and more often than not there will need
to be some manipulation of the differential or integrand in order to actually do the substitution.
Also, many integrals will require us to break them up so we can do multiple substitutions so be on
the lookout for those kinds of integrals/substitutions.
Area Problem
As noted in the first section of this section there are two kinds of integrals and to this point we've
looked at indefinite integrals. It is now time to start thinking about the second kind of integral :
Definite Integrals. However, before we do that we're going to take a look at the Area Problem.
The area problem is to definite integrals what the tangent and rate of change problems are to
derivatives.
The area problem will give us one of the interpretations of a definite integral and it will lead us to
the definition of the definite integral.
To start off we are going to assume that we've got a function ( ) f x that is positive on some
interval [a,b]. What we want to do is determine the area of the region between the function and
the x-axis.
It's probably easiest to see how we do this with an example. So let's determine the area between
( )
2
1 f x x = + on [0,2]. In other words, we want to determine the area of the shaded region
below.
Now, at this point, we can't do this exactly. However, we can estimate the area. We will
estimate the area by dividing up the interval into n subintervals each of width,
b a
x
n
−
Δ =
Then in each interval we can form a rectangle whose height is given by the function value at a
specific point in the interval. We can then find the area of each of these rectangles, add them up
and this will be an estimate of the area.
Note that by choosing the height as we did each of the rectangles will over estimate the area since
each rectangle takes in more area than the graph each time. Now let's estimate the area. First,
the width of each of the rectangles is
1
2
. The height of each rectangle is determined by the
function value at the right endpoint and so the height of each rectangle is nothing more that the
function value at the right endpoint. Here is the estimated area.
Of course taking the rectangle heights to be the function value at the right endpoint is not our only
option. We could have taken the rectangle heights to be the function value at the left endpoint.
Using the left endpoints as the heights of the rectangles will give the following graph and
estimated area.
In this case we can see that the estimation will be an underestimation since each rectangle misses
some of the area each time.
There is one more common point for getting the heights of the rectangles that is often more
accurate. Instead of using the right or left endpoints of each sub interval we could take the
midpoint of each subinterval as the height of each rectangle. Here is the graph for this case.
So, it looks like each rectangle will over and under estimate the area. This means that the
approximation this time should be much better than the previous two choices of points. Here is
the estimation for this case.
So, both the right and left endpoint estimation did not do all that great of a job at the estimation.
The midpoint estimation however did quite well.
Be careful to not draw any conclusion about how choosing each of the points will affect our
estimation. In this case, because we are working with an increasing function choosing the right
endpoints will overestimate and choosing left endpoint will underestimate.
If we were to work with a decreasing function we would get the opposite results. For decreasing
functions the right endpoints will underestimate and the left endpoints will overestimate.
Also, if we had a function that both increased and decreased in the interval we would, in all
likelihood, not even be able to determine if we would get an overestimation or underestimation.
Now, let's suppose that we want a better estimation, because none of the estimations above really
did all that great of a job at estimating the area. We could try to find a different point to use for
the height of each rectangle but that would be cumbersome and there wouldn't be any guarantee
that the estimation would in fact be better. Also, we would like a method for getting better
approximations that would work for any function we would chose to work with and if we just
pick new points that may not work for other functions.
The easiest way to get a better approximation is to take more rectangles (i.e. increase n). Let's
double the number of rectangles that we used and see what happens. Here are the graphs showing
the eight rectangles and the estimations for each of the three choices for rectangle heights that we
used above.
Notice, that unlike the first area we looked at, the choosing the right endpoints here will both over
and underestimate the area depending on where we are on the curve. This will often be the case
with a more general curve that the one we initially looked at. The area estimation using the right
endpoints of each interval for the rectangle height is,
We will use summation notation or sigma notation at this point to simplify up our notation a
little. If you need a refresher on summation notation check out the section devoted to this in the
Extras chapter.
Using summation notation the area estimation is,
( )
*
1
n
i
i
A f x x
=
≈ Δ
∑
The summation in the above equation is called a Riemann Sum.
To get a better estimation we will take n larger and larger. In fact, if we let n go out to infinity we
will get the exact area. In other words,
of ( )
2
4 f x x = − on [0,2]. If we use 8 n = and the midpoints for the rectangle height we get the
following graph,
In this case let's notice that the function lies completely below the x-axis and hence is always
negative. If we ignore the fact that the function is always negative and use the same ideas above
to estimate the area between the graph and the x-axis we get,
Our answer is negative as we might have expected given that all the function evaluations are
negative.
So, using the technique in this section it looks like if the function is above the x-axis we will get a
positive area and if the function is below the x-axis we will get a negative area. Now, what about
a function that is both positive and negative in the interval? For example, ( )
2
2 f x x = − on
[0,2]. Using 8 n = and midpoints the graph is,
Some of the rectangles are below the x-axis and so will give negative areas while some are above
the x-axis and will give positive areas. Since more rectangles are below the x-axis than above it
looks like we should probably get a negative area estimation for this case. In fact that is correct.
Here the area estimation for this case.
In cases where the function is both above and below the x-axis the technique given in the section
will give the net area between the function and the x-axis with areas below the x-axis negative
and areas above the x-axis positive. So, if the net area is negative then there is more area under
the x-axis than above while a positive net area will mean that more of the area is above the x-axis.
The Definition of the Definite Integral
In this section we will formally define the definite integral and give many of the properties of
definite integrals. Let's start off with the definition of a definite integral.
Definite Integral
Given a function ( ) f x that is continuous on the interval [a,b] we divide the interval into n
subintervals of equal width, x Δ , and from each interval choose a point,
*
i
x . Then the definite
integral of f(x) from a to b is
( ) ( )
*
1
lim
n
b
i
a n
i
f x dx f x x
→∞
=
= Δ
∑
∫
The definite integral is defined to be exactly the limit and summation that we looked at in the last
section to find the net area between a function and the x-axis. Also note that the notation for the
definite integral is very similar to the notation for an indefinite integral. The reason for this will
be apparent eventually.
There is also a little bit of terminology that we should get out of the way here. The number "a"
that is at the bottom of the integral sign is called the lower limit of the integral and the number
"b" at the top of the integral sign is called the upper limit of the integral. Also, despite the fact
that a and b were given as an interval the lower limit does not necessarily need to be smaller than
the upper limit. Collectively we'll often call a and b the interval of integration.
Let's work a quick example. This example will use many of the properties and facts from the
brief review of summation notation in the Extras chapter.
Example 1 Using the definition of the definite integral compute the following.
2
2
0
1 x dx +
∫
Solution
First, we can't actually use the definition unless we determine which points in each interval that
well use for
*
i
x . In order to make our life easier we'll use the right endpoints of each interval.
From the previous section we know that for a general n the width of each subinterval is,
Now, we are going to have to take a limit of this. That means that we are going to need to
"evaluate" this summation. In other words, we are going to have to use the formulas given in the
summation notation review to eliminate the actual summation and get a formula for this for a
general n.
To do this we will need to recognize that n is a constant as far as the summation notation is
concerned. As we cycle through the integers from 1 to n in the summation only i changes and so
anything that isn't an i will be a constant and can be factored out of the summation. In particular
any n that is in the summation can be factored out if we need to.
We've seen several methods for dealing with the limit in this problem so I'll leave it to you to
verify the results.
Wow, that was a lot of work for a fairly simple function. There is a much simpler way of
evaluating these and we will get to it eventually. The main purpose to this section is to get the
main properties and facts about the definite integral out of the way. We'll discuss how we
compute these in practice starting with the next section.
So, let's start taking a look at some of the properties of the definite integral.
Properties
1. ( ) ( )
b a
a b
f x dx f x dx = −
∫ ∫
. We can interchange the limits on any definite integral, all that
we need to do is tack a minus sign onto the integral when we do.
2. ( ) 0
a
a
f x dx =
∫
. If the upper and lower limits are the same then there is no work to do, the
integral is zero.
3. ( ) ( )
b b
a a
cf x dx c f x dx =
∫ ∫
, where c is any number. So, as with limits, derivatives, and
indefinite integrals we can factor out a constant.
5. ( ) ( ) ( )
b c b
a a c
f x dx f x dx f x dx = +
∫ ∫ ∫
where c is any number. This property is more
important than we might realize at first. One of the main uses of this property is to tell us
how we can integrate a function over the adjacent intervals, [a,c] and [c,b]. Note however
that c doesn't need to be between a and b.
6. ( ) ( )
b b
a a
f x dx f t dt =
∫ ∫
. The point of this property is to notice that as long as the function
and limits are the same the variable of integration that we use in the definite integral won't
affect the answer.
See the Proof of Various Integral Properties section of the Extras chapter for the proof of
properties 1 – 4. Property 5 is not easy to prove and so is not shown there. Property 6 is not
really a property in the full sense of the word. It is only here to acknowledge that as long as the
function and limits are the same it doesn't matter what letter we use for the variable. The answer
will be the same.
Now notice that the limits on the first integral are interchanged with the limits on the given
integral so switch them using the first property above (and adding a minus sign of course). Once
this is done we can plug in the known values of the integrals.
Solution
This example is mostly an example of property 5 although there are a couple of uses of property 1
in the solution as well.
We need to figure out how to correctly break up the integral using property 5 to allow us to use
the given pieces of information. First we'll note that there is an integral that has a "-5" in one of
the limits. It's not the lower limit, but we can use property 1 to correct that eventually. The other
limit is 100 so this is the number c that we'll use in property 5.
( ) ( ) ( )
12 100 12
5 5 100
f x dx f x dx f x dx
− −
= +
∫ ∫ ∫
We'll be able to get the value of the first integral, but the second still isn't in the list of know
integrals. However, we do have second limit that has a limit of 100 in it. The other limit for this
second integral is -10 and this will be c in this application of property 5.
( ) ( ) ( ) ( )
12 100 10 12
5 5 100 10
f x dx f x dx f x dx f x dx
−
− − −
= + +
∫ ∫ ∫ ∫
At this point all that we need to do is use the property 1 on the first and third integral to get the
limits to match up with the known integrals. After that we can plug in for the known integrals.
See the Proof of Various Integral Properties section of the Extras chapter for the proof of these
properties.
Interpretations of Definite Integral
There are a couple of quick interpretations of the definite integral that we can give here.
First, as we alluded to in the previous section one possible interpretation of the definite integral is
to give the net area between the graph of ( ) f x and the x-axis on the interval [a,b]. So, the net
area between the graph of ( )
2
1 f x x = + and the x-axis on [0,2] is,
2
2
0
14
1
3
x dx + =
∫
If you look back in the last section this was the exact area that was given for the initial set of
problems that we looked at in this area.
is the net change in ( ) f x on the interval [a,b]. In other words, compute the definite integral of
a rate of change and you'll get the net change in the quantity. We can see that the value of the
definite integral, ( ) ( ) f b f a − , does in fact give use the net change in ( ) f x and so there really
isn't anything to prove with this statement. This is really just an acknowledgment of what the
definite integral of a rate of change tells us.
It is important to note here that the Net Change Theorem only really makes sense if we're
integrating a derivative of a function.
Fundamental Theorem of Calculus, Part I
As noted by the title above this is only the first part to the Fundamental Theorem of Calculus.
We will give the second part in the next section as it is the key to easily computing definite
integrals and that is the subject of the next section.
The first part of the Fundamental Theorem of Calculus tells us how to differentiate certain types
of definite integrals and it also tells us about the very close relationship between integrals and
derivatives.
This one needs a little work before we can use the Fundamental Theorem of Calculus. The first
thing to notice is that the FToC requires the lower limit to be a constant and the upper limit to be
the variable. So, using a property of definite integrals we can interchange the limits of the
integral we just need to remember to add in a minus sign after we do that. Doing this gives,
The next thing to notice is that the FToC also requires an x in the upper limit of integration and
we've got x
2
. To do this derivative we're going to need the following version of the chain rule.
( ) ( ) ( ) ( ) ( ) where
d d du
g u g u u f x
dx du dx
= =
Next, we can get a formula for integrals in which the upper limit is a constant and the lower limit
is a function of x. All we need to do here is interchange the limits on the integral (adding in a
minus sign of course) and then using the formula above to get,
We can use pretty much any value of a when we break up the integral. The only thing that we
need to avoid is to make sure that ( ) f a exists. So, assuming that ( ) f a exists after we break
up the integral we can then differentiate and use the two formulas above to get,
Computing Definite Integrals
In this section we are going to concentrate on how we actually evaluate definite integrals in
practice. To do this we will need the Fundamental Theorem of Calculus, Part II.
To see the proof of this see the Proof of Various Integral Properties section of the Extras chapter.
Recall that when we talk about an anti-derivative for a function we are really talking about the
indefinite integral for the function. So, to evaluate a definite integral the first thing that we're
going to do is evaluate the indefinite integral for the function. This should explain the similarity
in the notations for the indefinite and definite integrals.
Also notice that we require the function to be continuous in the interval of integration. This was
also a requirement in the definition of the definite integral. We didn't make a big deal about this
in the last section. In this section however, we will need to keep this condition in mind as we do
our evaluations.
Next let's address the fact that we can use any anti-derivative of ( ) f x in the evaluation. Let's
take a final look at the following integral.
The constant that we tacked onto the second anti-derivative canceled in the evaluation step. So,
when choosing the anti-derivative to use in the evaluation process make your life easier and don't
bother with the constant as it will only end up canceling in the long run.
Also, note that we're going to have to be very careful with minus signs and parenthesis with these
problems. It's very easy to get in a hurry and mess them up.
Let's start our examples with the following set designed to make a couple of quick points that are
very important.
This is the only indefinite integral in this section and by now we should be getting pretty good
with these so we won't spend a lot of time on this part. This is here only to make sure that we
understand the difference between an indefinite and a definite integral. The integral is,
Remember that the evaluation is always done in the order of evaluation at the upper limit minus
evaluation at the lower limit. Also be very careful with minus signs and parenthesis. It's very
easy to forget them or mishandle them and get the wrong answer.
Notice as well that, in order to help with the evaluation, we rewrote the indefinite integral a little.
In particular we got rid of the negative exponent on the second term. It's generally easier to
evaluate the term with positive exponents.
[Return to Problems]
(c)
2
2 2
1
y y dy
−
−
+
∫
This integral is here to make a point. Recall that in order for us to do an integral the integrand
must be continuous in the range of the limits. In this case the second term will have division by
zero at 0 y = and since 0 y = is in the interval of integration, i.e. it is between the lower and
upper limit, this integrand is not continuous in the interval of integration and so we can't do this
integral.
Note that this problem will not prevent us from doing the integral in (b) since 0 y = is not in the
interval of integration.
[Return to Problems]
So what have we learned from this example?
First, in order to do a definite integral the first thing that we need to do is the indefinite integral.
So we aren't going to get out of doing indefinite integrals, they will be in every integral that we'll
be doing in the rest of this course so make sure that you're getting good at computing them.
point(s) of discontinuity occur between the limits of integration or at the limits themselves. If the
point of discontinuity occurs outside of the limits of integration the integral can still be evaluated.
In the following sets of examples we won't make too much of an issue with continuity problems,
or lack of continuity problems, unless it affects the evaluation of the integral. Do not let this
convince you that you don't need to worry about this idea. It arises often enough that it can cause
real problems if you aren't on the lookout for it.
Finally, note the difference between indefinite and definite integrals. Indefinite integrals are
functions while definite integrals are numbers.
Also, don't get excited about the fact that the lower limit of integration is larger than the upper
limit of integration. That will happen on occasion and there is absolutely nothing wrong with
this.
[Return to Problems]
(c)
2
5
2
1
2 3 w w
dw
w
− +
⌠
⎮
⌡
First, notice that we will have a division by zero issue at 0 w = , but since this isn't in the interval
of integration we won't have to worry about it.
Next again recall that we can't integrate quotients as a quotient of integrals and so the first step
that we'll need to do is break up the quotient so we can integrate the function.
Compare this answer to the previous answer, especially the evaluation at zero. It's very easy to
get into the habit of just writing down zero when evaluating a function at zero. This is especially
a problem when many of the functions that we integrate involve only x's raised to positive
integers and in these cases evaluate is zero of course. After evaluating many of these kinds of
definite integrals it's easy to get into the habit of just writing down zero when you evaluate at
zero. However, there are many functions out there that aren't zero when evaluated at zero so be
careful.
