Split (1)

train · 15k rows

text stringlengths 235 3.08k
, given each function f, evaluate f (−1), f (0), f (2), and f (4). { 49. f (x) = 7x + 3 if x < 0 7x + 6 if x ≥ 0 { 50. f (x) = if \| if x ≥ 2 51. f (x) = { 5x if 3 x 2 if x < 0 if 0 ≤ x ≤ 3 x > 3 For the following exercises, write the domain for the piecewise function in interval notation. { 52. f (x) = x + 1 if x < −2 −2x − 3 if x ≥ −2 { 53. f (x) = x2 − 2 if x < 1 −x2 + 2 if x > 1 { 54. f (x) = 2x − 3 if x < 0 if x ≥ 2 −3x2 TeCHnOlOGY 55. Graph y = 1 __ x 2 on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs. 1 _ x on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the 56. Graph y = viewing window. Show the graphs. exTenSIOn 57. Suppose the range of a function f is [−5, 8]. What is the range of \| f (x) \|? 58. Create a function in which the range is all nonnegative real numbers. 59. Create a function in which the domain is x > 2. ReAl-WORlD APPlICATIOnS 60. The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function h(t) = −16t 2 + 96t. What is the domain of the function? What does the domain mean in the context of the problem? 61. The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items are produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, C(x)? 38 CHAPTER 1 Functions
leARnInG OBjeCTIVeS In this section, you will: • • • • Find the average rate of change of a function. Use a graph to determine where a function is increasing, decreasing, or constant. Use a graph to locate local maxima and local minima. Use a graph to locate the absolute maximum and absolute minimum. 1. 3 RATeS OF CHAnGe AnD BeHAVIOR OF GRAPHS Gasoline costs have experienced some wild fluctuations over the last several decades. Table 1[5] lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year. y C(y) 2005 2.31 2006 2.62 2007 2.84 2008 3.30 2009 2.41 2010 2.84 2011 3.58 2012 3.68 Table 1 If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these. Finding the Average Rate of Change of a Function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table 1 did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. Average rate of change = Change in output __ Change in input ∆y _ ∆x y2 − y1 _ x2 − x1 f (x2) − f (x1) __ − x1 = = = x2 The Greek letter ∆ (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in y divided by the change in x.” Occasionally we write ∆ f instead of ∆y, which still represents the change in the function’s output value resulting from a change to its input value
. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was ∆y _ ∆x = $1.37 _ 7 years ≈ 0.196 dollars per year On average, the price of gas increased by about 19.6 each year. Other examples of rates of change include: • A population of rats increasing by 40 rats per week • A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) • A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) • The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage • The amount of money in a college account decreasing by $4,000 per quarter 5 http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. SECTION 1.3 rates oF change and Behavior oF graphs 39 rate of change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.” The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. ∆y _ ∆x = f (x2) − f (x1) __ x2 − x1 How To… Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values x1 and x2. 1. Calculate the difference y2 − y1 = ∆y. 2. Calculate the difference x2 − x1 = ∆x. ∆y _. 3. Find the ratio ∆x Example 1 Computing an Average Rate of Change Using the data in Table 1, find the average rate of change of the price of gasoline between 2007 and 2009. Solution In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is ∆y ___ ∆x = y2 − y1 _ x2 − x1 $2.41 − $2.84 __ 2009 − 2007 −$0.43 ______ 2 years
= = Analysis Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases. = −$0.22 per year Try It #1 Using the data in Table 1 at the beginning of this section, find the average rate of change between 2005 and 2010. Example 2 Computing Average Rate of Change from a Graph Given the function g (t) shown in Figure 1, find the average rate of change on the interval [−1, 2]. g(t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 1 40 CHAPTER 1 Functions Solution At t = −1, Figure 2 shows g (−1) = 4. At t = 2, the graph shows g (2) = 1. g(t) ∆g(t) = –3 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 ∆t = 3 t Figure 2 The horizontal change ∆ t = 3 is shown by the red arrow, and the vertical change ∆ g(t) = −3 is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of 1 − 4 _______ = 2 − (−1) −3 ___ 3 = −1 Analysis Note that the order we choose is very important. If, for example, we use answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as ( x1, y1, we will not get the correct ). ) and ( x2, y2 y2 − y1 _ x1 − x2 Example 3 Computing Average Rate of Change from a Table After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in Table 2. Find her average speed over the first 6 hours. t (hours) D(t) (miles) 0 10 1 55 2 90 3 153 4 214 5 240 6 292 7 300 Table 2 Solution Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of The average speed is 47 miles per hour. 292 − 10 _______ = 6 − 0