[Return to Problems]
This integral can't be done. There is division by zero in the third term at 0 t = and 0 t = lies in
the interval of integration. The fact that the first two terms can be integrated doesn't matter. If
even one term in the integral can't be integrated then the whole integral can't be done.
[Return to Problems]
So, we've computed a fair number of definite integrals at this point. Remember that the vast
majority of the work in computing them is first finding the indefinite integral. Once we've found
that the rest is just some number crunching.
There are a couple of particularly tricky definite integrals that we need to take a look at next.
Actually they are only tricky until you see how to do them, so don't get too excited about them.
The first one involves integrating a piecewise function.
The graph reveals a problem. This function is not continuous at 1 x = and we're going to have to
watch out for that.
(a) ( )
22
10
f x dx
∫
For this integral notice that 1 x = is not in the interval of integration and so that is something that
we'll not need to worry about in this part.
Also note the limits for the integral lie entirely in the range for the first function. What this
means for us is that when we do the integral all we need to do is plug in the first function into the
integral.
In this part 1 x = is between the limits of integration. This means that the integrand is no longer
continuous in the interval of integration and that is a show stopper as far we're concerned. As
noted above we simply can't integrate functions that aren't continuous in the interval of
integration.
Let's first address the problem of the function not beginning continuous at 1 x = . As we'll see,
in this case, if we can find a way around this problem the second problem will also get taken care
of at the same time.
In the previous examples where we had functions that weren't continuous we had division by
zero and no matter how hard we try we can't get rid of that problem. Division by zero is a real
problem and we can't really avoid it. In this case the discontinuity does not stem from problems
with the function not existing at 1 x = . Instead the function is not continuous because it takes on
different values on either sides of 1 x = . We can "remove" this problem by recalling Property 5
from the previous section. This property tells us that we can write the integral as follows,
( ) ( ) ( )
3 1 3
2 2 1
f x dx f x dx f x dx
− −
= +
∫ ∫ ∫
On each of these intervals the function is continuous. In fact we can say more. In the first
integral we will have x between -2 and 1 and this means that we can use the second equation for
( ) f x and likewise for the second integral x will be between 1 and 3 and so we can use the first
function for ( ) f x . The integral in this case is then,
So, to integrate a piecewise function, all we need to do is break up the integral at the break
point(s) that happen to occur in the interval of integration and then integrate each piece.
Next we need to look at is how to integrate an absolute value function.
Example 5 Evaluate the following integral.
3
0
3 5 t dt −
∫
Solution
Recall that the point behind indefinite integration (which we'll need to do in this problem) is to
determine what function we differentiated to get the integrand. To this point we've not seen any
functions that will differentiate to get an absolute value nor will we ever see a function that will
differentiate to get an absolute value.
The only way that we can do this problem is to get rid of the absolute value. To do this we need
to recall the definition of absolute value.
if 0
if 0
x x
x
x x
≥ ⎧
=
⎨
− <
⎩
Once we remember that we can define absolute value as a piecewise function we can use the
work from Example 4 as a guide for doing this integral.
What we need to do is determine where the quantity on the inside of the absolute value bars is
negative and where it is positive. It looks like if
5
3
t > the quantity inside the absolute value is
positive and if
5
3
t < the quantity inside the absolute value is negative.
Next, note that
5
3
t = is in the interval of integration and so, if we break up the integral at this
point we get,
5
3 3
3
5
0 0
3
3 5 3 5 3 5 t dt t dt t dt − = − + −
∫ ∫ ∫
Now, in the first integrals we have
5
3
t < and so 3 5 0 t − < in this interval of integration. That
means we can drop the absolute value bars if we put in a minus sign. Likewise in the second
integral we have
5
3
t > which means that in this interval of integration we have 3 5 0 t − > and so
we can just drop the absolute value bars in this integral.
Integrating absolute value functions isn't too bad. It's a little more work than the "standard"
definite integral, but it's not really all that much more work. First, determine where the quantity
inside the absolute value bars is negative and where it is positive. When we've determined that
point all we need to do is break up the integral so that in each range of limits the quantity inside
the absolute value bars is always positive or always negative. Once this is done we can drop the
absolute value bars (adding negative signs when the quantity is negative) and then we can do the
integral as we've always done.
Substitution Rule for Definite Integrals
We now need to go back and revisit the substitution rule as it applies to definite integrals. At
some level there really isn't a lot to do in this section. Recall that the first step in doing a definite
integral is to compute the indefinite integral and that hasn't changed. We will still compute the
indefinite integral first. This means that we already know how to do these. We use the
substitution rule to find the indefinite integral and then do the evaluation.
There are however, two ways to deal with the evaluation step. One of the ways of doing the
evaluation is the probably the most obvious at this point, but also has a point in the process where
we can get in trouble if we aren't paying attention.
Let's work an example illustrating both ways of doing the evaluation step.
Example 1 Evaluate the following definite integral.
0
2 3
2
2 1 4 t t dt
−
−
∫
Solution
Let's start off looking at the first way of dealing with the evaluation step. We'll need to be
careful with this method as there is a point in the process where if we aren't paying attention
we'll get the wrong answer.
Solution 1 :
We'll first need to compute the indefinite integral using the substitution rule. Note however, that
we will constantly remind ourselves that this is a definite integral by putting the limits on the
integral at each step. Without the limits it's easy to forget that we had a definite integral when
we've gotten the indefinite integral computed.
Notice that we didn't do the evaluation yet. This is where the potential problem arises with this
solution method. The limits given here are from the original integral and hence are values of t.
We have u's in our solution. We can't plug values of t in for u.
in the substitution process, but it is often forgotten when doing definite integrals. Note as well
that in this case, if we don't go back to t's we will have a small problem in that one of the
evaluations will end up giving us a complex nu | 677.169 | 1 |
Fall Review - Variables and Expressions
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Algebra that Functions
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Section 1.2 Translating Verbal Phrases into Mathematical Operations
A1.1.6 Simplify numerical expressions, including those involving radicals and absolute value; A1.1.4 Solve simple equations in one variable using inverse relationships between operations such as addition and subtraction (taking the opposite), multiplication and division (multiplying by the reciprocal), raising to a power and taking a root; A1.3.6 Represent linear relationships graphically, algebraically (including the slope-intercept form) and verbally and relate a change in the slope or the y-intercept to its effect on the various representations; A1.3.11 Apply and use linear equations and/or inequalities as mathematical representations of proportional relationships to solve problem; including rate problems, work problems, and percent mixture problems; | 677.169 | 1 |
Find an EncinoThe math sections measure a student's ability to reason quantitatively, solve mathematical problems, and interpret data presented in graphical form. These sections focus on four areas of mathematics that are typically covered in the first three years of American high school education: Arithmetic | 677.169 | 1 |
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Australian Signpost Maths 3 has been updated and redesigned to reflect current best practice in the teaching and learning of maths concepts, activities and digital techlogy. Written by Alan McSeveny and his experienced author team, Australian Signpost Maths provides a complete year's work and addresses all aspects of the Australian Mathematics Curriculum, including content and proficiency strands. All activities in this student book have been matched to the Australian Curriculum content strands and develop students' conceptual understanding, logical reasoning and problem solving. Worked examples and explanations are given throughout the student book where new ideas are introduced, and a maths pictorial dictionary is provided to help students digest new vocabulary. Open-ended problem solving and inquiry-based investigations and activities are designed to meet differentiated needs and learning styles. Exercises are carefully graded and colour-coded by content strand. | 677.169 | 1 |
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Algebra is much broader than elementary algebra and can be generalized. In addition to working directly with numbers, algebra covers working with symbols, variables, and setelements. Addition and multiplication are viewed as general operations, and their precise definitions lead to structures such as groups, ringsfields. and
The Greek mathematicians Hero of Alexandria and Diophantus[2] continued the traditions of Egypt and Babylon, but Diophantus's book Arithmetica is on a much higher level.[3] Later, Arab and Muslim mathematicians developed algebraic methods to a much higher degree of sophistication. Although Diophantus and the Babylonians used mostly special ad hoc methods to solve equations, Al-Khowarazmi was the first to solve equations using general methods. He solved the linear indeterminate equations, quadratic equations, second order indeterminate equations and equations with multiple variable.
The word "algebra" is named after the Arabic word "al-jabr , الجبر" from the title of the book al-Kitāb al-muḫtaṣar fī ḥisāb al-ğabr wa-l-muqābala , الكتاب المختصر في حساب الجبر والمقابلة, meaning The book of Summary Concerning Calculating by Transposition and Reduction, a book written by the Islamic Persian mathematician, Muhammad ibn Mūsā al-Khwārizmī (considered the "father of algebra"), in 820. The word Al-Jabr means "reunion". The Hellenistic mathematician Diophantus has traditionally been known as the "father of algebra" but in more recent times there is much debate over whether al-Khwarizmi, who founded the discipline of al-jabr, deserves that title instead.[4] Those who support Diophantus point to the fact that the algebra found in Al-Jabr is slightly more elementary than the algebra found in Arithmetica and that Arithmetica is syncopated while Al-Jabr is fully rhetorical.[5] Those who support Al-Khwarizmi point to the fact that he introduced the methods of "reduction" and "balancing" (the transposition of subtracted terms to the other side of an equation, that is, the cancellation of like terms on opposite sides of the equation) which the term al-jabr originally referred to,[6] and that he gave an exhaustive explanation of solving quadratic equations,[7] supported by geometric proofs, while treating algebra as an independent discipline in its own right.[8] His algebra was also no longer concerned "with a series of problems to be resolved, but an exposition which starts with primitive terms in which the combinations must give all possible prototypes for equations, which henceforward explicitly constitute the true object of study." He also studied an equation for its own sake and "in a generic manner, insofar as it does not simply emerge in the course of solving a problem, but is specifically called on to define an infinite class of problems."[9]
In some directions of advanced study, axiomatic algebraic systems such as groups, rings, fields, and algebras over a field are investigated in the presence of a geometric structure (a metric or a topology) which is compatible with the algebraic structure. The list includes a number of areas of functional analysis:
Elementary algebra
Elementary algebra is the most basic form of algebra. It is taught to students who are presumed to have no knowledge of mathematics beyond the basic principles of arithmetic. In arithmetic, only numbers and their arithmetical operations (such as +, −, ×, ÷) occur. In algebra, numbers are often denoted by symbols (such as a, x, or y). This is useful because:
It allows the general formulation of arithmetical laws (such as a + b = b + aa and b), and thus is the first step to a systematic exploration of the properties of the real number system. for all
It allows the reference to "unknown" numbers, the formulation of equationsx such that 3x + 1 = 10"). and the study of how to solve these (for instance, "Find a number
It allows the formulation of functional relationships (such as "If you sell xx − 10 dollars, or f(x) = 3x − 10, where f is the function, and x is the number to which the function is applied."). tickets, then your profit will be 3
Polynomials
A polynomial is an expression that is constructed from one or more variables and constants, using only the operations of addition, subtraction, and multiplication (where repeated multiplication of the same variable is standardly denoted as exponentiation with a constant non-negative whole number exponent). For example, x2 + 2x − 3 is a polynomial in the single variable x.
An important class of problems in algebra is factorization of polynomials, that is, expressing a given polynomial as a product of other polynomials. The example polynomial above can be factored as (x − 1)(x + 3). A related class of problems is finding algebraic expressions for the roots of a polynomial in a single variable.
Abstract algebra
Abstract algebra extends the familiar concepts found in elementary algebra and arithmetic of numbers to more general concepts.
Sets: Rather than just considering the different types of numbers, abstract algebra deals with the more general concept of sets: a collection of all objects (called elements) selected by property, specific for the set. All collections of the familiar types of numbers are sets. Other examples of sets include the set of all two-by-two matrices, the set of all second-degree polynomials (ax2 + bx + c), the set of all two dimensional vectors in the plane, and the various finite groups such as the cyclic groups which are the group of integers modulon. Set theory is a branch of logic and not technically a branch of algebra.
Binary operations: The notion of addition (+) is abstracted to give a binary operation, ∗ say. The notion of binary operation is meaningless without the set on which the operation is defined. For two elements a and b in a set S, a ∗ b is another element in the set; this condition is called closure. Addition (+), subtraction (-), multiplication (×), and division (÷) can be binary operations when defined on different sets, as is addition and multiplication of matrices, vectors, and polynomials.
Identity elements: The numbers zero and one are abstracted to give the notion of an identity element for an operation. Zero is the identity element for addition and one is the identity element for multiplication. For a general binary operator ∗ the identity element e must satisfy a ∗ e = a and e ∗ a = a. This holds for addition as aa and 0 + a = a and multiplication a × 1 = a and 1 × a = a. Not all set and operator combinations have an identity element; for example, the positive natural numbers (1, 2, 3, ...) have no identity element for addition. + 0 =
Inverse elements: The negative numbers give rise to the concept of inverse elements. For addition, the inverse of a is −a, and for multiplication the inverse is 1/a. A general inverse element a−1 must satisfy the property that a ∗ a−1 = e and a−1 ∗ a = e.
Associativity: Addition of integers has a property called associativity. That is, the grouping of the numbers to be added does not affect the sum. For example: (2 + 3) + 4 = 2 + (3 + 4). In general, this becomes (a ∗ b) ∗ c = a ∗ (b ∗ c). This property is shared by most binary operations, but not subtraction or division or octonion multiplication.
Commutativity: Addition of integers also has a property called commutativity. That is, the order of the numbers to be added does not affect the sum. For example: 2+3=3+2. In general, this becomes a ∗ b = b ∗ a. Only some binary operations have this property. It holds for the integers with addition and multiplication, but it does not hold for matrix multiplication or quaternion multiplication .
Groups – structures of a set with a single binary operation
Combining the above concepts gives one of the most important structures in mathematics: a group. A group is a combination of a set S and a single binary operation ∗, defined in any way you choose, but with the following properties:
An identity element e exists, such that for every member a of S, e ∗ a and a ∗ ea. are both identical to
Every element has an inverse: for every member a of S, there exists a member a−1 such that a ∗ a−1 and a−1 ∗ a are both identical to the identity element.
The operation is associative: if a, b and c are members of S, then (a ∗ b) ∗ c is identical to a ∗ (b ∗ c).
If a group is also commutative—that is, for any two members a and b of S, a ∗ b is identical to b ∗ a—then the group is said to be Abelian.
For example, the set of integers under the operation of addition is a group. In this group, the identity element is 0 and the inverse of any element a is its negation, −a. The associativity requirement is met, because for any integers a, b and c, (a + b) + c = a + (b + c)
The nonzero rational numbers form a group under multiplication. Here, the identity element is 1, since 1 × a = a × 1 = a for any rational number a. The inverse of a is 1/a, since a × 1/a = 1.
The integers under the multiplication operation, however, do not form a group. This is because, in general, the multiplicative inverse of an integer is not an integer. For example, 4 is an integer, but its multiplicative inverse is ¼, which is not an integer.
Semigroups, quasigroups, and monoids are structures similar to groups, but more general. They comprise a set and a closed binary operation, but do not necessarily satisfy the other conditions. A semigroup has an associative binary operation, but might not have an identity element. A monoid is a semigroup which does have an identity but might not have an inverse for every element. A quasigroup satisfies a requirement that any element can be turned into any other by a unique pre- or post-operation; however the binary operation might not be associative.
All groups are monoids, and all monoids are semigroups.
Rings and fields—structures of a set with two particular binary operations, (+) and (×)
Groups just have one binary operation. To fully explain the behaviour of the different types of numbers, structures with two operators need to be studied. The most important of these are rings, and fields.
Distributivity generalised the distributive law for numbers, and specifies the order in which the operators should be applied, (called the precedence). For the integers (a + b) × c = a × c + b × c and c × (a + b) = c × a + c × b, and × is said to be distributive over +.
A ring has two binary operations (+) and (×), with × distributive over +. Under the first operator (+) it forms an Abelian group. Under the second operator (×) it is associative, but it does not need to have identity, or inverse, so division is not allowed. The additive (+) identity element is written as 0 and the additive inverse of a is written as −a.
The integers are an example of a ring. The integers have additional properties which make it an integral domain.
A field is a ring with the additional property that all the elements excluding 0 form an Abelian group under ×. The multiplicative (×) identity is written as 1 and the multiplicative inverse of a is written as a−1.
The rational numbers, the real numbers and the complex numbers are all examples of fields.
^(Boyer 1991, "The Arabic Hegemony" p. 229) "It is not certain just what the terms al-jabr and muqabalah mean, but the usual interpretation is similar to that implied in the translation above. The word al-jabr presumably meant something like "restoration" or "completion" and seems to refer to the transposition of subtracted terms to the other side of an equation; the word muqabalah is said to refer to "reduction" or "balancing" - that is, the cancellation of like terms on opposite sides of the equation."
^(Boyer 1991, "The Arabic Hegemony" p. 230) "The six cases of equations given above exhaust all possibilities for linear and quadratic equations having positive root. So systematic and exhaustive was al-Khwarizmi's exposition that his readers must have had little difficulty in mastering the solutions."
^ Gandz and Saloman (1936), The sources of al-Khwarizmi's algebra, Osiris i, p. 263–277: "In a sense, Khwarizmi is more entitled to be called "the father of algebra" than Diophantus because Khwarizmi is the first to teach algebra in an elementary form and for its own sake, Diophantus is primarily concerned with the theory of numbers".
^(Boyer 1991, "The Arabic Hegemony" p. 239) "Abu'l Wefa was a capable algebraist as well as a trigonometer. [...] His successor al-Karkhi evidently used this translation to become an Arabic disciple of Diophantus - but without Diophantine analysis! [...] In particular, to al-Karkhi is attributed the first numerical solution of equations of the form ax2n + bxn = c (only equations with positive roots were considered)," | 677.169 | 1 |
Essentials of Engineering Mathematics|2nd Edition
Product Description:
First published in 1992, Essentials of Engineering Mathematics is a widely popular reference ideal for self-study, review, and fast answers to specific questions. While retaining the style and content that made the first edition so successful, the second edition provides even more examples, new material, and most importantly, an introduction to using two of the most prevalent software packages in engineering: Maple and MATLAB. Specifically, this edition includes:
Introductory accounts of Maple and MATLAB that offer a quick start to using symbolic software to perform calculations, explore the properties of functions and mathematical operations, and generate graphical output
New problems involving the mean value theorem for derivatives
Extension of the account of stationary points of functions of two variables
The concept of the direction field of a first-order differential equation
Introduction to the delta function and its use with the Laplace transform
The author includes all of the topics typically covered in first-year undergraduate engineering mathematics courses, organized into short, easily digestible sections that make it easy to find any subject of interest. Concise, right-to-the-point exposition, a wealth of examples, and extensive problem sets at the end each chapter--with answers at the end of the book--combine to make Essentials of Engineering Mathematics, Second Edition ideal as a supplemental textbook, for self-study, and as a quick guide to fundamental concepts and techniques.
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Introduction to Mathematical Thinking
Introduction to Mathematical Thinking
Introduction to Mathematical Thinking
Stanford University
About this course: Learn how to think the way mathematicians do – a powerful cognitive process developed over thousands of years.
Mathematical thinking is not the same as doing mathematics – at least not as mathematics is typically presented in our school system. School math typically focuses on learning procedures to solve highly stereotyped problems. Professional mathematicians think a certain way to solve real problems, problems that can arise from the everyday world, or from science, or from within mathematics itself. The key to success in school math is to learn to think inside-the-box. In contrast, a key feature of mathematical thinking is thinking outside-the-box – a valuable ability in today's world. This course helps to develop that crucial way of thinking.
Who is this class for: This ten-week course is designed with two particular audiences in mind. First, people who want to develop or improve mathematics-based, analytic thinking for professional or general life purposes. Second, high school seniors contemplating a mathematics or math-related major at college or university, or first-year students at college or university who are thinking of majoring in mathematics or a math-dependent subject. To achieves this aim, the first part of the course has very little traditional mathematical content, focusing instead on the thinking processes required for mathematics. The more mathematical examples are delayed until later, when they are more readily assimilated.
Created by:Stanford University
Taught by:Dr. Keith Devlin, Co-founder and Executive Director
H-STAR institute
Level
Intermediate
Commitment
Expect to require at least 10 hours of study per week to complete this course satisfactorily.
START with the Welcome lecture. It explains what this course is about. (It comes with a short Background Reading assignment, to read before you start the course, and a Reading Supplement on Set Theory for use later in the course, both in downloadable PDF forma...
6 videos
Graded: Problem Set 1
WEEK 2
Week 2
In Week 2 we continue our discussion of formalized parts of language for use in mathematics. By now you should have familiarized yourself with the basic structure of the course: 1. Watch the first lecture and answer the in-lecture quizzes; tackle each of the p...
6 videos
Graded: Problem Set 2
WEEK 3
Week 3
This week we continue our analysis of language for use in mathematics. Remember, while the parts of language we are focusing have particular importance in mathematics, our main interest is in the analytic process itself: How do we formalize concepts from every...
4 videos
Graded: Problem Set 3
WEEK 4
Week 4
This week we complete our analysis of language, putting into place the linguistic apparatus that enabled, mathematicians in the 19th Century to develop a formal mathematical treatment of infinity, thereby finally putting Calculus onto a firm footing, three hun...
4 videos
Graded: Problem Set 4
WEEK 5
Week 5
This week we take our first look at mathematical proofs, the bedrock of modern mathematics.
4 videos
Graded: Problem Set 5
WEEK 6
Week 6
This week we complete our brief look at mathematical proofs
4 videos
Graded: Problem Set 6
WEEK 7
Week 7
The topic this week is the branch of mathematics known as Number Theory. Number Theory, which goes back to the Ancient Greek mathematicians, is a hugely important subject within mathematics, having ramifications throughout mathematics, in physics, and in some ...
4 videos
Graded: Problem Set 7
WEEK 8
Week 8
In this final week of instruction, we look at the beginnings of the important subject known as Real Analysis, where we closely examine the real number system and develop a rigorous foundation for calculus. This is where we really benefit from our earlier analy...
5 videos
Graded: Problem Set 8
WEEK 9
Weeks 9 & 10: Test Flight
Test Flight provides an opportunity to experience an important aspect of "being a mathematician": evaluating real mathematical arguments produced by others. There are three stages. It is important to do them in order, and to not miss any steps. STAGE 1: You co...
4 videos
Graded: Test Flight Peer Assessments
Graded: Evaluation Exercise 1
Graded: Evaluation Exercise 2
Graded: Evaluation Exercise 3What is the passing grade for this course
Ratings and Reviews
Rated 4.9 out of 5 of 89 ratings
JL
With consistent effort, I have found this course to re-ignite my passion for mathematics. Keith Devlin, the instructor, does an excellent job of helping us to develop our ability to think in mathematical and logical ways. In short, I have found a new joy and excitement for doing mathematics, even stuff I consider hard. I have a new-found passion and new "eyes" for maths that I didn't have before. Excellent course! | 677.169 | 1 |
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Transforms and Applications Primer for Engineers with Examples and MATLAB®is required reading for engineering and science students, professionals, and anyone working on problems involving transforms. This invaluable primer contains the most essential integral transforms that both practicing engineers and students need to understand. It provides a large number of examples to explain the use of transforms in different areas, including circuit analysis, differential equations, signals and systems, and mechanical vibrations.
Includes an appendix with suggestions and explanations to help you optimize your use of MATLAB
Laplace and Fourier transforms are by far the most widely used and most useful of all integral transforms, so they are given a more extensive treatment in this book, compared to other texts that include them. Offering numerous MATLAB functions created by the author, this comprehensive book contains several appendices to complement the main subjects. Perhaps the most important feature is the extensive tables of transforms, which are provided to supplement the learning process. This book presents advanced material in a format that makes it easier to understand, further enhancing its immense value as a teaching tool for engineers and research scientists in academia and industry, as well as students in science and engineering.
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Chapter 11: The Derivative and Graphs 225 11.1 Extrema of Functions 225 11.1.1 Global and local extrema 225 11.1.2 The Extreme Value Theorem 227 11.1.3 How to find global maxima and minima 228 11.2 Rolle's Theorem 230 11.3 The Mean Value Theorem 233 11.3.1 Consequences of the Mean Value Theorem 235 11.4 The Second Derivative and Graphs 237 11.4.1 More about points of inection 238 11.5 Classifying Points Where the Derivative Vanishes 239 11.5.1 Using the first derivative 240 11.5.2 Using the second derivative 242
Chapter 12: Sketching Graphs 245 12.1 How to Construct a Table of Signs 245 12.1.1 Making a table of signs for the derivative 247 12.1.2 Making a table of signs for the second derivative 248 12.2 The Big Method 250 12.3 Examples 252 12.3.1 An example without using derivatives 252 12.3.2 The full method: example 1 254 12.3.3 The full method: example 2 256 12.3.4 The full method: example 3 259 12.3.5 The full method: example 4 262
Chapter 21: Improper Integrals: How to Solve Problems 451 21.1 How to Get Started 451 21.1.1 Splitting up the integral 452 21.1.2 How to deal with negative function values 453 21.2 Summary of Integral Tests 454 21.3 Behavior of Common Functions near 8 and -8 456 21.3.1 Polynomials and poly-type functions near 8 and -8 456 21.3.2 Trig functions near 8 and -8 459 21.3.3 Exponentials near 8 and -8 461 21.3.4 Logarithms near 8 465 21.4 Behavior of Common Functions near 0 469 21.4.1 Polynomials and poly-type functions near 0 469 21.4.2 Trig functions near 0 470 21.4.3 Exponentials near 0 472 21.4.4 Logarithms near 0 473 21.4.5 The behavior of more general functions near 0 474 21.5 How to Deal with Problem Spots Not at 0 or 8 475
Chapter 23: How to Solve Series Problems 501 23.1 How to Evaluate Geometric Series 502 23.2 How to Use the nth Term Test 503 23.3 How to Use the Ratio Test 504 23.4 How to Use the Root Test 508 23.5 How to Use the Integral Test 509 23.6 Comparison Test, Limit Comparison Test, and p-test 510 23.7 How to Deal with Series with Negative Terms 515
Chapter 26: Taylor and Power Series: How to Solve Problems 551 26.1 Convergence of Power Series 551 26.1.1 Radius of convergence 551 26.1.2 How to find the radius and region of convergence 554 26.2 Getting New Taylor Series from Old Ones 558 26.2.1 Substitution and Taylor series 560 26.2.2 Differentiating Taylor series 562 26.2.3 Integrating Taylor series 563 26.2.4 Adding and subtracting Taylor series 565 26.2.5 Multiplying Taylor series 566 26.2.6 Dividing Taylor series 567 26.3 Using Power and Taylor Series to Find Derivatives 568 26.4 Using Maclaurin Series to Find Limits 570 | 677.169 | 1 |
Linear Functions Technology Lesson
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This lesson is a great way to incorporate technology in the middle school math classroom. Students use the interactive and convenient online graphing calculator website Desmos.com to develop a deeper conceptual understanding of linear functions. Students will specifically draw connections between the relationships of the slope and y-intercept of a linear function and the effect it has on the shape of the function.
This is a very student-centered activity. Students may work together in pairs or individually. The lesson is created in a way where questioning guides student work, scaffolds student thinking, encourages students to make conjectures, and eventually draws students to make the mathematical connections intended for the lesson.
This activity requires access to internet for every student.
This activity is intended for students who have basic or no prior knowledge of linear equations and how slope and y-intercept affect the behavior of a linear function.
This product also includes a section that can be used either for a lesson wrap up in class or homework. This section also requires access to the internet. Students write down their final generalizations from the lesson and test out their theories using the slider function in Desmos | 677.169 | 1 |
takes an international perspective in exploring the role of mathematics in everyday life and is intended for high school age readers. As (RLM) is intended for a younger and less mathematically experienced audience, the authors and editors faced unique challenges in selecting and preparing entries. The articles in the book are meant to be understandable by anyone with a curiosity about mathematical topics. is intended to serve all students of math such that an 8th- or 9th-grade student just beginning their study of higher maths can at least partially comprehend and appreciate the value of courses to be taken in future years. Accordingly, articles were constructed to contain material that might serve all students. To be of maximum utility to students and teachers, most of the 80 topics found herein - arranged alphabetically by theory or principle - were predesigned to correspond to commonly studied fundamental mathematical concepts as stated in high school level curriculum objectives. However, as high school level maths generally teach concepts designed to develop skills toward higher maths of greater utility, this format sometimes presented a challenge with regard to articulating understandable or direct practical applications for fundamental skills without introducing additional concepts to be studied in more advanced math classes. | 677.169 | 1 |
Basic Math Skills Review #2: Algebraic Fractions
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The second in a series of reference pages detailing basic mathematical skills necessary for success in college mathematics. Algebraic Fractions covers how to simplify, add, subtract, multiply, and, divide fractions which have variables, as well as how to simplify complex fractions. Step-by-step instructions are given, followed by worked out examples. For those students with weak factoring skills, there is also a Factoring Methods review in this series.
This is a great reference for students who need to brush up on their basic skills in order to be successful in algebra-based courses | 677.169 | 1 |
Product Description:
Normal 0 false false false MicrosoftInternetExplorer4 Basic College Mathematics with Early Integers is a new addition to the Martin-Gay worktext series. This text is designed for a 1-semester basic math courses in which an "early "introduction of integers is desired. Integers are introduced in chapter 2, and students continue to work with them throughout the text. This gives students ample opportunity to practice operations with integers and to become comfortable with them, prior to being introduced to algebra in chapter 7, Equations. The Whole Numbers; Integers and Introduction to Variables; Fractions; Decimals; Ratio, Proportion, and Measurement; Percent; Statistics and Probability; Equations; Geometry; Tables; The Bigger Picture; Exponents and Polynomials For all readers interested in basic college mathematics.
REVIEWS for Basic College Mathematics with Early Integ | 677.169 | 1 |
Description of the book "The Manga Guide to Linear Algebra":
Reiji wants two things in life: a black belt in karate and Misa, the girl of his dreams. Luckily, Misa's big brother is the captain of the university karate club and is ready to strike a deal: Reiji can join the club if he tutors Misa in linear algebra. Follow along in The Manga Guide to Linear Algebra as Reiji takes Misa from the absolute basics of this tricky subject through mind-bending operations like performing linear transformations, calculating determinants, and finding eigenvectors and eigenvalues. With memorable examples like miniature golf games and karate tournaments, Reiji transforms abstract PDF concepts into something concrete, understandable, and even fun. As you follow Misa through her linear algebra crash course, you'll learn about: * Basic vector and matrix operations such as addition, subtraction, and multiplication * Linear dependence, independence, and bases * Using Gaussian elimination to calculate inverse matrices * Subspaces, dimension, and linear span * Practical applications of linear algebra in fields like computer graphics, cryptography, and engineering But Misa's brother may get more than he bargained for as sparks start to fly between student and tutor.
Will Reiji end up with the girl-or just a pummeling from her oversized brother? Real math, real ePub romance, and real action come together like never before in The Manga Guide to Linear Algebra.
Reviews of the The Manga Guide to Linear Algebra
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9781609781026
160978101.09
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Summary
Hundreds of realistic practice questions and exercises to prepare you for the Math portion of the GRE.
Kaplan's Math Workbook for the GRE, 9th Edition, comprehensively addresses the math section of the GRE Revised General Test. This workbook is a highly effective way to prepare for the math section of the GRE Revised General Test.
Kaplan's Math Workbook for the GRE, 9th Edition includes:
• 6 full-length Quantitative Reasoning practice sets • Diagnostic tool for even more targeted Quantitative practice • Review of crucial math skills and concepts (including arithmetic, algebra, data interpretation, geometry, and probability) • Key strategies for all Quantitative Reasoning question types on the revised GRE Kaplan is dedicated to helping our students score higher. We guarantee that students will raise their scores–or get their money back.
Author Biography
Celebrating 75+ | 677.169 | 1 |
Input and Output Function Tables and Coordinate Plane
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This PowerPoint guides students in determining relationships between numbers using an input/output function table. It also guides students into how to determine a one step pattern/rule. After this, there are there tasks where students must follow a rule and analyze the results of the rule | 677.169 | 1 |
Algebra - Laws of Exponents - Note Packet
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2.43 MB | 102 pages
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In this packet, there are examples, definitions, and "on your own" problems that can be used in a Pre-Algebra or Algebra classroom. The packet thoroughly explains the laws of exponents along with scientific notation. Difficulty increases as the examples increase. After each rule, a quiz review is provided. This packet can be used to give notes or for extra practice. The packet is saved as a word document so you have the option to edit the notes to fit your needs. It can be easily saved as a pdf then imported into ActivSoftware. ActivSoftware is the program that links to our Promethean Boards. If you have any questions, please ask. Thanks for looking | 677.169 | 1 |
Freshman and sophomore life sciences students respond well to the modeling approach to calculus, difference equations, and differential equations presented in this book. Examples of population dynamics, pharmacokinetics, and biologically relevant physical processes are introduced in Chapter 1, and these and other life sciences topics are developed throughout the text.