282 ___ 6 = 47 Analysis Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2, 3], the average speed is 63 miles per hour. Computing Average Rate of Change for a Function Expressed as a Formula Example 4 Compute the average rate of change of f (x) = x 2 − 1 __ on the interval [2, 4]. x Solution We can start by computing the function values at each endpoint of the interval. 1 __ f (2) = 22 − 2 1 __ f(4) = 42 − 4 1 __ = 4 − 2 7 __ = 2 1 __ = 16 − 4 = 63 __ 4 Now we compute the average rate of change. SECTION 1.3 rates oF change and Behavior oF graphs 41 Average rate of change = f (4) − f (2) _ 4 − 2 7 63 __ __ − 4 2 _ 4 − 2 49 __ 4 _ 2 49 __ 8 = = = Try It #2 Find the average rate of change of f (x) = x − 2 √ — x on the interval [1, 9]. Example 5 Finding the Average Rate of Change of a Force The electrostatic force F, measured in newtons, between two charged particles can be related to the distance between 2 _ the particles d, in centimeters, by the formula F(d) = d 2. Find the average rate of change of force if the distance between the particles is increased from 2 cm to 6 cm. Solution We are computing the average rate of change of F (d) = 2 __ d 2 on the interval [2, 6]. Average rate of change = = F(6) − F(2) __ 6 − 2 2 2 _ _ 62 − 22 _ 6 − 2 2 − 2 __ __ 36 4 _ 4 − 16 __ 36 _ 4 = − 1 __ 9 = = Simplify. Combine numerator terms. Simplify. The average rate of change is − 1 __ newton per centimeter. 9 Example 6 Finding an Average Rate of Change as an Expression Find the average rate of change of g(t) = t2 + 3t + 1 on the interval [0, a]. The answer will be an expression involving a. Solution We use the average rate of change formula. Average rate of change = g(a) − g(0) _ a − 0 Evaluate. = (a2 + 3a + 1) − (02 + 3(0) + 1
) ___ a − 0 Simplify. = a2 + 3a + 1 − 1 _____________ a a(a + 3) _______ a = a + 3 = Simplify and factor. Divide by the common factor a. This result tells us the average rate of change in terms of a between t = 0 and any other point t = a. For example, on the interval [0, 5], the average rate of change would be 5 + 3 = 8. 42 CHAPTER 1 Functions Try It #3 Find the average rate of change of f (x) = x2 + 2x − 8 on the interval [5, a]. Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function. Increasing –5 –4 –3 –2 f(x) 20 16 12 8 4 –1 –4 –8 –12 –16 –20 Decreasing 21 3 4 5 x Increasing f(b) > f(a) where b > a f(b) < f(a) where b > a f(b) > f(a) where b > a Figure 3 The function f(x) = x3 − 12x is increasing on (−∞, −2) ∪ (2, ∞) and is decreasing on (−2, 2). While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is “local minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The
singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local. Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain. For the function whose graph is shown in Figure 4, the local maximum is 16, and it occurs at x = −2. The local minimum is −16 and it occurs at x = 2. Local maximum = 16 occurs at x = –2 f(x) Increasing (–2, 16) 20 16 12 8 4 Decreasing –5 –4 –3 –2 –1 –4 –8 –12 –16 –20 21 3 4 5 x Increasing (2, –16) Local minimum = –16 occurs at x = 2 Figure 4 SECTION 1.3 rates oF change and Behavior oF graphs 43 To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum. f(x) f(b) Local maximum Increasing function Decreasing function a b c Figure 5 Definition of a local maximum x These observations lead us to a formal definition of local extrema. local minima and local maxima A function f is an increasing function on an open interval if f (b) > f (a) for every two input values a and b in the interval where b > a. A function f is a decreasing function on an open interval if f (b) < f (a) for every two input values a and b in the interval where b > a. A function f has a local maximum at a point b in an open interval (a, c) if f (b) ≥ f (x) for every point x (x does not equal b) in the interval. f has a local minimum at a point b in the interval (a, c) if f (b) ≤ f
(x) for every point x (x does not equal both) in the interval. Example 7 Finding Increasing and Decreasing Intervals on a Graph Given the function p(t) in Figure 6, identify the intervals on which the function appears to be increasing. p 5 4 3 2 1 –1 1 2 3 4 5 6 t –1 –2 Figure 6 Solution We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from t = 1 to t = 3 and from t = 4 on. In interval notation, we would say the function appears to be increasing on the interval (1, 3) and the interval (4, ∞). 44 CHAPTER 1 Functions Analysis Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at t = 1, t = 3, and t = 4. These points are the local extrema (two minima and a maximum). Example 8 Finding Local Extrema from a Graph x 2 __ __ + Graph the function f (x) =. Then use the graph to estimate the local extrema of the function and to determine 3 x the intervals on which the function is increasing. Solution Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between x = 2 and x = 3, and a mirror-image high point, or local maximum, somewhere between x = −3 and x = −2. f(x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 7 Analysis Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 provides screen images from two different technologies, showing the estimate for the local maximum and minimum. y 6 4 2 0 –2 2.4494898, 1.6329932 2 4 6 x Maximum X = –2.449491 Y = –1.632993 (a) (b) Figure 8 Based on these estimates, the function is increasing on the interval (−∞, −2.449) and (2.449, ∞). Notice that, while we expect the extrema to