The ultimate goal of calculus for many life sciences students primarily involves modeling living systems with difference and differential equations. Understanding the concepts of derivative and integral is crucial, but the ability to compute a large array of derivatives and integrals is of secondary importance.
Students should have studied algebra, geometry and trigonometry, but may be life sciences students because they have not enjoyed their previous mathematics courses. This text can help them understand the relevance and importance of mathematics to their world. It is not a simplistic approach, however, and indeed is written with the belief that the mathematical depth of a course in calculus for the life sciences should be comparable to that of the traditional course for physics and engineering students.
This textbook is for students who have successfully experienced an introductory calculus course in high school. College Calculus begins with a brief review of some of the content of the high school calculus course, and proceeds to give students a thorough grounding in the remaining topics in single variable calculus, including integration techniques, applications of the definite integral, separable and linear differential equations, hyperbolic functions, parametric equations and polar coordinates, L'Hôpital's rule and improper integrals, continuous probability models, and infinite series. Each chapter concludes with several "Explorations," extended discovery investigations to supplement that chapter's material.
The text is ideal as the basis of a course focused on the needs of prospective majors in the STEM disciplines (science, technology, engineering, and mathematics). A one-term course based on this text provides students with a solid foundation in single variable calculus and prepares them for the next course in college level mathematics, be it multivariable calculus, linear algebra, a course in discrete mathematics, statistics, etc | 677.169 | 1 |
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Unformatted text preview: Calculus I for Science and Engineering (Math 2250) Fall 2011 1 Technicalities Instructor: Clay Shonkwiler ( clayton@math.uga.edu ) Office: Boyd 436 Course web page: Text: University Calculus: Early Transcendentals (2 nd edition), by Joel Hass, Maurice D. Weir, and George B. Thomas, Jr. Times/Locations: 1:252:15 MWF in Boyd 222, 2:003:15 R in Boyd 303 Office Hours: Wednesdays 2:304:30. 2 Summary of the Course This course provides an introduction to single variable calculus. You will develop a deep under- standing of the three most important concepts of the calculus: limits, derivatives, and integrals. Limits allow us to understand the behavior of a function as its inputs approach some specified value and provide the foundation for the other two concepts. Derivatives arise as slopes of tangent lines, rates of change, and linear approximations. Integrals provide a way of finding the area under a curve and the total change of a rate of change. You should understand each of these concepts theoretically, geometrically, and heuristically and be able to compute effectively enough to apply them appropriately. In order to do so you will need to develop your abilities to think mathematically and communicate effectively. 3 Homework There will be weekly homework assignments which will typically consist of a mix of WebWork problems (which you complete online; see below for instructions) and written problems (which you turn in to me). Homework is an important part of any math class, as it is impossible to learn mathematics without actually doing mathematics. The goal of the assignments is to deepen your understanding of the concepts, tools and techniques discussed in class, as well as to give you the opportunity to practice explaining your mathematical thinking. The importance of effective communication is vital: knowledge without the ability to communicate that knowledge is of limited value. As such, to get full credit on a problem your solution must be clear and well-written. The written portion of your homework must be stapled with your name clearly written at the top. What you turn in should be a final copy: it should be neat, legible, and well-organized. If I cant read or understand your work you wont receive any credit. Late homework will not (and, in the case of WebWork, cannot ) be accepted, so you should turn in whatever you have completed on the due date in order to get credit for it. Your lowest homework grade will be dropped from the calculation of your final grade....
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This note was uploaded on 02/20/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Spring '08 term at University of Georgia Athens. | 677.169 | 1 |
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Introduce solving systems of equations using the substitution method with this bundle of goodies!
Contents:
- PowerPoint Presentation (20 animated slides)
Use this to accompany discussion and note-taking.
- Notes (multiple formats/options both foldable and traditional included)
This is a very versatile set of notes. There are various options for use and they will work well in interactive notebooks as well as with more traditional note-taking. All of the notes are coordinated with the PowerPoint presentation, but could also be used alone.
- Two winter themed worksheets of ten problems each (Version A is easier than Version B | 677.169 | 1 |
Free Study Material For RRB Examination: Mathematics (Basic Algebra)
Free Study Material For RRB Examination
Subject : Mathematics (Basic Algebra)
A Civil Servant should be well-versed in basics of Algebra.
In the Civil Services Aptitude Test Paper 2, in Basic Numeracy, certainly there
will be asked some questions based on equations and their roots. | 677.169 | 1 |
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Archimedes & The Law of The Lever
Important information
Course
Online
Duration: 6Mathematics
Maths
Law
Teachers
Course programme
This is a Professional Development course for high school maths and science teachers developed at the School of Mathematics and Statistics, Faculty of Science, UNSW.
This purely online course will introduce teachers to Archimedes' law of the lever, centres of mass, expected values of probability distributions, barycentric coordinates, and interesting applications via online videos.
There will also be fun activities and challenges that will develop and deepen your mathematical understanding, and give you ideas and materials for engaging your students.
You will be able to interact with fellow educators from around the country. | 677.169 | 1 |
Mathematics Proficiencies
The Australian Curriculum: Mathematics is organised around the interaction of three content strands and four proficiency strands. The content strands are Number and Algebra, Measurement and Geometry, and Statistics and Probability. They describe what is to be taught and learnt.
The proficiency strands are Understanding, Fluency, Problem-Solving and Reasoning. They describe how content is explored or developed, that is, the thinking and doing of mathematics. They provide the language to build in the developmental aspects of the learning of mathematics and have been incorporated into the content descriptions of the three content strands. This approach has been adopted to ensure students' proficiency in mathematical skills develops throughout the curriculum and becomes increasingly sophisticated over the years of schooling.
The Australian Curriculum: Mathematics aims to be relevant and applicable to the 21st century. The inclusion of the proficiencies of understanding, fluency, problem-solving and reasoning in the curriculum is to ensure that student learning and student independence are at the centre of the curriculum. The curriculum focuses on developing increasingly sophisticated and refined mathematical understanding, fluency, reasoning, and problem-solving skills. These proficiencies enable students to respond to familiar and unfamiliar situations by employing mathematical strategies to make informed decisions and solve problems efficiently.
The proficiency strands describe the actions in which students can engage when learning and using the content of the Australian Curriculum: Mathematics.
Understanding
Students build a robust knowledge of adaptable and transferable mathematical concepts. They make connections between related concepts and progressively apply the familiar to develop new ideas. They develop an understanding of the relationship between the 'why' and the 'how' of mathematics. Students build understanding when they connect related ideas, when they represent concepts in different ways, when they identify commonalities and differences between aspects of content, when they describe their thinking mathematically and when they interpret mathematical information.
Fluency
Students develop skills in choosing appropriate procedures; carrying out procedures flexibly, accurately, efficiently and appropriately; and recalling factual knowledge and concepts readily. Students are fluent when they calculate answers efficiently, when they recognise robust ways of answering questions, when they choose appropriate methods and approximations, when they recall definitions and regularly use facts, and when they can manipulate expressions and equations to find solutions.
Problem-solving
Students develop the ability to make choices, interpret, formulate, model and investigate problem situations, and communicate solutions effectively. Students formulate and solve problems when they use mathematics to represent unfamiliar or meaningful situations, when they design investigations and plan their approaches, when they apply their existing strategies to seek solutions, and when they verify that their answers are reasonable.
Reasoning
Students develop an increasingly sophisticated capacity for logical thought and actions, such as analysing, proving, evaluating, explaining, inferring, justifying and generalising. Students are reasoning mathematically when they explain their thinking, when they deduce and justify strategies used and conclusions reached, when they adapt the known to the unknown, when they transfer learning from one context to another, when they prove that something is true or false, and when they compare and contrast related ideas and explain their choices | 677.169 | 1 |
Gcse level maths coursework
Study Maths GCSE with Oxford Open Learning, a Not for Profit Educational Trust Home Study | Qualified Tutors | Foundation and Higher | Easy Payments. Describes the new A-level and GCSE exams Including subjects, time-scales, key features of new exams and likely affect on results. GCSE 2016 : GCSE Examination is Grade level Conducting by the University of UKThis website Gives the info about GCSE Results 2016,GCSE 2016 Result day and.
Gcse level maths coursework
Extracts from this document Introduction GCSE PE coursework - circuit training Introduction I will be performing a circuit over the next several weeks. OCR GCSE English Literature (9-1) (from 2015) qualification information including specification, exam materials, teaching resources, learning resources. The Maths GCSE Guide is divided up into 6 subsections In each subsection are the topics with the necessary formulae and examples of their use. Subject content for GCSE in single science for teaching from 2016. Study Maths GCSE with Oxford Open Learning, a Not for Profit Educational Trust Home Study | Qualified Tutors | Foundation and Higher | Easy Payments.
IST students may choose between two types of courses to meet the individual needs of the students and their families: tutor supported and independent study. A level, GCSE or IGCSE distance learning courses! The Oxford Open Learning Trust provides Qualified Tutors, High Quality Courses and Easy Payments. The General Certificate of Secondary Education (GCSE) is an academically rigorous, internationally recognised qualification (by Commonwealth countries with. Distance learning courses and resources from National Extension College Home study qualifications including GCSEs, IGCSEs, A levels, Childcare, Management. GCSE learning resources for adults, children, parents and teachers organised by level, subject and topic.
PSA! DoSomethingorg Has a TON of Scholarship Opportunities Right Now SPOILER: college is crazy-expensive Sorry Did we spoil it? There are. Study Guides Tough GCSE topics broken down and explained by out team of expert teachers Learn more. GCSE Maths (pre-2015) learning resources for adults, children, parents and teachers organised by topic.
Science GCSE from 2016 Here are the main points: Totally exam-based There is no practical exam, although you have to do certain practicals in school. From GCSE to A-level, AQA History helps students study a range of periods with both national and international perspectives.
The Maths GCSE Guide is divided up into 6 subsections In each subsection are the topics with the necessary formulae and examples of their use. AQA provides qualifications that enable students to progress to the next stage in their lives We also support teachers to develop their professional skills. GCSE Maths (pre-2015) learning resources for adults, children, parents and teachers organised by topic. | 677.169 | 1 |
This Illustrator Graphic Style is a Chalk Graphic Style. There is an outlined and a filled in version of the style. It is vector based so it can be sized as needed without losing detail. This could also be used for objects/shapes.
Course contains over 28 hours of video.The table of contents listed below has been pulled from an older version of this course, however, it is largely representative of the material uploaded. For a wholely accurate listing feel free to consult the DVD menu once you've finished downloading.
Chalk Dust Productions are like Math Tutor DVD. They are specially commissioned by Houghton Mifflin to supplement their various textbooks, but it is not necessary at all to possess the textbooks in order to benefit from these videos. 5th edition: FULL SET is 11 DVDs, hardbound text, student solutions guide, and technical support. This program includes College Algebra, Trigonometry and Limits | 677.169 | 1 |
104 pgs, non-consumable and non-reproducible. Paperback.
I grew up in Japan and Korea as a kid and love their education systems. As a homeschool parent, I am always looking for affordable curriculum. We are in middle school math now and have had to switch from Singapore Math not because of aging out, but because the books lack the instructions for how to do the harder math it assigns. We use the Textbooks, Workbooks, and Tests books as supplements now, and are using Lifepac math now, but we are still looking for any other math curriculum, our goal for University is Engineering so we want to be prepared.
We have used several other math programs before coming to Singapore. This program is just what my daughter needs. It's fast-moving, with little review. It has challenging word problems that keep her thinking about number and how to manipulate them to get the answer.
She found most of the other programs dreadfully boring, but is excited every day when we pull out her math book.
Excellent resource. We would give all the Singapore Math books 5 stars! The books are visually clear and easy to understand. I recommend taking the placement test to make accurate book selection. We did and ended up starting in a lower book that I thought we would, but quickly gained speed. Now our child is faster than a calculator and loves math!
My daughter, who struggles learning her math concepts, loves this book. The lessons are short and very easy. I am so grateful to have found this book. She will be testing next month so this has come at just the right time. | 677.169 | 1 |
Linear Algebra
Description
Beginning with the basic concepts of vector spaces such as linear independence, basis and dimension, quotient space, linear transformation and duality with an exposition of the theory of linear operators on a finite dimensional vector space, this book includes the concepts of eigenvalues and eigenvectors, diagonalization, triangulation and Jordan and rational canonical forms. Inner product spaces which cover finite dimensional spectral theory, and an elementary theory of bilinear forms are also discussed. | 677.169 | 1 |
equations is used in all branches of engineering and science. In essence, once a student begins to study more complex problems, nature usually obeys a differential equation which means that the equation involves one or more derivatives of the unknown variable | 677.169 | 1 |
This is an introductory computer science text that includes detailed discussions of several types of mathematical algorithms. There are wide-ranging chapters on matrix operations, linear programming, polynomials and Fast Fourier Transforms, number theory, and computational geometry. There's also a discussion of the "master method" for finding asymptotic behavior of sequences defined by an approximate first-order recurrence. There's a lengthy appendix of mathematical background, but this is not very thorough and is intended for review.
Most of the book deals with straight computer-science topics, such as data structures, sorting & searching, recursive approaches, and graphs. The book takes a practical approach and emphasizes useful algorithms rather than a thorough study of all algorithms. It generally avoids numerical considerations and error analyses, but does estimate running time of the algorithms.
The matrix chapter deals primarily with LU decomposition and its uses, but also has some material on least-squares approximations. The linear programming chapter is quite thorough and covers how to cast problems as a linear program, a good bit on the simplex method, some discussion of duality, and methods of finding an initial feasible solution. It includes some analysis of typical classes of problems that these methods can be used on. The polynomials chapter is not as thorough as the others but does cover storing and computing with polynomials, and the closely-related topics of discrete and fast Fourier transforms. The number theory chapter covers several aspects of divisibility and modular arithmetic, the RSA cryptosystem, primality testing, and factorization (primarily through Pollard's rho method). The computational geometry chapter covers a variety of topics, mostly convex hulls and intersections of objects.
Bottom line: a good quick reference for math students and mathematicians, as well as a thorough text for computer science students. | 677.169 | 1 |
Pages
4 January 2016
Five things you might not know about the new GCSE content #1
This is the first of two posts listing changes to GCSE content that aren't widely known. If you're really familiar with the new GCSE specification then you probably already know all of this. But if you're worried that you might have missed something, read on.
1. Graph Transformations
No more stretching of functions. I missed this one so thanks to @MrBenWard for pointing it out today. This is an odd change - I can't see the rationale. Here's what's in and what's out:
I'm going to have to edit a lot of resources...
Don't forget that the featured transformations (ie reflections and translations) may also be applied to the graphs of y = sinx and y = cosx.
2. Trial and Improvement
Contrary to popular belief, trial and improvement hasn't gone! I thought it had until I saw these tweets from AQA:
AQA's Teaching Guidance says that "students should be able to use systematic trial and improvement to find approximate solutions of equations where there is no simple analytical method".
Any trial and improvement questions in the new GCSE may be more challenging than what we're used to, so we can't rely on past exam questions and resources. Here's an example of a new specification trial and improvement question from AQA:
A solid has volume V = x3 + 2x2
The volume of the solid is 90cm3
Work out the value of x to one decimal place.
You must show your working.
Note that the question doesn't explicitly tell the student to use trial and improvement, which is a notable change from previous exam questions.
3. Compound Measures
The new specification lists pressure under compound measures - this is new to maths GCSE. The specification says, "Use compound units such as speed, rates of pay, unit pricing, density and pressure" - see Question 3 in this set of questions from justmaths.co.uk for an example.
In general students should be taught the importance of looking at units in compound measure questions. If speed is measured in km/h then we can see from the units that we need to divide distance by time to get speed. Using this approach students should be able to figure out how to answer any compound measure question even if they're not familiar with the context (eg if a question asks for population density in people/km2 then they simply divide population by area - they know this by looking at the units, no formula required).
It's worth noting that students will need to know the equivalence of the notation m/s and ms-1.
It was Mel's post 'GCSE 9-1 New content – Error Intervals' that first alerted me to error intervals and truncating so if you haven't read it then please do. These topics are fairly straightforward to teach, but a few of my top set Year 10 got a truncating question wrong when I tested them on it only a week later... They've been rounding for years and suddenly we're telling them not to round!
"How many numbers between 100 and 200 have at least one digit that is 7?"
I'm not sure this kind of thing needs direct teaching - it just needs a logical approach. This second question from AQA requires the product rule - I think it might be worth spending a lesson or two on this (perhaps more if we extend to combinations and permutations - I will try to find out whether this necessary):
"Daniel has 10 shirts, 8 pairs of trousers and 5 pairs of socks.
a) How many different combinations of shirt, trousers and socks does he have?
b) He chooses a shirt. How many combinations are there now?"
Apart from this lovely PowerPoint from newmrsc on TES, I don't have much in the way of resources for this topic so please send me anything that you have!
Formulae
Finally, just a reminder that your students will need to memorise more formulae than they used to. When the new GCSE was first announced this made the headlines:
"The new curriculum for GCSE maths will see pupils in England spending more classtime studying the subject and memorising mathematical formulae, such as Pythagoras's theorem."
Our students have always had to memorise Pythagoras' Theorem and plenty of other facts, theorems and formulae. I think that the impact of this change was exaggerated. Saying that, I will be annoyed when my students lose marks because they forget the quadratic formula...
I'm pleased that students will no longer be required to memorise imperial to metric conversions because (unlike times tables, Ms Blower), it's entirely practical to look these up online if ever needed.
I hope that this post has been helpful. Don't forget I have a page of new GCSE resources and I've written various posts about new GCSE topics:
7 comments:
Jo, this is an incredibly useful blog and highlights many areas I wasn't aware of, so many thanks on behalf of my whole maths department who have to read this as a compulsory task this evening. You really are making our lives easier. | 677.169 | 1 |
Course Description: Students will discover a whole new way of doing andviewing math. The familiar formulas of perimeter, area,and volume will be covered, as well as the basics oftriangle-based trigonometry. Other topics will includemeasurement, and the relationships of points, lines, angles, and figures in space. Applications of algebra skills will helpstudents understand the connectedness of underlying math principles. | 677.169 | 1 |
We're Ready, Are You?
The team at StudyForge has been hard at work, adapting their courses and preparing
for the launch of the new BC Ed Curriculum Plan for Grades 7, 8 and 9. Our
curriculum is designed by British Columbia teachers, for British Columbia teachers.
If you're not familiar with StudyForge's amazing digital curriculaware, check out
these short videos explaining a bit about who we are.
Choose your experience:
ONLINE
TEACHER
ONLINE
TEACHER
What Can We Offer You?
All our courses are aligned to the new curriculum and include everything you
currently have in a StudyForge course and more! Our high quality, informative,
interactive and intriguing videos will continue to be there, including all new
content from the BCEd. plan. You will continue to receive these videos aligned to
robust practice questions with detailed interactive solutions. We are also adding
inquiry videos and applets to guide students through the questions they have along
the way, and allow them to discover the methods themselves. In addition we are
creating projects throughout the courses that allow teacher and student choice to
create an engaging experience, giving teachers an opportunity to see students grow
in the curricular competencies.
Engaging StudyForge Videos
New content is being added to our video library almost daily to
cover the new outcomes required by the BC Ed Curriculum Plan. To
view a sample of a StudyForge video lesson click here.
Practice Questions with Detailed Solutions
In addition to the new video content, we are adding lots and lots more practice
questions, not only for the new curriculum but also to round out some of our current
curriculum, and all with detailed, step by step solutions.
Mastery Assignments
We have created chapter assignments throughout these three courses to follow the
StudyForge pedagogy of mastery. Take a look at the updated look or download a
sample here.
Inquiry Videos
StudyForge is hard at work building 'inquiry' pieces through video
that hint at areas where mathematical topics are relevant in the
student's every-day world. And, as we all know, not all topics are
relevant. We aim to inspire students to realize that although they
may not use polynomials every day, that there are applications
that stem from the basics surrounding them.
GeoGebra
We are adding interactive GeoGebra applets all over the place
so that students can explore the mathematical ideas with a hands on approach, and be
allowed to ask and answer questions that will lead them down the road of discovery.
Project Based Learning
In over 50% of the chapters in each course, StudyForge is preparing amazing student
projects. Projects that require student inquiry and deep thought to apply the things
they've learned and allow them to extend those skills into other areas. Teachers
will have choice in how to apply these to their courses, how many they want to use
or whether or not to allow student choice, all while still providing an opportunity
for learners to grow in the curricular competencies, and teachers to assess that
growth. Click on the images to take a look at the sample's provided here.
Randomized Multiple Choice Assessments
With new projects and a focus on assessing competencies, there is
always the worry of teacher time being spread thin. Because of this
we will be adding multiple choice, computer graded assessments to
every chapter of every course, so that more teacher grading time is
freed up for the formative opportunities provided by the projects and
mastery assignments. These test banks will contain multiple versions
of each question, so that teachers can allow retests for students, or
have multiple students taking the tests at the same time. | 677.169 | 1 |
Using Geometer's Sketchpad
Geometry is one of the most visual courses in high school mathematics. The Geometer's Sketchpad is an electronic ruler and compass for today's technological geometry classroom. This module is designed to familiarize teachers with this new tool through interactive tutorials and then to demonstrate how it can be used effectively in the teaching of geometry. Credit: 1 grad. sem. hr.
Common Core Standardsfor Mathemtical Practice that are emphasized include:
4. Model with mathematics.
5. Use appropriate tools strategically.
7. Look for and make use of structure.
Using the Geometer's Sketchpad in the Mathematics Classroom was written by Jeremy Bartusch of the University of Illinois in November 1996. It was revised by Tony Peressini in January 1998, and by Tom Anderson in March 2004 and again in October 2011 for Sketchpad Version 5. | 677.169 | 1 |
This student-friendly textbook for the Statistics 2 Module of A-Level Maths comprehensively covers the AQA exam specification. It contains straightforward, accessible notes explaining all the theory, backed up with useful step-by-step examples. There are practice questions throughout the book to test understanding, with recap and exam-style questions at the end of each section (detailed answers to all the questions are included at the back). Finally, there's a CD-ROM containing two complete Statistics 2 practice exam papers - ideal to print out for realistic practice before the final tests.
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Mathematics & Statistics
Math for the Modern World MATH 128–your future career and courses included. Mathematics is both an art and a tool created by humans. The common bond is a way of thinking and a way of reasoning to describe and solve problems of many types.
This and use math to solve problems and pose new problems. Topics include scientific notation, basic financial math, linear, exponential and polynomial models and an introduction to probability. 4 credits.
Calculus for Management & Social Sciences MATH 134
A one-semester introduction to differential and integral calculus. Theory is presented informally and topics and techniques are limited to polynomials, rational functions, logarithmic and exponential functions. (This course cannot be used to satisfy core or complementary requirements by students majoring in biology, chemistry, computer science, engineering, mathematics or physics). 4 credits. | 677.169 | 1 |
Find the exact information you need to solve a problem on the fly, or go deeper to master
the technologies and skills you need to succeed
Chapter 6. Calculus
In This Chapter
Overview
Covering the mathematics of limits, otherwise known as calculus, this chapter is the final chapter of Part I on the fundamentals of mathematics. While the demand for calculus in the programming that is dealt with in this book is limited, with respect to several of the problems concerning physics, it is necessary. Needless to say, calculus is an involved topic, and the discussion in this chapter covers only selected and essential topics that anticipate problems dealt with in subsequent chapters.
Differentiation and Integration
The two principal techniques in calculus extend the work discussed in Chapter 3, on graphs, and in ...
Find the exact information you need to solve a problem on the fly, or go deeper to master
the technologies and skills you need to succeed | 677.169 | 1 |
This comprehensive and engaging textbook introduces the basic principles and techniques of signal processing, from the fundamental ideas of signals and systems theory to real-world applications. Students are introduced to the powerful foundations of modern signal processing, including the basic geometry of Hilbert space, the mathematics of Fourier transforms, and essentials of sampling, interpolation, approximation and compression The authors discuss real-world issues and hurdles to using these tools, and ways of adapting them to overcome problems of finiteness and localization, the limitations of uncertainty, and computational costs. It includes over 160 homework problems and over 220 worked examples, specifically designed to test and expand students' understanding of the fundamentals of signal processing, and is accompanied by extensive online materials designed to aid learning, including Mathematica® resources and interactive demonstrations.
Includes numerous examples and exercises, specifically designed to test and expand students' understanding of the fundamentals of signal processing
Breadth of coverage and variety of supporting material allows for use in different course settings and for self-study
Written by leading experts in modern signal processing techniques
Standard engineering notation is used throughout, making mathematical examples easy for students to follow, understand and apply
Reviews & endorsements
"This is a major book about a serious subject - the combination of engineering and mathematics that goes into modern signal processing: discrete time, continuous time, sampling, filtering, and compression. The theory is beautiful and the applications are so important and widespread."
Gil Strang, Massachusetts Institute of Technology
"A refreshing new approach to teaching the fundamentals of signal processing. Starting from basic concepts in algebra and geometry, [the authors] bring the reader to deep understandings of modern signal processing. Truly a gem!"
Rico Malvar, Microsoft Research
"A wonderful book that connects together all the elements of modern signal processing … it's all here and seamlessly integrated, along with a summary of history and developments in the field. A real tour-de-force, and a must-have on every signal processor's shelf!"
Robert D. Nowak, University of Wisconsin, Madison
"Finally a wonderful and accessible book for teaching modern signal processing to undergraduate students."
Stéphane Mallat, École Normale Supérieure
"Most introductory signal processing textbooks focus on classical transforms, and study how these can be used. Instead, Foundations of Signal Processing encourages readers to think of signals first. It develops a 'signal-centric' view, one that focuses on signals, their representation and approximation, through the introduction of signal spaces. Unlike most entry-level signal processing texts, this general view, which can be applied to many different signal classes, is introduced right at the beginning. From this, starting from basic concepts, and placing an emphasis on intuition, this book develops mathematical tools that give the readers gets a fresh perspective on classical results, while providing them with the tools to understand many state of the art signal representation techniques."
Antonio Ortega, University of Southern California
"Foundations of Signal Processing … is a pleasure to read. Drawing on the authors' rich experience of research and teaching of signal processing and signal representations, it provides an intellectually cohesive and modern view of the subject from the geometric point of view of vector spaces. Emphasizing Hilbert spaces, where fine technicalities can be relegated to backstage, this textbook strikes an excellent balance between intuition and mathematical rigor, that will appeal to both undergraduate and graduate engineering students. The last two chapters, on sampling and interpolation, and on localization and uncertainty, take full advantage of the machinery developed in the previous chapters to present these two very important topics of modern signal processing, that previously were only found in specialized monographs. The explanations of advanced topics are exceptionally lucid, exposing the reader to the ideas and thought processes behind the results and their derivation. Students will learn … why things work, at a deep level, which will equip them for independent further reading and research. I look forward to using this text in my own teaching."
Yoram Bresler, University of Illinois, Urbana-Champaign
Resources for
Foundations of Signal Processing
Martin Vetterli, Jelena Kovačević, Vivek K GoyalAuthors
Martin Vetterli, École Polytechnique Fédérale de Lausanne Martin Vetterli is a Professor of Computer and Communication Sciences at the École Polytechnique Fédérale de Lausanne, and the President of the Swiss National Science Foundation. He has formerly held positions at Columbia University and the University of California, Berkeley, and has received the IEEE Signal Processing Society Technical Achievement Award (2001) and Society Award (2010). He is a Fellow of the ACM, EURASIP and the IEEE, and is a Thomson Reuters Highly Cited Researcher in Engineering.
Jelena Kovačević, Carnegie Mellon University, Pennsylvania Jelena Kovačević is the David Edward Schramm Professor and Head of Electrical and Computer Engineering, and a Professor of Biomedical Engineering, at Carnegie Mellon University. She has been awarded the Belgrade October Prize (1986), the E. I. Jury Award (1991) from Columbia University, and the 2010 Philip L. Dowd Fellowship at Carnegie Mellon University. She is a former Editor-in-Chief of IEEE Transactions on Image Processing, and a Fellow of the IEEE.
Vivek K Goyal, Boston University Vivek K Goyal is an Assistant Professor of Electrical and Computer Engineering at Boston University, and a former Esther and Harold E. Edgerton Associate Professor of Electrical Engineering at the Massachusetts Institute of Technology. He has been awarded the IEEE Signal Processing Society Magazine Award (2002), and the Eliahu Jury Award (1998) from the University of California, Berkeley, for outstanding achievement in systems, communications, control and signal processing. He is a Fellow of the IEEE | 677.169 | 1 |
Of special note is the inclusion of a series mascot--Math Mutt--who provides young students with tips and encouragement as they tackle the diverse exercises designed to provide a fundamental understanding of basic applied math.
A decade ago, the Applied Math and Science Academy labs were created to teach students about robotics and other technology - subjects they would no doubt need to know about to compete for jobs in this high-tech world.
Occupations in the applied math skills category include those in which workers need to understand mathematical concepts and be able to apply them to their work; in these occupations, knowledge of statistics and trigonometry may also be | 677.169 | 1 |
Triangle AlMark O.
...Precalculus is the hinge of mathematics. Until now, math has been about numbers and shapes, and how they relate to one another. In precalculus, we continue to use the tools that got us this far, but we start dealing with the relationships among numbers (and shapes) as things in themselves. | 677.169 | 1 |
TOPICS IN MATHEMATICS Documents
Showing 1 to 15 of 15
Section 3.4
Kyle Matthews
A penny saved is a penny earned.
When saving money we generally deposit relatively small quantities over
long periods of time. An account in which a xed amount of money deposited
over regular intervals is called a systematic savi
Homework 8.6
Homework 8.6 goals:
Know what the Casear Cipher is and how to use it.
Know what the Ane Cipher is and how to use it.
Know what an inverse is mod n and how to nd it.
1. Do these problems in section 8.6 of the book: 1-6, 9, 10.
2. Find the i
Homework 8.7
Homework 8.7 goals:
Know what the Hill Cipher is and how to use it.
1. Do these problems in section 8.7 of the book: 1, 2, 3, 4, 5, 6, 7, 8
2. Suppose a computer hacker gets a hold of a public key of 1,024 bits;
roughly a number that is at m
Homework 8.2
Homework 8.2 goals:
Understand what modulo means
Understand what it means for two numbers to be congruent modulo n.
Know what the properties of modulo arithmetic are, and how to use them.
1. Do these problems from section 8.2 in the book:
Homework 8.1
Homework 8.1 goals:
Know what it means when something divides something.
Know the dierence between prime and composite numbers.
Be able to nd the prime factorization of simple composite numbers.
Know how to execute the Sieve of Eratosthen
Homework 3.2
Homework 3.2 goals:
Know what a simple interest is and how to calculate it.
1. Do these problems in section 3.2 in the book: 1,2, 7, 8, 11, 12, 19-22
2. Create your own word problem that uses the simple interest formula.
Write a solution to
Homework 3.3
Homework 3.3 goals:
Understand what a compounding period is.
Understand how compounding interest is derived.
Understand how to use the compounding interest formula.
Be able to solve for F, P, r, and t in the compounding interest formula.
Section 3.1
Ken Ueda
If you want to know what God thinks about money, just look at the people he
gave it to. - Dorothy Parker
The story of money is one that is long and goes back before history. The
rst minting of coins dates back to the 7th and 6th centu
Section 3.2
Kyle Matthews
Take no usury or interest from him; but fear your God, that your brother may
live with you. You shall not lend him your money for usury, nor lend him your
food at a prot. -Leviticus 25:36-37
We begin our study of interest with th
Section 8.6
Ken Ueda
The Magic Words are Squeamish Ossifrage.
Cryptology or cryptography is the study of codes and encryptions. In this
case, one is not a pseudoscientic version of the other. Secret codes are encryption methods that are used in order to p
Section 8.7
Kyle Matthews
There are two types of encryption: one that will prevent your sister from reading your diary and one that will prevent your government. -Bruce Schneier
In the previous section we saw how to encrypt plaintext using ane encryption.