be symmetric, the two different technologies agree only up to four decimals due to the differing — approximation algorithms used by each. (The exact location of the extrema is at ± √ 6, but determining this requires calculus.) Try It #4 Graph the function f (x) = x3 − 6x2 − 15x + 20 to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing. SECTION 1.3 rates oF change and Behavior oF graphs 45 Example 9 Finding Local Maxima and Minima from a Graph For the function f whose graph is shown in Figure 9, find all local maxima and minima. y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –5 –4 –3 –2 21 3 4 5 x f Figure 9 Solution Observe the graph of f. The graph attains a local maximum at x = 1 because it is the highest point in an open interval around x = 1. The local maximum is the y-coordinate at x = 1, which is 2. The graph attains a local minimum at x = −1 because it is the lowest point in an open interval around x = −1. The local minimum is the y-coordinate at x = −1, which is −2. Analyzing the Toolkit Functions for Increasing or Decreasing Intervals We will now return to our toolkit functions and discuss their graphical behavior in Figure 10, Figure 11, and Figure 12. Function Increasing/Decreasing Example Constant Function f(x) = c Neither increasing nor decreasing Identity Function f(x) = x Increasing Quadratic Function f(x) = x2 Increasing on (0, ∞) Decreasing on (−∞, 0) Minimum at x = 0 Figure 10 y y y x x x 46 CHAPTER 1 Functions Function Increasing/Decreasing Example Cubic Function f(x) = x3 Increasing Reciprocal f(x) = 1 __ x Decreasing (−∞, 0) ∪ (0, ∞) y y y Reciprocal Squared f(x) = 1 _ x2 Increasing on (−∞, 0) Decreasing on (0, ∞) Figure 11 Function Increasing/Decreasing Example y Cube Root — x f(x) = 3 √ Square Root x f(x) = √ — Increasing Increasing on (0, ∞) y
y Absolute Value f(x) = ∣ x ∣ Increasing on (0, ∞) Decreasing on (−∞, 0) Figure 12 x x x x x x SECTION 1.3 rates oF change and Behavior oF graphs 47 Use A Graph to locate the Absolute Maximum and Absolute Minimum There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The y-coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 13. y 5 4 3 2 1 f Absolute maximum is f (2) = 2 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 21 3 4 5 x Absolute minimum is f (0) = −2 Figure 13 Not every function has an absolute maximum or minimum value. The toolkit function f (x) = x3 is one such function. absolute maxima and minima The absolute maximum of f at x = c is f (c) where f (c) ≥ f (x) for all x in the domain of f. The absolute minimum of f at x = d is f (d) where f (d) ≤ f (x) for all x in the domain of f. Example 10 Finding Absolute Maxima and Minima from a Graph For the function f shown in Figure 14, find all absolute maxima and minima. y 20 16 12 8 4 –1 –4 –8 –12 –16 –20 –5 –4 –3 –2 f 21 3 4 5 x Figure 14 Solution Observe the graph of f. The graph attains an absolute maximum in two locations, x = −2 and x = 2, because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at x = −2 and x = 2, which is 16. The graph attains an absolute minimum at x = 3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x = 3, which is −10. Access this online resource for additional instruction and
practice with rates of change. • Average Rate of Change (http://openstaxcollege.org/l/aroc) 48 CHAPTER 1 Functions 1.3 SeCTIOn exeRCISeS VeRBAl 1. Can the average rate of change of a function be constant? 3. How are the absolute maximum and minimum similar to and different from the local extrema? 2. If a function f is increasing on (a, b) and decreasing on (b, c), then what can be said about the local extremum of f on (a, c)? 4. How does the graph of the absolute value function compare to the graph of the quadratic function, y = x 2, in terms of increasing and decreasing intervals? AlGeBRAIC For the following exercises, find the average rate of change of each function on the interval specified for real numbers b or h. 5. f (x) = 4x 2 − 7 on [1, b] 7. p(x) = 3x + 4 on [2, 2 + h] 9. f (x) = 2x 2 + 1 on [x, x + h] 6. g (x) = 2x 2 − 9 on [4, b] 8. k(x) = 4x − 2 on [3, 3 + h] 10. g(x) = 3x 2 − 2 on [x, x + h] 11. a(t) = 1 ____ t + 4 13. j(x) = 3x 3 on [1, 1 + h] on [9, 9 + h] 12. b(x) = 1 _____ x + 3 14. r(t) = 4t 3 on [2, 2 + h] on [1, 1 + h] 15. f (x + h) − f(x) __ h given f(x) = 2x2 − 3x on [x, x + h] GRAPHICAl For the following exercises, consider the graph of f shown in Figure 15. 16. Estimate the average rate of change from x = 1 to x = 4. 17. Estimate the average rate of change from x = 2 to x = 5 Figure 15 For the following exercises, use the graph of each function to estimate the intervals on which the function is increasing or decreasing. 18. –5 –4 –3 –2 y 5 4 3 2 1 –1
–1 –2 –3 –4 –5 19. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 20. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x SECTION 1.3 section exercises 49 21. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x For the following exercises, consider the graph shown in Figure 16. 22. Estimate the intervals where the function is increasing or decreasing. 23. Estimate the point(s) at which the graph of f has a local maximum or a local minimum. For the following exercises, consider the graph in Figure 17. 24. If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing. 25. If the complete graph of the function is shown, estimate the absolute maximum and absolute minimum. y 100 80 60 40 20 –2 –1 –20 –40 –60 –80 –100 –5 –4 –3 21 3 4 5 x Figure 16 y 250 200 150 100 50 –4 –2 –50 –100 –150 –200 –250 –10 –8 –6 42 6 8 10 x Figure 17 nUMeRIC 26. Table 3 gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales (a) between 2001 and 2002, and (b) between 2001 and 2004? Year Sales (millions of dollars) 1998 201 1999 219 2000 233 Table 3 2001 243 2002 249 2003 251 2004 249 2005 243 2006 233 27. Table 4 gives the population of a town (in thousands) from 2000 to 2008. What was the average rate of change of population (a) between 2002 and 2004, and (b) between 2002 and 2006? Year Population (thousands) 2000 87 2001 84 2002 83 2003 80 2004 77 2005 76 2006 78 2007 81 2008 85 Table 4 50 CHAPTER 1 Functions For the following exercises, find the average rate of change of each function on the interval specified. 28. f (x) = x 2 on [1, 5] 29. h(x) = 5 − 2x 2 on [−2, 4] 30
. q(x) = x3 on [−4, 2] 1 _ x on [1, 3] 32. y = 4 __ t3 on [−1, 3] 34. k (t) = 6t2 + 31. g (x) = 3x 3 − 1 on [−3, 3] 33. p(t) = (t 2 − 4)(t + 1) ____________ t 2 + 3 on [−3, 1] TeCHnOlOGY For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increasing and decreasing. 35. f(x) = x 4 − 4x 3 + 5 — 37. g(t) = t √ 39. m(x) = x 4 + 2x 3 − 12x 2 − 10x + 4 t + 3 36. h(x) = x 5 + 5x 4 + 10x 3 + 10x 2 − 1 38. k(t) = 3 t 2 _ − t 3 40. n(x) = x 4 − 8x 3 + 18x 2 − 6x + 2 exTenSIOn 41. The graph of the function f is shown in Figure 18. Based on the calculator screen shot, the point (1.333, 5.185) is which of the following? a. a relative (local) maximum of the function b. the vertex of the function c. the absolute maximum of the function d. a zero of the function Maximum X = 1.3333324 Y = 5.1851852 Figure 18 1 __ 42. Let f (x) =. Find a number c such that the average x rate of change of the function f on the interval (1, c) is − 1 __ 4 43. Let f(x) = 1 __. Find the number b such that the x average rate of change of f on the interval (2, b) 1 __. is − 10 ReAl-WORlD APPlICATIOnS 44. At the start of a trip, the odometer on a car read 21,395. At the end of the trip, 13.5 hours later, the odometer read 22,125. Assume the scale on the odometer is in miles. What is the average speed the car traveled during this trip? 45. A driver of a car stopped at a gas station
to fill up his gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank? 46. Near the surface of the moon, the distance that 47. The graph in Figure 19 illustrates the decay of a an object falls is a function of time. It is given by d(t) = 2.6667t2, where t is in seconds and d(t) is in feet. If an object is dropped from a certain height, find the average velocity of the object from t = 1 to t = 2. radioactive substance over t days 16 14 12 10 8 6 0 15 5 10 Time (days) Figure 19 t 20 Use the graph to estimate the average decay rate from t = 5 to t = 15. SECTION 1.4 composition oF Functions 51 leARnInG OBjeCTIVeS In this section, you will: • • • • • Combine functions using algebraic operations. Create a new function by composition of functions. Evaluate composite functions. Find the domain of a composite function. Decompose a composite function into its component functions. 1. 4 COMPOSITIOn OF FUnCTIOnS Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day. Using descriptive variables, we can notate these two functions. The function C(T) gives the cost C of heating a house for a given average daily temperature in T degrees Celsius. The function T(d) gives the average daily temperature on day d of the year. For any given day, Cost = C(T(d)) means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would
write C(T(5)). Cost for the temperature C(T(5)) Temperature on day 5 By combining these two relationships into one function, we have performed function composition, which is the focus of this section. Combining Functions Using Algebraic Operations Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function. Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If w(y) is the wife’s income and h(y) is the husband’s income in year y, and we want T to represent the total income, then we can define a new function. T(y) = h(y) + w(y) If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write T = h + w Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions. 52 CHAPTER 1 Functions f _ For two functions f (x) and g(x) with real number outputs, we define new functions f + g, f − g, fg, and g by the relations (f + g)(x) = f (x) + g(x) (f − g)(x) = f (x) − g(x) (fg)(x) = f (x)g(x) f g  (x) =  _ f (x) _ g(x) Example 1 Performing Algebraic Operations on Functions g _  (x), given f (x) = x − 1 and g(x) = x
2 − 1. Are they the same Find and simplify the functions (g − f )(x) and  f function? Solution Begin by writing the general form, and then substitute the given functions. (g − f )(x) = g(x) − f (x) (g − f )(x) = x 2 − 1 − (x − 1) (g − f )(x) = x 2 − x (g − f )(x) = x(x − 1) x2 − 1 _ x − 1 g(x) _ f (x) g _  (x) =  f g _  (x) =  f g _  (x) =  f g _  (x) = x + 1  f (x + 1)(x − 1) ____________ x − 1 where x ≠ 1 where x ≠ 1 where x ≠ 1 No, the functions are not the same. g _  (x), the condition x ≠ 1 is necessary because when x = 1, the denominator is equal to 0, which makes Note: For  f the function undefined. Try It #1 Find and simplify the functions (fg)(x) and (f − g)(x). f (x) = x − 1 and g(x) = x 2 − 1 Are they the same function? Create a Function by Composition of Functions Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation: ( f ∘ g)(x) = f (g(x)) SECTION 1.4 composition oF Functions 53 We read the left-hand side as “ f composed with g at x,” and the right-hand side as “ f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ is called the composition operator. We use this operator mainly when we
wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases f (g(x)) ≠ f (x)g(x). It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g(x). Then the function f takes g(x) as an input and yields an output f (g(x)). g(x), the output of g is the input of f (f ° g)(x) = f (g(x)) x is the input of g In general, f ∘ g and g ∘ f are different functions. In other words, in many cases f ( g(x)) ≠ g(f (x)) for all x. We will also see that sometimes two functions can be composed only in one specific order. For example, if f (x) = x2 and g(x) = x + 2, then but f (g(x)) = f (x + 2) = (x + 2)2 = x2 + 4x + 4 g(f (x)) = g(x2) = x2 + 2 These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value x = − 1 _. 2 Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs. composition of functions When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input x and functions f and g, this action defines a composite function, which we write as f ∘ g such that (f ∘ g)(x) = f (g(x)) The domain of the composite function f ∘ g is all x such that x is in the domain of g and g(x) is in the domain of