Section 8.2
Kyle Matthews
Mathematicians do not study objects, but the relations between objects; to
them it is a matter of indierence if these objects are replaced by others, provided that the relations do not change. Matter does not engage their attenti
Section 8.1
Ken Ueda
Mathematics is the queen of the sciences and number-theory the queen of
mathematics. -Carl Friedrick Gauss
Let us begin with a brief history of elementary number theory. Elementary
number theory, as the eld has adopted this rudimentar
Section 3.3
Ken Ueda
Dont think money does everything or you are going to end up doing everything for money. - Voltaire
Many kinds of investments earn interest that is compounded at regular intervals. The interest is paid on the initial deposit as well as
Homework 3.1
Homework 3.1 goals:
Know how to algebraically manipulate exponents.
Understand what a logarithm is.
Be able to algebraically manipulate logarithms.
1. Do these problems in section 3.1 of the book: 1-14 Calculators may be
used to check but | 677.169 | 1 |
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Unformatted text preview: MA366 Sathaye Final Thoughts First, some general advice. Review notes (class and on line), quizzes and old exams. You will be allowed three sheets of notes to be used during the exam. Be sure to have these hand written or typed, but not Xerox copies of printed material. Make sure that your notes are accurate and understandable. No calculators are allowed. Be sure to review your integration skills and algebraic manipulations. If you tend to forget some of the formulas, write them down in your notes. Verify the time an place of the exam carefully. We now describe how to prepare for the final exam. 1. Basic Theory. Review the basic definitions of linear/ nonlinear equations, order of equations and concept of a solution. Review the fundamental theorems of existence and uniqueness for linear as well as nonlinear equations of order 1. Also review how the theorems are modified for higher order equations and linear systems. 2. Equations of order 1. Study the precise method to solve a first order linear equation. Review necessary integration skills. Learn how to recognize a separable equation and how to solve it. Note that in general, it is difficult to solve in terms of y , but you should be able to solve for y if needed. A special case of separable equations is the case of autonomous equation y = f ( y ). Review how the phase line is constructed and how to sketch sample solution curves to illustrate the nature of solutions. Be sure to review the concepts of critical points (equilibrium solutions).nature of solutions....
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Explore how technology can help students build algebraic reasoning as they model and solve contextualized problems using a variety of representations, including graphs, tables, equations, and words. Engaging, open-construct problems provide opportunities for students to explain their reasoning and processes.
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Mathematics / Statistics
Assuming minimal background on the part of students, this text gradually develops the principles of basic real analysis and presents the background necessary to understand applications used in such disciplines as statistics, operations research and engineering. Learn More
This book discusses special modifications and extensions of designs that arise in certain fields of application such as genetics, bioinformatics, agriculture, medicine, manufacturing, marketing, etc. Well-known and highly-regarded contributors have written individual chapters that have been extensively reviewed by the Editor to ensure that each individual contribution relates to material found in Volumes 1 and 2 of this book series. Learn More
This book is the first of two volumes that update Oscar Kempthorne's groundbreaking 1952 classic of the same name. This first volume is concerned primarily with the philosophical basis for experimental design and a mathematical-statistical framework within which to discuss the subject. Learn More
The Third Edition features various additions as well as improvements that have been developed over the last decade, and the most significant addition to the text involves technology. LT Assistant, a computer application for learning the simplex method, has been developed by co-author Gerard Keough and was designed to be used with this book. Learn More
This Tenth Edition of Biostatistics maintains its predecessors' comprehensive approach to Biostatistics as it is used in the biological sciences. Successfully utilized by both statisticians and practitioners, Biostatistics is an algebra-based text geared towards the advanced undergraduate and graduate student. text provides the fundamental concepts and techniques of real analysis for students in all of these areas. It helps one develop the ability to think deductively, analyse mathematical situations and extend ideas to a new context Learn More
This version of Advanced Engineering Mathematics by Prof. Erwin Kreyszig, globally the most popular textbook on the subject, is restructured to present the content in a concise and easy-to-understand manner. It fulfills the need for a book that not only effectively explains the concepts but also aids in visualizing the underlying geometric interpretation. Learn More
This book provides an introduction to the theory and applications of modeling and simulation with a multidisciplinary perspective and the authors offer a concise look at the key concepts that make up the field of modeling and simulation. The book is organized into the following three parts: Principles of Modeling and Simulation (Part One) provides a brief history of modeling and simulation, lists the many uses or applications of modeling and simulation, speaks to the advantages and disadvantages of using models in problem solving and covers the two main reasons to employ modeling and simulation by solving a specific problem and using related applications to gain insight into complex concepts. Learn More | 677.169 | 1 |
Axia College Material Appendix A Final Cumulative Test Overview and Timeline Final Cumulative Test Overview Ch. 1-3 1. Real numbers and algebraic expressions 2. Solving algebraic equations and inequalities 3. Graphing linear equations Ch. 7-9 1. Functions 2. Graphs of functions 3. Systems of equations 4. Systems of inequalities Final Exam Timeline You should budget your time wisely and prepare for the final exam throughout the course. The CheckPoints and assignments in the course are designed to assist you in preparing for the Final Cumulative Exam on Ch. 1-3 & 7-9. If you complete your course activities and use the feedback provided by the instructor, you will be on the right track to successfully complete your final exam. Due in Week Two : Expressions and Equations—Part 2 in MyMathLab ® . This quiz assesses the skills learned in Weeks One and Two.
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Solution set of any inequality or compound inequality, which has one-variable, lies in the real line which is one dimensional. So a difficulty appears when computer assisted graphical representation is intended to use for teaching these topics. Sketching a one-dimensional graph by using computer software is not a straightforward work. In this paper, an innovative approach is suggested by using GeoGebra, the free dynamic mathematics software. This approach was theoretically based on Sackur's suggestion, which is created by semiotic register theorem of Duval. The dynamic application, described in this paper, also provides an opportunity to observe more complex, even impossible to solve analytically, inequalities' solution sets visually. At the end of the paper, it is also explained that how this approach also has the potential of allowing the students to make detailed reasoning towards finding solutions of inequalities and compound inequalities. (Contains 2 tables and 12 figures.) | 677.169 | 1 |
Course Summary
Study the topics you'll find on the NY Regents Integrated Algebra exam, including linear equations, quadratics and probability, with this fun review course. Our self-paced learning materials can be used as a comprehensive study guide to prepare for the test at any time, from any mobile device.
Who's it for?
Anyone who needs help understanding material from integrated algebra will benefit from taking this course. You will be able to grasp the subject matter faster, retain critical knowledge longer and earn better grades. You're in the right place if you | 677.169 | 1 |
prepare students for any mathematics course at the college algebra level. Many key topics have been rewritten in response to user and reviewer feedback and have made significant improvements in design and pedagogy. DLC: Algebra | 677.169 | 1 |
Geometry by John R. Silvester
Book Description
This book is a guided tour of geometry, from Euclid through to algebraic geometry. It shows how mathematicians use a variety of techniques to tackle problems, and it links geometry to other branches of mathematic.s It is a teaching text, with a large number of exercises woven into the exposition. Topics covered: ruler and compasses constructions, transformations, triangle and circle theorems, classification of isometries and groups of isometries in dimensions 2 and 3, Platonic solids, conics, similarities, affine, projective and Mobius transformations, non-Euclidean geometry, projective geometry, the beginnings of algebraic geometry.
Buy Geometry book by John R. Silvester from Australia's Online Bookstore, Boomerang Books.
Books By Author John R. Silvester
For students who studied no geometry at school . This problem-based course starts with some history and moves on to constructions, plane geometry, circles and conics to end with an introduction to algebraic geometry | 677.169 | 1 |
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3 0507
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Changes for version 1.1 included the insertion of a new chapter 5 for the 2007 specification and some minor mathematical corrections in chapters 2 and 4. Please note that these are not side-barred. Changes for version 1.2 include further corrections to chapter 4 and a correction to the numbering of the exercises in chapter 5. Changes for version 1.3 include mathematical corrections to exercises and answers.
Introduction The general complex number The modulus and argument of a complex number The polar form of a complex number Addition, subtraction and multiplication of complex numbers of the form x + iy 1.6 The conjugate of a complex number and the division of complex numbers of the form x +iy 1.7 Products and quotients of complex numbers in their polar form 1.8 Equating real and imaginary parts 1.9 Further consideration of |z2 – z1| and arg(z2 –z1) 1.10 Loci on Argand diagrams
Chapter 3: Summation of finite series 3.1 3.2 3.3 3.4 Introduction Summation of series by the method of differences Summation of series by the method of induction Proof by induction extended to other areas of mathematics
Chapter 4: De Moivre's theorem and its applications 4.1 4.2 4.3 4.4 4.5 4.6 4. , where α is a non-real number
know how to add.8 1.6 Introduction The general complex number The modulus and argument of a complex number The polar form of a complex number
Further Pure 2 (MFP2) Textbook
Addition.klmGCE Further Mathematics (6370)
Chapter 1: Complex Numbers
1. you will:
• • • • • •
know what is meant by a complex number. know how to solve equations using real and imaginary parts.
4
.3 1.10 Loci on Argand diagrams
This chapter introduces the idea of a complex number.7 1.2 1. multiply and divide complex numbers. understand what an Argand diagram is.5 1. know how to sketch loci on Argand diagrams.4 1. know what is meant by the modulus and argument of a complex number.1 1. subtraction and multiplication of complex numbers of the form x + iy The conjugate of a complex number and the division of complex numbers of the form x + iy Products and quotients of complex numbers in their polar form Equating real and imaginary parts Further consideration of z2 − z1 and arg( z2 − z1 )
1. subtract.9
1. When you have completed it.
" . This is because real numbers are actually numbers of the form x + 0i. both (+8) 2 and (−8) 2 are equal to 64. ! (with which you are familiar). The term x + iy is a complex number with x being the real part and iy the imaginary part. is really a subset of the set of complex numbers.
−1 = i
It follows that
i 2 = −1
−64 = 64 × −1 = 64 × −1 = 8i. For example.
5
. if you use a calculator to evaluate −64 you get an error message. The set of real numbers. a symbol is used to denote it – the symbol used is i. where x and y are real numbers. As
−1 cannot be evaluated. So.klmGCE Further Mathematics (6370)
1. both 2 + 3i and −1 − 4i are complex numbers. This is because squaring every real number gives a positive value.2
The general complex number
The most general number that can be written down has the form x + iy .
1.1 Introduction
Further Pure 2 (MFP2) Textbook
You will have discovered by now that some problems cannot be solved in terms of real numbers.
3
Further Pure 2 (MFP2) Textbook
The modulus and argument of a complex number
Just as real numbers can be represented by points on a number line. are negative.
6
. –π). The argument of a complex number z is given y by arg z = θ . The point P(x.klmGCE Further Mathematics (6370)
1. arg z. y) r θ O x
If the complex number x + iy is denoted by z. Thus z = OP = r. y) in the plane of coordinates with axes Ox and Oy represents the complex number x + iy and the number is uniquely represented by that point. and hence z = x + iy. where tan θ = x You must be careful when x or y. The modulus of a complex number z is given by z = x 2 + y 2 The argument of z. The diagram of points in Cartesian coordinates representing complex numbers is called an Argand diagram. is defined as the angle between the line OP and the positive x-axis – usually in the range (π. y
P(x. or both. complex numbers can be represented by points in a plane. z ('mod zed') is defined as the distance from the origin O to the point P representing z.
it follows that z may be written in the form r cos θ + ir sin θ . where z = r and arg z = θ For brevity. 3 3
(
)
8
. θ). Write the complex numbers given in Exercise 1A in polar coordinate form.
Further Pure 2 (MFP2) Textbook y P(x. r (cosθ + i sin θ ) can be written as (r.klmGCE Further Mathematics (6370)
1. or modulus–argument. 4 4
(
)
(b) 4 cos − 2π + i sin − 2π . Find. This is called the polar.4 The polar form of a complex number
In the diagram alongside. 2. If P is the point representing the complex number z = x + iy. the complex numbers given in polar coordinate form by: (a) z = 2 cos 3π + i sin 3π . in the form x + iy. y) r θ O x
A complex number may be written in the form z = r (cosθ + i sin θ ). form of a complex number. x = r cos θ and y = r sin θ .
Exercise 1B
1.
that is to say z2 − z1 is the length AB in the Argand diagram. Similarly arg( z2 − z1 ) is the angle between OC and the positive direction of the x-axis. This in turn is the angle between AB and the positive x direction. If the complex number z1 is represented by the point A, and the complex number z2 is represented by the point B in an Argand diagram, then
z2 − z1 = AB, and arg( z2 − z1 ) is the angle between AB and the positive
A locus is a path traced out by a point subjected to certain restrictions. Paths can be traced out by points representing variable complex numbers on an Argand diagram just as they can in other coordinate systems. Consider the simplest case first, when the point P represents the complex number z such that z = k . This means that the distance of P from the origin O is constant and so P will trace out a circle.
z = k represents a circle with centre O and radius k
If instead z − z1 = k , where z1 is a fixed complex number represented by the point A on an Argand diagram, then (from Section 1.9) z − z1 represents the distance AP and is constant. It follows that P must lie on a circle with centre A and radius k.
z − z1 = k represents a circle with centre z1 and radius k
Note that if z − z1 ≤ k , then the point P representing z can not only lie on the circumference of the circle, but also anywhere inside the circle. The locus of P is therefore the region on and within the circle with centre A and radius k. Now consider the locus of a point P represented by the complex number z subject to the conditions z − z1 = z − z2 , where z1 and z2 are fixed complex numbers represented by the points A and B on an Argand diagram. Again, using the result of Section 1.9, it follows that AP = BP because z − z1 is the distance AP and z − z2 is the distance BP. Hence, the locus of P is a straight line.
z − z1 = z − z2 represents a straight line – the perpendicular
bisector of the line joining the points z1 and z2 Note also that if z − z1 ≤ z − z2 the locus of z is not only the perpendicular bisector of AB, but also the whole half plane, in which A lies, bounded by this bisector. y All the loci considered so far have been related to distances – there are also simple loci in Argand diagrams involving angles. O α The simplest case is the locus of P subject to the condition that arg z = α , where α is a fixed angle.
P
x
15
klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
This condition implies that the angle between OP and Ox is fixed (α ) so that the locus of P is a straight line. arg z = α represents the half line through O inclined at an angle α to the positive direction of Ox Note that the locus of P is only a half line – the other half line, shown dotted in the diagram above, would have the equation arg z = π + α , possibly ±2π if π + α falls outside the specified range for arg z. In exactly the same way as before, the locus of a point P satisfying arg( z − z1 ) = α , where z1 is a fixed complex number represented by the point A, is a line through A. arg( z − z1 ) = α represents the half line through the point z1 inclined at an angle α to the positive direction of Ox
y P A O α
x
Note again that this locus is only a half line – the other half line would have the equation arg( z − z1 ) = π + α , possibly ±2π. Finally, consider the locus of any point P satisfying α ≤ arg( z − z1 ) ≤ β . This indicates that the angle between AP and the positive x-axis lies between α and β , so that P can lie on or within the two half lines as shown shaded in the diagram below.
y
(b) (i) Write down the modulus and argument of each of the complex numbers 4 + 2i and 3 − i. where a and b are real. show that the triangle OPQ is right-angled. (a) Indicate on an Argand diagram the region of the complex plane in which 0 ≤ arg ( z + 1) ≤ 2π . (i) Calculate the value of z A . (a) The complex numbers z and w are such that z = ( 4 + 2i )( 3 − i ) and w = 4 + 2i . Give each modulus in an exact surd form and each argument in radians between − π and π. 3 (b) The complex number z is such that 0 ≤ arg ( z + 1) ≤ 2π 3 π ≤ arg z + 3 ≤ π. respectively. 3−i
Express each of z and w in the form a + ib. (ii) Express z A in the form a + ib. 4 + 2i and 3 − i. (ii) The points O. and ( ) 6 (i) Sketch another Argand diagram showing the region R in which z must lie. (c) At the point A defined in part (b)(ii).klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
8. (ii) Mark on this diagram the point A belonging to R at which z has its least possible value.
[AEB June 1997]
20
. Find the exact length of PQ and hence. z = z A .