f. It is important to realize that the product of functions fg is not the same as the function composition f (g(x)), because, in general, f (x)g(x) ≠ f (g(x)). Example 2 Determining whether Composition of Functions is Commutative Using the functions provided, find f (g(x)) and g(f (x)). Determine whether the composition of the functions is commutative. f (x) = 2x + 1 g(x) = 3 − x 54 CHAPTER 1 Functions Solution Let’s begin by substituting g(x) into f (x). Now we can substitute f (x) into g(x). f (g(x)) = 2(3 − x) + 1 = 6 − 2x + 1 = 7 − 2x g(f (x)) = 3 − (2x + 1) = 3 − 2x − 1 = − 2x + 2 We find that g(f (x)) ≠ f (g(x)), so the operation of function composition is not commutative. Example 3 Interpreting Composite Functions The function c(s) gives the number of calories burned completing s sit-ups, and s(t) gives the number of sit-ups a person can complete in t minutes. Interpret c(s(3)). Solution The inside expression in the composition is s(3). Because the input to the s-function is time, t = 3 represents 3 minutes, and s(3) is the number of sit-ups completed in 3 minutes. Using s(3) as the input to the function c(s) gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups). Example 4 Investigating the Order of Function Composition Suppose f (x) gives miles that can be driven in x hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f (g(y)) or g(f (x))? Solution The function y = f (x) is a function whose output is the number of miles driven corresponding to the number of hours driven. number of miles = f (number of hours) The function g(y) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means
: number of gallons = g (number of miles) The expression g(y) takes miles as the input and a number of gallons as the output. The function f (x) requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression f (g(y)) is meaningless. The expression f (x) takes hours as input and a number of miles driven as the output. The function g(y) requires a number of miles as the input. Using f (x) (miles driven) as an input value for g(y), where gallons of gas depends on miles driven, does make sense. The expression g(f (x)) makes sense, and will yield the number of gallons of gas used, g, driving a certain number of miles, f (x), in x hours. Q & A… Are there any situations where f (g(y)) and g(f (x)) would both be meaningful or useful expressions? Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order. Try It #2 The gravitational force on a planet a distance r from the sun is given by the function G(r). The acceleration of a planet subjected to any force F is given by the function a(F). Form a meaningful composition of these two functions, and explain what it means. SECTION 1.4 composition oF Functions 55 evaluating Composite Functions Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function. Evaluating Composite Functions Using Tables When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function. Example 5 Using a Table to Evaluate a Composite Function Using Table 1, evaluate f (g(3)) and g(f (3)). x 1 2 3 4 f (x) 6 8 3 1 Table
1 g(x) 3 5 2 7 Solution To evaluate f (g(3)), we start from the inside with the input value 3. We then evaluate the inside expression g(3) using the table that defines the function g: g(3) = 2. We can then use that result as the input to the function f, so g(3) is replaced by 2 and we get f (2). Then, using the table that defines the function f, we find that f (2) = 8. g(3) = 2 f (g(3)) = f (2) = 8 To evaluate g(f (3)), we first evaluate the inside expression f (3) using the first table: f (3) = 3. Then, using the table for g, we can evaluate g(f (3)) = g(3) = 2 Table 2 shows the composite functions f ∘ g and g ∘ f as tables. x 3 g(x) 2 f (g(x)) 8 Table 2 f (x) 3 g(f (x)) 2 Try It #3 Using Table 1, evaluate f (g(1)) and g(f (4)). Evaluating Composite Functions Using Graphs When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the x- and y-axes of the graphs. 56 CHAPTER 1 Functions How To… Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs. 1. Locate the given input to the inner function on the x-axis of its graph. 2. Read off the output of the inner function from the y-axis of its graph. 3. Locate the inner function output on the x-axis of the graph of the outer function. 4. Read the output of the outer function from the y-axis of its graph. This is the output of the composite function. Example 6 Using a Graph to Evaluate a Composite Function Using Figure 1, evaluate f (g(1)). g(x) f(x) 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 321 1 –2 –3 –4 –5 321 4 5 6 7 x (a) (b) Figure 1 Solution To evaluate f (g(1)), we start with the inside evaluation. See Figure 2
. g(x) f (x) 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 (1, 3) 321 4 5 6 7 x g(1) = 3 Figure 2 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 (3, 6) 321 4 5 6 7 x f (3) = 6 We evaluate g(1) using the graph of g(x), finding the input of 1 on the x-axis and finding the output value of the graph at that input. Here, g(1) = 3. We use this value as the input to the function f. f (g(1)) = f (3) We can then evaluate the composite function by looking to the graph of f (x), finding the input of 3 on the x-axis and reading the output value of the graph at this input. Here, f (3) = 6, so f ( g(1)) = 6. SECTION 1.4 composition oF Functions 57 Analysis Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value. g(x) f(x) –10 –8 –6 –4 10 8 6 4 2 –2–2 –4 –6 –8 –10 10 8 6 4 2 –2–2 –4 –6 –8 –10 42 6 8 10 x 42 6 8 10 x –10 –8 –6 –4 Figure 3 Try It #4 Using Figure 1, evaluate g(f (2)). Evaluating Composite Functions Using Formulas When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression. While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition f ( g(x)). To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like f (t) = t2 − t, we substitute the value inside the parentheses into the formula wherever we see the input variable. How To… Given a formula for a composite function, evaluate the function. 1. Evaluate the inside function using the input value or variable provided. 2. Use the resulting output
as the input to the outside function. Example 7 Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input Given f (t) = t2 − t and h(x) = 3x + 2, evaluate f (h(1)). Solution Because the inside expression is h(1), we start by evaluating h(x) at 1. Then f (h(1)) = f (5), so we evaluate f (t) at an input of 5. h(1) = 3(1) + 2 h(1) = 5 f (h(1)) = f (5) f (h(1)) = 52 − 5 f (h(1)) = 20 Analysis specific numerical values. It makes no difference what the input variables t and x were called in this problem because we evaluated for 58 CHAPTER 1 Functions Try It #5 Given f (t) = t 2 − t and h(x) = 3x + 2, evaluate a. h(f (2)) b. h(f (−2)) Finding the Domain of a Composite Function As we discussed previously, the domain of a composite function such as f ∘ g is dependent on the domain of g and the domain of f. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as f ∘ g. Let us assume we know the domains of the functions f and g separately. If we write the composite function for an input x as f (g(x)), we can see right away that x must be a member of the domain of g in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that g(x) must be a member of the domain of f, otherwise the second function evaluation in f (g(x)) cannot be completed, and the expression is still undefined. Thus the domain of f ∘ g consists of only those inputs in the domain of g that produce outputs from g belonging to the domain of f. Note that the domain of f composed with g is the set of all x such that x is in the domain of g and g(x) is in the domain of f. domain of a composite function The domain of a composite function f (g(x)) is the set of those inputs x in the domain of g for which g(x) is in the domain of f. How To
… Given a function composition f (g(x)), determine its domain. 1. Find the domain of g. 2. Find the domain of f. 3. Find those inputs x in the domain of g for which g(x) is in the domain of f. That is, exclude those inputs x from the domain of g for which g(x) is not in the domain of f. The resulting set is the domain of f ∘ g. Example 8 Finding the Domain of a Composite Function Find the domain of (f ∘ g)(x) where f (x) = and g(x) = 5 ____ x − 1 4 _____ 3x − 2 2 __ Solution The domain of g(x) consists of all real numbers except x =, since that input value would cause us to divide by 3 0. Likewise, the domain of f consists of all real numbers except 1. So we need to exclude from the domain of g(x) that value of x for which g(x) = 1. 4 _____ 3x − 2 = 1 4 = 3x − 2 6 = 3x x = 2 2 __ So the domain of f ∘ g is the set of all real numbers except and 2. This means that 3 We can write this in interval notation as 2 __ or x ≠ 2 x ≠ 3 2 2 __ __, 2  ∪ (2, ∞)  ∪   − ∞, 3 3 SECTION 1.4 composition oF Functions 59 Finding the Domain of a Composite Function Involving Radicals Find the domain of (f ∘ g)(x) where f (x) = √ — x + 2 and g(x) = √ — 3 − x Solution Because we cannot take the square root of a negative number, the domain of g is (−∞, 3]. Now we check the domain of the composite function (f ∘ g)(x) = √ ___________ — 3 − x +2 √ ___________ — For (f ∘ g)(x) = √ √ roots are positive, √ — 3 − x +2, √ 3 − x ≥ 0, 3 − x ≥ 0, which gives a domain of (−∞, 3]. — 3 − x +2 ≥ 0, since the radicand of a square root must be positive. Since the square Analysis This example shows that knowledge of the
range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of f ∘ g can contain values that are not in the domain of f, though they must be in the domain of g. Try It #6 Find the domain of (f ∘ g)(x) where f (x) = 1 ____ x − 2 and g(x) = √ — x + 4 Decomposing a Composite Function into its Component Functions In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient. Example 9 Decomposing a Function Write f (x) = √ 5 − x 2 as the composition of two functions. — Solution We are looking for two functions, g and h, so f (x) = g(h(x)). To do this, we look for a function inside a function in the formula for f (x). As one possibility, we might notice that the expression 5 − x2 is the inside of the square root. We could then decompose the function as h(x) = 5 − x2 and g(x) = √ — x We can check our answer by recomposing the functions. g(h(x)) = g(5 − x2) = √ — 5 − x2 Try It #7 Write f (x) = as the composition of two functions. 4 __ — 4 + x2 3 − √ Access these online resources for additional instruction and practice with composite functions. • Composite Functions (http://openstaxcollege.org/l/compfunction) • Composite Function notation Application (http://openstaxcollege.org/l/compfuncnot) • Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph) • Decompose Functions (http://openstaxcollege.org/l/decompfunction) • Composite Function Values (http://openstaxcollege.org/l/compfuncvalue) 60 CHAPTER 1 Functions 1.4 SeCTIOn exeRCISeS VeRBAl 1. How does one find the domain of the quotient of 2. What is the composition of two functions, f ∘ g?