[AQA March 1999]
9. P and Q in the complex plane represent the complex numbers 0 + 0i. or otherwise.
know how to extend these results to polynomials of higher degree.1 2. be able to form cubic equations with related roots. know that there is a relationship between the number of real roots and form of a polynomial equation.
21
.2 2. know the relationship between the roots of a cubic equation and its coefficients. and be able to sketch graphs. When you have completed it.klmGCE Further Mathematics (6370)
2.8 Introduction Quadratic equations Cubic equations
Further Pure 2 (MFP2) Textbook
Chapter 2: Roots of Polynomial Equations
Relationship between the roots of a cubic equation and its coefficients Cubic equations with related roots An important result Polynomial equations of degree n Complex roots of polynomial equations with real coefficients
This chapter revises work already covered on roots of equations and extends those ideas.6 2.7 2. know that complex conjugates are roots of polynomials with real coefficients.4 2.5 2. you will: • • • • • • know how to solve any quadratic equation.3 2.
2. even if this section is familiar to you it provides a suitable base from which to move on to equations of higher degree. by the quadratic formula. However.klmGCE Further Mathematics (6370)
2. 0)
x
(0. In this chapter you are going to study the properties of the roots of these equations and investigate methods of solving them. which are x = 2 and − 4. You will know. one with x 4 as the highest power of x is called a quartic equation. –8)
The roots of this quadratic equation are those of ( x − 2)( x + 4) = 0. Similarly. a polynomial equation of degree 3 has x 3 as the highest power of x and is called a cubic equation. 0)
(2. a sketch of part of y = x 2 + 2 x − 8 is shown below.
22
. one with x 2 as the highest power of x. For example. that quadratic equations of the type you have met have two roots (which may be coincident). A polynomial equation of degree 2.
y
(–4.2
Quadratic equations
You should be familiar with quadratic equations and their properties from your earlier studies of pure mathematics. There are normally two ways of solving a quadratic equation – by factorizing and.e. the roots of the equation ax 2 + bx + c = 0 are the points of intersection of the curve y = ax 2 + bx + c and the line y = 0 (i. for example. the x-axis).1 Introduction
Further Pure 2 (MFP2) Textbook
You should have already met the idea of a polynomial equation. in cases where this is impossible. Graphically. is called a quadratic equation.
(0. using ideas from 2 4 ± 2i or 2 ± i. The quadratic equation ax 2 + bx + c = 0 . It follows that the equation x 2 − 4 x + 5 = 0 does have Chapter1. 0)
x
In this case.klmGCE Further Mathematics (6370)
y
Further Pure 2 (MFP2) Textbook
A sketch of part of the curve y = x 2 − 4 x + 4 is shown below. the curve touches the x-axis.
y
(0. 4)
O
(2. 2 ⎛ ⎞ Certainly. has complex roots if b 2 − 4ac < 0
23
. 1)
O x
This curve does not touch the x-axis so the equation x 2 − 4 x + 5 = 0 cannot have real roots. this becomes 2 two roots. You may also have observed that whether a quadratic equation has real or complex roots depends on the value of the discriminant b 2 − 4ac. where a. This leads to x = 4 ± 16 − 20 and. The equation x 2 − 4 x + 4 = 0 may be written as ( x − 2) 2 = 0 and x = 2. 5) (2. In fact the two roots are complex conjugates. b and c are real. Not all quadratic equations are as straightforward as the ones considered so far. a repeated root. A sketch of part of the curve y = x 2 − 4 x + 5 is shown below. but they are both complex numbers. x 2 − 4 x + 5 will not factorize so the quadratic formula ⎜ x = −b ± b − 4ac ⎟ has to 2a ⎝ ⎠
be used to solve this equation.
3
Cubic equations
As mentioned in the introduction to this chapter. Solve the equations (a) x 2 + 6 x + 10 = 0. Also. Hence the curve must cross the line y = 0 at least once.
24
. If a < 0. as x → ∞. (b) x 2 + 10 x + 26 = 0.
2. the term ax3 becomes the dominant part of the expression and ax3 → ∞ (if a > 0) .klmGCE Further Mathematics (6370)
Exercise 2A
Further Pure 2 (MFP2) Textbook
1. whilst ax3 → −∞ when x → −∞. All cubic equations have at least one real root – and this real root is not always easy to locate. equations of the form ax 3 + bx 2 + cx + d = 0 are called cubic equations. The reason for this is that cubic curves are continuous – they do not have asymptotes or any other form of discontinuity. and ax3 → ∞ as x → −∞ and this does not affect the result. then ax3 → −∞ as x → ∞.
When you have completed it.klmGCE Further Mathematics (6370)
3.1 3.2 3. to other kinds of series. be able to apply the method of induction in circumstances other than in the summation of series. understand an important method known as the method of induction.
38
. know which method is appropriate for the summation of a particular series. with which you are familiar from earlier studies.3 3. you will: • • • • know new methods of summing series.4 Introduction Summation of series by the method of differences Summation of series by the method of induction
Further Pure 2 (MFP2) Textbook
Chapter 3: Summation of Finite Series
Proof by induction extended to other areas of mathematics
This chapter extends the idea of summation of simple series.
2 + 6 + 18 + 54 + 162 + 486
is a series of 6 terms. Thus. For instance. The two series above are examples of finite series. the sum of an arithmetic progression is a series. That is.
39
. with common ratio 3.
2 + 5 + 8 + 11 + 14
is a series of 5 terms. you multiply by a fixed number (called the common ratio). with common difference 3.1 Introduction
Further Pure 2 (MFP2) Textbook
You should already be familiar with the idea of a series – a series is the sum of the terms of a sequence.klmGCE Further Mathematics (6370)
3. in geometric progression. A finite series is a series with a finite number of terms. the sum of a number of terms where the terms follow a definite pattern. The sum of a geometric progression is also a series. Instead of adding a fixed number to find the next consecutive number in the series. In this case each term is bigger than the preceeding term by a constant number – this constant number is usually called the common difference. Thus. in arithmetic progression.
the sum has exactly the same form as S(n) but with n replaced by k + 1. the next term in the series. then you are assuming that the sum of the first k terms is S(k). You may think that this rather begs the question but it must be understood that the result is assumed to be true for only one value of n. the examples worked in Section 3. say. To summarise: 1 Assume that the result of the summation is true for n = k and prove that it is true for n = k + 1 2 Prove that the result is true for n = 1 Statement 1 shows that.3
Further Pure 2 (MFP2) Textbook
Summation of a series by the method of induction
The method of induction is a method of summing a series of. it can be said that the summation result is true for all positive integers n. it is demonstrated that the result is true for n = 1. Suppose you have to show that the sum of n terms of a series is S(n). by putting k = 1 (which is known to be true from Statement 2). and Statement 1 shows that by putting k = 2 the result must be true for n = 3. 1 + 1 + 1 +… + 1 1 1 . and so on. where k < n.2 are used again here. namely n = k . say k. the result must be true for n = 2. Finally.klmGCE Further Mathematics (6370)
3. + = k + 1× 2 2 × 3 3 × 4 k ( k + 1) ( k + 1)( k + 2 ) k + 1 ( k + 1)( k + 2 ) Then k +1 1 k 1 ∑ r r +1 = k +1 + k +1 k + 2 ( ) ( )( ) r =1
= k ( k + 2) + 1 ( k + 1)( k + 2 )
2 = k + 2k + 1 ( k + 1)( k + 2 )
45
. By building up the result. ( r + 1) n + 1
Solution
Assume that the result is true for n = k . If you assume that the summation is true for one particular integer. You then use this assumption to prove that the sum of the series to k + 1 terms is S ( k + 1) – that is to say that by adding one extra term. There is a formal way of writing out the method of induction which is shown in the examples below. and comparison. that is to say 1 + 1 + 1 +… + 1 = k . For convenience.
Example 3.1
Show that
r =1
∑r
n
1 = n .3. n terms when the sum is given in terms of n. 1× 2 2 × 3 3 × 4 k ( k + 1) k + 1 Adding the next term to both sides.
The method of induction is certainly useful in the summation of series but it is not confined to this area of mathematics. This chapter concludes with a look at its use in three other connections – sequences, divisibility and de Moivre's theorem for positive integers.
which is of the same form as uk but with k + 1 replacing k. Hence, if the result is true for
1+1 n = k , it is true for n = k + 1. But when k = 1, u1 = 2 1 − 1 = 3 as given. Therefore the 2 −1 result is true for all positive integers n ≥ 1 by induction.
Prove by induction that if n is a positive integer, 32 n + 7 is divisible by 8.
Solution
The best approach is a little different to that used so far. Assume that the result is true for n = k , in other words that
2 k +1 2 k +1 When n = k + 1 the expression is 3 ( ) + 7. Consider 3 ( ) + 7 − 32 k + 7 , the difference
. in particular.
53
.6 4. know how to work out the nth roots of unity and. be able to find shorter ways of working out powers of complex numbers. be able to solve certain types of polynomial equations. know a new way of expressing complex numbers. When you have completed it.5 4.2 4.1 4. where α is a non-real number
This chapter introduces de Moivre's theorem and many of its applications.4 4. the cube roots.klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
Chapter 4: De Moivre's Theorem and its Applications
4.3 4. discover alternative methods for establishing some trigonometric identities. you will: • • • • • • know the basic theorem.
(
cos π + i sin π 6 6
) = cos 3π + i sin 3π 6 6
3
= cos π + i sin π 2 2 = 0+i = i. 6 6
(
)
3
Solution
It would. be possible to multiply cos π + i sin π by itself three times. The method for doing this will be illustrated through examples.
Example 4.2. Instead. but this would 6 6 be laborious and time consuming – even more so had the power been greater than 3. of course.1
Simplify cos π + i sin π .
56
.klmGCE Further Mathematics (6370)
4.2
n
Further Pure 2 (MFP2) Textbook
Using de Moivre's theorem to evaluate powers of complex numbers
One very important application of de Moivre's theorem is in the addition of complex numbers of the form ( a + ib ) .
(c) Express each of these four roots in the form a + ib and show. with the aid of a diagram. (c) Deduce the exact values of sin π and sin 2π . show that sin 5θ ≡ sin θ 16sin 4 θ − 20sin 2 θ + 5 . where r > 0 and − π < θ ≤ π. (a) By considering z = cos θ + i sin θ and using de Moivre's theorem. where r > 0 and − π < θ ≤ π. (b) Hence solve the equation z 4 + 64 = 0 giving your answers in the form r ( cos θ + i sin θ ) . where a and b are real numbers to be determined to two decimal places. (a) Write down the modulus and argument of the complex number −64.
[AEB June 1996]
2. (b) Using your answers to part (a). (b) Find the exact values of the solutions of the equation 16 x 4 − 20 x 2 + 5 = 0. 2 2
z 3 = (1 + i )
(
3 −i
)
giving your answers in the form a + ib.
[AQA January 2002]
(
)
81
. (a) Express each of the complex numbers 1+ i and 3 −i in the form r ( cos θ + i sin θ ) . (i) show that (ii) solve the equation
(
3 −i
)
5
(1 + i )
10
= − 1 + 3 i. that the points in the complex plane which represent them form the vertices of a square. explaining clearly the reasons for your 5 5 answers.klmGCE Further Mathematics (6370)
Miscellaneous exercises 4
Further Pure 2 (MFP2) Textbook
1.
[AQA June 2001]
3.
Show that these roots lie on a circle. (iii) Express p and q as integer multiples of cos 2π and cos 4π .
82
. 5 5
[NEAB June 1998]
2πi 5 . (ii) Write down the quadratic equation. (c) Mark on an Argand diagram the points corresponding to the five roots of the equation. with integer coefficients. Find the possible values of a. where w = e (i) Show that p + q = −1 and pq = −1. w3 and w4 . respectively. where a and b are integers to be determined. whose roots are p and q.
(ii) Show that the other fifth roots of unity are 1. (a) (i) Show that w =
2πi e5
is one of the fifth roots of unity. (a) Verify that is a root of the equation
z1 = 1 + e 5
πi
( z − 1)
5
= −1. (d) By considering the Argand diagram. 2 2 2 a −ω +ω a +ω −ω where ω is one of the non-real cube roots of unity. (b) Let p = w + w4 and q = w2 + w3 .
[AQA June 2000]
5. 5 5 (iv) Hence obtain the values of cos 2π and cos 4π in surd form.
(b) Find the other four roots of the equation. w2 .
Further Pure 2 (MFP2) Textbook
4. and state the centre and radius of the circle.klmGCE Further Mathematics (6370)
z 2 + z + 1 = 0. find (i) arg z1 in terms of π. (ii) z1 in the form a cos π . b
[AQA Specimen]
6. (a) Show that the non-real cube roots of unity satisfy the equation
(b) The real number a satisfies the equation 1 1 + = 1.
(d) Find the area of the triangle ABC. α . B and C corresponding to the three roots found in part (b). where r is a surd and − π < θ ≤ π. and find the other two roots giving your answers in the form reiθ . B and C. (e) The point P lies on the circle through A. β and γ the complex numbers represented by P. A. show that
πi
( w − α )2 + ( w − β )2 + ( w − γ ) 2
[AQA June 1999]
= 6. Denoting by w. (c) Indicate on an Argand diagram points A. giving your answer in surd form. (a) Express the complex number 2 + 2i in the form reiθ .klmGCE Further Mathematics (6370)
(b) Show that one of the roots of the equation z 3 = 2 + 2i
Further Pure 2 (MFP2) Textbook
9. where r > 0 and − π < θ ≤ π.
84
.
is 2 e12 . respectively. B and C.
be able to rewrite more complicated expressions in a form that can be reduced to standard integrals. be able to recognise algebraic expressions which integrate to standard integrals. you will: • • • • be able to recognise the derivatives of standard inverse trigonometrical functions. When you have studied it. be able to extend techniques already familiar to you to differentiate more complicated expressions.3 5.2 5.1 5.4 5.
85
.5 Introduction and revision The derivatives of standard inverse trigonometrical functions Applications to more complex differentiation Standard integrals integrating to inverse trigonometrical functions Applications to more complex integrals
This chapter revises and extends work on inverse trigonometrical functions.klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
Chapter 5: Inverse Trigonometrical Functions
5.
but for any given value of y there are infinitely many values of x . Note that sin −1 y is not cosec y which would normally be written as (sin y ) −1 when expressed in terms of sine.
86
. we write x = sin −1 y (or arc sin y ) . However. for a given value of x .klmGCE Further Mathematics (6370)
5. y = sin −1 x has infinitely many values. The use of the superscript -1 is merely the convention we use to denote an inverse in the same way as we say that f −1 is the inverse of the function f .
For any given value of x there is only one corresponding value of y . In order to overcome this obstacle.
As it stands. but if we wish to describe sin −1 x as a function. If y = sin x . The graph of y = sin −1 x being the inverse. we restrict the range of y to − π ≤ y ≤ π so that the 2 2 −1 sketch of y = sin x becomes the sketch shown. is the reflection of y = sin x in the line y = x and a sketch of it is as shown.1 Introduction and revision
Further Pure 2 (MFP2) Textbook
You should have already met the inverse trigonometrical functions when you were studying the A2 specification module Core 3. and for the sake of completeness some revision is included in this section. The sketch of y = sin x will be familiar to you and is shown below. in order to present a clear picture. we must make sure that the function has precisely one value.
klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
By doing this. The sketches of y = cos x and y = cos −1 x are shown below. 2 2
Notice that the gradient of y = sin −1 x is always greater than zero. This value is usually called the principal value.
87
.
sin −1 x is the angle between − 1 π and 1 π inclusive whose sine is x. we ensure that for any given value of x there is a unique value of y for which y = sin −1 x . We can define cos −1 x in a similar way but with an important difference.
klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
In this case.
When it comes to tan −1 x we can restrict the range to − π and π .
cos −1 x is the angle between 0 and π inclusive whose cosine is x. it would not be sensible to restrict y to values between − π and π since for 2 2 every value of x ≥ 0 there would be two values of y and for values of x < 0 there would be no value of y. Instead we choose the range 0 ≤ x ≤ π and the sketch is as shown. 2 2
The sketch of y = tan −1 x is shown below.
Exercise 5A
1. Express in terms of π the values of: (a) tan −1 1 (d) cos −1 0 (b) cos −1 3 2 ⎛ ⎞ (e) tan −1 ⎜ − 1 ⎟ 3⎠ ⎝ (c) sin −1 − 1 2
( )
(f) cos −1 (−1)
88
. 2 2
tan −1 x is the angle between − π and + π inclusive whose tangent is x.