f _ g? two functions, 3. If the order is reversed when composing two 4. How do you find the domain for the composition of functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not. two functions, f ∘ g? AlGeBRAIC 5. Given f (x) = x 2 + 2x and g(x) = 6 − x 2, find f + g, 6. Given f (x) = −3x2 + x and g(x) = 5, find f + g, f − g, f _ g. Determine the domain for each f − g, fg, and f _ g. Determine the domain for each function fg, and function in interval notation. in interval notation. 7. Given f (x) = 2x 2 + 4x and g(x) =, find f + g, f _ g. Determine the domain for each f − g, fg, and 1 _ 2x 8. Given f (x) = 1 _ and g(x) = x − 4 f _ g. Determine the domain for each f + g, f − g, fg, and 1 _ 6 − x, find function in interval notation. 9. Given f (x) = 3x 2 and g(x) = √ f _ g. Determine the domain for each f − g, fg, and x − 5, find f + g, — function in interval notation. function in interval notation. 10. Given f (x) = √ — g x and g(x) = \|x − 3\|, find _. f Determine the domain for each function in interval notation. 11. Given f (x) = 2x 2 + 1 and g(x) = 3x − 5, find the following: a. f ( g(2)) b. f ( g(x)) c. g( f (x)) d. ( g ∘ g)(x) e. ( f ∘ f )(−2) For the following exercises, use each pair of functions to find f (g(x)) and g(f (x)). Simplify your answers. 12. f (x) = x 2 + 1, g(x) = √ — x + 2 13. f (x) = �
� — x + 2, g(x) = x 2 + 3 14. f (x) = \|x\|, g(x) = 5x + 1 15. f (x) = 3 √ 16. f (x(x) = — x, g(x) = x + 1 _ x3, g(x) = 2 _ x + 4 1 ___ x−4 17. f (x) = For the following exercises, use each set of functions to find f (g(h(x))). Simplify your answers. 18. f (x) = x 4 + 6, g(x) = x − 6, and h(x) = √ — x 1 _ x, and h(x) = x + 3 19. f (x) = x 2 + 1, g(x) = 1 _ x, and g(x) = x − 3, find the 20. Given f (x) = following: a. ( f ∘ g)(x) b. the domain of ( f ∘ g)(x) in interval notation c. ( g ∘ f )(x) d. the domain of ( g ∘ f )(x) f _ g  x e.  — 2 − 4x and g(x) = − 3 21. Given f (x) = √ following: a. ( g ∘ f )(x) b. the domain of ( g ∘ f )(x) in interval notation _ x, find the SECTION 1.4 section exercises 61 22. Given the functions f (x) = and g(x + x2, find the following: a. ( g ∘ f )(x) b. ( g ∘ f )(2) 23. Given functions p(x) = 1 _ and m(x) = x 2 − 4, x √ state the domain of each of the following functions using interval notation: — a. p(x) _ m(x) b. p(m(x)) c. m(p(x)) 24. Given functions q(x) = 1 _ and h(x) = x 2 − 9, state — x √ the domain of each of the following functions using 1 _ x and g(x) = √ 25. For f (x) = — x − 1, write the domain of ( f ∘ g
)(x) in interval notation. interval notation. a. q(x) _ h(x) b. q(h(x)) c. h(q(x)) For the following exercises, find functions f (x) and g(x) so the given function can be expressed as h(x) = f (g(x)). 26. h(x) = (x + 2)2 27. h(x) = (x − 5)3 30. h(x) = 4 + 3 √ — x 3 √ 31. h(x) = _______ 1 ______ 2x − 3 28. h(x) = 3 _ x − 5 32. h(x) = 1 _ (3x 2 − 4)−3 29. h(x) = 4 _______ (x + 2)2 4 √ 33. h(x) = _______ 3x − 2 ______ x + 5 34. h(x  35. h(x) = √ — 2x + 6 36. h(x) = (5x − 1)3 37. h(x) = 3 √ — x − 1 38. h(x) = \|x 2 + 7\| 39. h(x) = 1 _ (x − 2)3 40. h(x) =  2 1  _ 2x − 3 41. h(x) = √ _______ 2x − 1 _ 3x + 4 GRAPHICAl For the following exercises, use the graphs of f, shown in Figure 4, and g, shown in Figure 5, to evaluate the expressions. f(x) 6 5 4 3 2 1 – –1 1(x) 6 5 4 3 2 1 – –1 1 Figure 4 Figure 5 42. f ( g(3)) 46. f ( f (5)) 43. f ( g(1)) 47. f ( f (4)) 44. g( f (1)) 48. g( g(2)) 45. g( f (0)) 49. g( g(0)) 62 CHAPTER 1 Functions For the following exercises, use graphs of f (x), shown in Figure 6, g(x), shown in Figure 7, and h(x), shown in Figure 8, to evaluate the expressions. f(x) f (x) 5 4 3 2 1 f (x) 5 4 3 2 1 g