∫ a 2 + x 2 = a tan
The second integral is
dx
1
−1
x +c a
∫
dx a − x2
2
This interval also requires a substitution Let x = a sin θ Then
∫
dx a −x
2 2
=
dx = a cos θ dθ a cos θ dθ
∫
= ∫ a cos θ d θ a cos θ =θ +c = sin −1 x + c a
a 2 − a 2 sin 2 θ
()
∫
dx a −x
2 2
= sin −1 x + c a
()
94
.klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
5. formulae from the Formulae and Statistical Tables Booklet supplied for each AS and A2 module apart from MPC1 can be quoted without proof. However. this does not preclude a question requiring a proof of a result from this booklet being set.4 Standard integrals integrating to inverse trigonometrical functions
Generally speaking. There are two standard results. the proofs of which are given here and the methods for these proofs should be committed to memory. as you have been taught. The first one is
∫ a 2 + x2
Let x = a tan θ so that dx = a sec2 θ dθ Then dx = a sec 2 θ dθ ∫ a 2 + x 2 ∫ a 2 + a 2 tan 2 θ 2 = ∫ a sec θ2dθ a 2 sec θ = ∫ 1 dθ a 1θ +c = a 1 tan −1 x + c = a a
dx
This integral requires a substitution.
so tanh x → 1 as x → ∞. ⎝ 1+ e ⎠ Now. from which it can be deduced that tanh x < 1. It would also be worthwhile committing the general shapes of y = sinh x. it follows that the numerator of this fraction is less than its denominator. 1+ e −2 x As e > 0 for all values of x. So the graph of y = tanh x has an asymptote at x = 1.
y
y
y = sinh x
0
x
y = cosh x
1 0
x
The sketch of y = tanh x requires a little more consideration.klmGCE Further Mathematics (6370)
6. it follows that the numerator in the bracketed expression above is less than its denominator. so that tanh x > −1. So the graph of y = tanh x has an asymptote at y = −1. y = cosh x and y = tanh x to memory. e 2 x → 0 and tanh x → − 1 = −1. In Section 5.1. and has y = ±1 as asymptotes. as x → −∞. Some sketches are given below but it would be a good exercise to make a table of values and confirm the general shapes for yourself.3 Graphs of hyperbolic functions
Further Pure 2 (MFP2) Textbook
The graphs of hyperbolic functions can be sketched easily by plotting points. as e 2 x > 0 for all values of x. It can also be deduced that as e −2 x → 0 as x → ∞.
y = tanh x
y
−2 x
1 0 –1
x
105
. Hence the curve y = tanh x lies between y = +1 and y = −1. if the numerator and denominator of tanh x are divided by e 2 x . tanh x = 1 − e −2 x . it was shown that tanh x could be written as 2x tanh x = e2 x − 1 e +1 2x ⎞ ⎛ = − ⎜ 1 − e2 x ⎟ . 1
()
Now. Also.
as the next example shows. It must not be used in a proof – for example.
cos 2 x = 1 − 2sin 2 x cosh 2 x = 1 + 2sinh 2 x. In fact the only differences are those of sign – for example.klmGCE Further Mathematics (6370)
6.
because sin 2 x is a product of two sines.
110
.5 Osborne's rule
Further Pure 2 (MFP2) Textbook
It should be clear that the results and identities for hyperbolic functions bear a remarkable similarity to the corresponding ones for trigonometric functions. To change a trigonometric function into its corresponding hyperbolic function.4 must be used for that. There is a rule for obtaining the identities of hyperbolic functions from those for trigonometric functions – it is called Osborne's rule. care must be exercised in using this rule. whereas cos 2 x + sin 2 x = 1. It is known that sec 2 x = 1 + tan 2 x but sech 2 x = 1 − tanh 2 x The reason that the sign has to be changed here is that a product of sines is implied because 2 tan 2 x = sin 2 x . cos x It should be noted that Osborne's rule is only an aid to memory. The method shown in Section 5. where a product of two sines appears change the sign of the corresponding hyperbolic term For example. because then Note also that because then
cos ( x + y ) = cos x cos y − sin x sin y cosh ( x + y ) = cosh x cosh y + sinh x sinh y. However. that cosh 2 x − sinh 2 x = 1. the corresponding hyperbolic identity is cosh 2 x − sinh 2 x = 1.
say. and hence an inverse. sinh −1 2 using a calculator.3) you will see that for every value of y > 1 there are two values of x. and the range for the inverse will be y ≥ 0. This is because the mapping f : x → cosh x is not a one-to-one mapping. so there are inverse hyperbolic functions.
y = tanh
y = sinh
−1
(
)
y
−1
x
y
x
0
x
–1
0
1
x
Note also that y = cosh x does not have an inverse. They are defined in a similar way to inverse trigonometric functions – so.klmGCE Further Mathematics (6370)
6. in the line y = x. respectively. If you look at the graph of y = cosh x (in Section 5. then y = sinh −1 x. These sketches are shown below. . The sketches of y = sinh −1 x and y = tanh −1 x are the reflections of y = sinh x and y = tanh x. cos −1 x. you use it in the same way as you would if it was a trigonometric function (pressing the appropriate buttons for hyperbolic functions).8 Inverse hyperbolic functions
Further Pure 2 (MFP2) Textbook
Just as there are inverse trigonometric functions sin −1 x. if the domain of y = cosh x is restricted to x ≥ 0 there will be a one-to-one mapping. etc.
115
. and likewise for the other five hyperbolic functions. if x = sinh y. However. Note that the curve y = tanh −1 x has asymptotes at x = ±1. To find the value of.
(a) Explain. show that 1 − x2 tanh −1 x = 1 ln 1 + x . (ii) Evaluate y at the stationary point. 1 and 2. (a) Prove that d tanh x = sech 2 x. 2 1− x
( )
(d) Show that
∫
[AQA June 1990]
1 2
0
tanh −1 x dx = a ln b 2 . or otherwise. where n is an integer. by means of a sketch.
129
.
[NEAB March 1998]
10. giving your answer in the form p − ln q.
[AQA March 2000]
9. where a is to be determined. (b) Given that cosh x = 17 and sinh y = 4 . dx 1 − x2
(
)
(c) By expressing
1 in partial fractions and integrating. prove that d tanh −1 x = 1 .klmGCE Further Mathematics (6370)
Further Pure 2 (MFP2) Textbook
8. k = 1 and k > 1 are 0. (a) State the values of x for which cosh −1 x is defined. (ii) show that one of the possible values of x + y is ln12 and find the other possible value in the form ln a. (b) A curve C is defined for these values of x by the equation y = x − cosh −1 x. respectively. ( ) 1 − x2
where a and b are numbers to be determined. where p and q are numbers to be determined. 8 3
(i) express y in the form ln n. ( ) dx (b) Hence. (i) Show that C has just one stationary point. why the numbers of distinct values of x satisfying the equation cosh x = k in the cases k < 1.
The diagram shows a region R in the x–y plane bounded by the curve y = sinh x. 2k
B
x
(ii) Show that the area of the region R is 2 . ⎦ 4⎣
(ii) Hence find. the volume swept out when the region R is rotated through an angle of 2π radians about the x-axis.klmGCE Further Mathematics (6370)
y
Further Pure 2 (MFP2) Textbook
11. 4
2 (b) (i) Show that cosh ( ln k ) = k + 1 . correct to three significant figures. show that OB = ln 3. the x-axis and the line AB which is perpendicular to the x-axis. A
R
O (a) Given that AB = 3 . 3
(c) (i) Show that
∫
ln 3 0
sinh 2 x dx = 1 ⎡sinh ( ln 9 ) − ln 9 ⎤ .
[NEAB June 1998]
130
.
know a formula which can be used to evaluate the length of an arc when the equation of the curve is given in parametric form.2 7. you will: • • • know a formula which can be used to evaluate the length of an arc when the equation of the curve is given in Cartesian form.3 Introduction Arc length Area of surface of revolution
Further Pure 2 (MFP2) Textbook
Chapter 7: Arc Length and Area of Surface of Revolution
This chapter introduces formulae which allow calculations concerning curves.klmGCE Further Mathematics (6370)
7.1 7. know methods of evaluating a curved surface area of revolution when the equation of the curve is given in Cartesian or in parametric form. When you have completed it.
131
.
the area under the curve y = f ( x ) above the x-axis and between the lines x = a and x = b is given by A =
Further Pure 2 (MFP2) Textbook
y
∫a
b
y dx.
∫
b a
πy 2 dx which gives the volume of the solid
132
. For example. differentiation and manipulation of algebraic. the skills needed to solve them do not concern the formulae themselves but involve integration. as with many problems.
O
a
b
x
You will also be familiar with the formula V =
of revolution when that part of the curve between the lines x = a and x = b is rotated about the x-axis. The formulae to be introduced in this chapter should be committed to memory.klmGCE Further Mathematics (6370)
7. You should also realise that.1 Introduction
You will probably already be familiar with some formulae to do with the arc length of a curve and the area of surface of revolution. trigonometric and hyperbolic functions – many of which have been introduced in earlier chapters.
it follows that 2πyδ s < δ A < 2π ( y + δ y ) δ s
2πy < δ A < 2π ( y + δ y ) . respectively. δs Now as δ x → 0. If the actual area generated by the rotation of arc PQ about the x-axis is denoted by δA. dA = 2πy ds
A= =
or. Using the formula S = 2πrh for the area of the curved surface of a cylinder. and the length of arc PQ is δs. are taken on the curve y = f ( x ) . P and Q. the area of the former is 2πyδ s and that of the latter is 2π ( y + δ y ) δ s. The area of this surface is known as the 'curved surface area' or 'area of surface of revolution'. Therefore. This arc is rotated about the x-axis by 2π radians. it forms a surface.3 Area of surface of revolution
Further Pure 2 (MFP2) Textbook
If an arc of a curve is rotated about an axis. but smaller than the area of the cylinder width δs obtained by rotating the point Q about the same axis. δ y → 0 and δ s → 0 so that the right-hand side of the inequality tends to 2πy. The coordinates of P and Q are ( x. Suppose two closely spaced points.
You can see from the diagram that the curved surface generated by the rotation is larger than that of the cylinder of width δs obtained by rotating the point P about the x-axis. y + δ y ) . y ) and
y P y O a b δs δx y x Q δy
( x + δ x.klmGCE Further Mathematics (6370)
7.2)
The area of surface of revolution obtained by rotating an arc of the curve y = f ( x ) through 2π radians about the x-axis between the points where x = a and x = b is given by A=
∫
⎛ dy ⎞ 2πy 1 + ⎜ ⎟ dx a ⎝ dx ⎠
b
2
137
. dividing by δs.
∫ ∫
b a
2πy ds
2
⎛ dy ⎞ 2πy 1 + ⎜ ⎟ dx a ⎝ dx ⎠
b
(from section 6. | 677.169 | 1 |
Speedstudy Algebra 1 is the fun way to learn first-degree equations, inequalities, operations with powers and exponents, linear equations with variables, polynomials, rational expressions, quadratic equations, and more. With easy-to-follow animated lessons, fast-paced game play, and a truly entertaining deep-space interface, this program provides a unique way to learn algebra.
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Speedstudy Pre-Calculus is the ideal preparation for calculus. This product is designed to provide a solid foundation in the concepts and skills necessary to succeed in calculus. The curriculum-based lessons are designed by educators to help students every step of the way. You'll learn about functions, continuity, independent and dependent variables, functions as variation, the slope of a chord, techniques of finding limits, and other lessons critical to preparation for calculus.
Speedstudy Calculus is your chance to take the mystery and confusion out of calculus, and gain a solid understanding of the subject. You'll learn the properties of derivatives, the derivatives of common functions, implicit and successive differentiation, maximum and minimum values, integrals of common function, numerical integration, and much more. You will be pleased with this product.
These programs are not just textbooks on CD-ROMs—they're like having a fun, personal tutor of your very own. These four Speedstudy programs provide a solid educational foundation designed to raise grades and test scores and improve skills in the classroom and beyond. Using step-by-step animations and a fun 3-D interface, these products give students the tools they need to master important concepts. | 677.169 | 1 |
Mathematics Options
Mathematics Options
Below are details and insights about the Math Department's course offerings, guidelines, requirements, curriculum, and pathways.
Common Core
Common Core changes are now the major focus of the math department and all of our math teachers are embracing the new changes. We are currently introducing them in our Algebra, Geometry and Algebra II classes. In mathematics, the Sequoia Union High School District is planning a three-year rollout. We will formally teach Common Core Algebra next year, introduce Common Core Geometry in 2015-16 and finally Common Core Algebra II in 2016-2017. The district school board has decided that we will continue teaching our traditional structure instead of changing to an international pathway (Integrated Math). The biggest change in content is that statistics will be heavily emphasized and integrated into the Algebra and Algebra II classes. Students will be engaged more in solving open-ended real-world problems. Another major difference is that testing will be computer based and include free response questions requiring a written response rather than a simple multiple-choice answer as in the CST. Testing will take place during a student's junior year. The California High School Exit Exam (CAHSEE), a requirement for graduation, will continue to be first administered to students in the spring of their sophomore year.
Requirements
M-A requires a minimum of two years of math to graduate. The UC system and the California state colleges require a minimum of three years of math, including Algebra II. Any student planning to attend a four-year college should successfully complete Algebra II with a C or higher. Students can improve their college eligibility by taking as much math as possible in the traditional sequence of Algebra I, Geometry, Algebra II, Pre-calculus, and Calculus. Students who have difficulty with Algebra I may decide to take Integrated Math, a bridge course, before moving onto Geometry.
Course Explanations
Algebra is considered the gateway course to college; without it, college success is unlikely. Algebra I is required for M-A graduation and is part of the math content of the California High School Exit Exam. Beyond that, algebra skills can significantly affect a student's calculus performance throughout the year. Calculus teachers have a saying: "Many times when solving a calculus problem, the first step is calculus and the rest is all algebra."
M-A has two options for the third year of math required by the California schools: AS Algebra II and Algebra II. The prerequisite for AS Algebra II is a minimum of a B in geometry, while Algebra II requires only a C in Geometry. AS Algebra II is intended primarily for sophomores and freshman who intend to complete calculus. Algebra II is intended for seniors and juniors who have found math challenging but who still want rigorous preparation for college math courses. Students should know these differences when enrolling in either AS or regular Algebra II. The math department strongly recommends that juniors and seniors enroll in Algebra II, not AS Algebra II. Juniors leaving Algebra II can take Statistics or Algebra II/trigonometry, which will prepare them for Pre-calculus or other math courses required in college.
Another option for juniors or seniors is AP Statistics or Statistics. AP statistics requires completion of or concurrent enrollment in Pre-calculus, while Statistics requires completion of or concurrent enrollment in Algebra II. The two courses differ in pacing and content. AP Statistics is a fun, hands-on course, but it moves at an intense pace, twice as fast as Statistics. Statistics allows students more time to practice and master the material. The Math Department encourages students to take two math courses concurrently so that they can complete Calculus and AP Statistics. The majority of students taking AP statistics are also enrolled in AP Calculus or Pre-calculus. At M-A, approxiamtely 160 students are taking AP Statistics or Statistics.
Students also have the option of taking BC Calculus or AB Calculus. The complicated difference between these courses is explained in detail at our annual January meeting of parents and students. Please note for future planning that BC calculus students participate in a four-week summer program that typically runs from the Monday after graduation until the week of the fourth of July, immediately following their pre-calculus course.
If you have questions about the Math Department offerings, please contact Jen Payne at jpayne@seq.org. | 677.169 | 1 |
8.F.1 Tables, Graphs, and Equations
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
2.12 MB | 9 pages
PRODUCT DESCRIPTION
It is easy for teachers to overlook that the relationships between tables, graphs, and equations are not always immediately apparent to students. In this introduction to functions, students see different functions in context and create alternate representations of the same function. Each function has a real-life story behind it, which helps students obtain a better grasp of the math. Included are steps for using the Sage 'N Scribe cooperative learning structure to increase engagement and maximize student progress.
Included in this product:
•Eight problems on 2 pages formatted for Sage 'N Scribe
•The same problems on a 2-page regular worksheet
•Teacher guide for cooperative learning
•How to differentiate with this activity
• | 677.169 | 1 |
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