(x) h(x) f (x1 1 2 3 4 x x x Figure 6 Figure 7 Figure 8 50. g( f (1)) 51. g( f (2)) 52. f ( g(4)) 53. f ( g(1)) 54. f (h(2)) 55. h( f (2)) 56. f ( g(h(4))) 57. f ( g( f (−2))) nUMeRIC For the following exercises, use the function values for f and g shown in Table 3 to evaluate each expression. x f (x) g(x Table 58. f ( g(8)) 62. f ( f (4)) 59. f ( g(5)) 63. f ( f (1)) 60. g( f (5)) 64. g( g(2)) 61. g( f (3)) 65. g( g(6)) For the following exercises, use the function values for f and g shown in Table 4 to evaluate the expressions. −3 −2 x f (x) 9 11 g(x) −8 −3 −1 7 0 Table 1 −3 −8 66. ( f ∘ g)(1) 69. ( g ∘ f )(3) 67. ( f ∘ g)(2) 70. ( g ∘ g )(1) 68. ( g ∘ f )(2) 71. ( f ∘ f )(3) For the following exercises, use each pair of functions to find f (g(0)) and g(f (0)). 72. f (x) = 4x + 8, g(x) = 7 − x2 73. f (x) = 5x + 7, g(x) = 4 − 2x2 74. f (x) = √ — x + 4, g(x) = 12 − x3 75. f (x) =, g(x) = 4x + 3 1 _ x + 2 For the following exercises, use the functions f (x) = 2x2 + 1 and g(x) = 3x + 5 to evaluate or find the composite function as indicated. 76. f ( g(2)) 77. f ( g(x)) 78. g( f ( − 3)) 79. ( g ∘ g )(x) SECTION 1.4 section exercises 63 exTenSIOnS For the following exercises, use
f (x) = x3 + 1 and g(x) = 3 √ — x − 1. 80. Find ( f ∘ g)(x) and ( g ∘ f )(x). Compare the two answers. 81. Find ( f ∘ g)(2) and ( g ∘ f )(2). 82. What is the domain of ( g ∘ f )(x)? 83. What is the domain of ( f ∘ g)(x)? 1 __ 84. Let f (x) =. x a. Find ( f ∘ f )(x). b. Is ( f ∘ f )(x) for any function f the same result as the answer to part (a) for any function? Explain. For the following exercises, let F (x) = (x + 1)5, f (x) = x5, and g(x) = x + 1. 85. True or False: ( g ∘ f )(x) = F (x). 86. True or False: ( f ∘ g )(x) = F (x). For the following exercises, find the composition when f (x) = x2 + 2 for all x ≥ 0 and g(x) = √ — x − 2. 87. ( f ∘ g)(6) ; ( g ∘ f )(6) 88. ( g ∘ f )(a) ; ( f ∘ g )(a) 89. ( f ∘ g )(11) ; (g ∘ f )(11) ReAl-WORlD APPlICATIOnS 90. The function D(p) gives the number of items that will be demanded when the price is p. The production cost C(x) is the cost of producing x items. To determine the cost of production when the price is $6, you would do which of the following? a. Evaluate D(C(6)). b. Evaluate C(D(6)). c. Solve D(C(x)) = 6. d. Solve C(D(p)) = 6. 92. A store offers customers a 30 % discount on the price x of selected items. Then, the store takes off an additional 15 % at the cash register. Write a price function P(x) that computes the final price of the item in terms of the original price x. (
Hint: Use function composition to find your answer.) 94. A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula r(t) = 2t + 1, express the area burned as a function of time, t (minutes). 96. The radius r, in inches, of a spherical balloon is ___ 3V ___ 4π 3 √ related to the volume, V, by r(V) = pumped into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t. a. Find the composite function r(V(t)). b. Find the exact time when the radius reaches. Air is 10 inches. 91. The function A(d) gives the pain level on a scale of 0 to 10 experienced by a patient with d milligrams of a pain- reducing drug in her system. The milligrams of the drug in the patient’s system after t minutes is modeled by m(t). Which of the following would you do in order to determine when the patient will be at a pain level of 4? a. Evaluate A(m(4)). b. Evaluate m(A(4)). c. Solve A(m(t)) = 4. d. Solve m(A(d)) = 4. 93. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in — t + 2, find the area minutes according to r(t) = 25 of the ripple as a function of time. Find the area of the ripple at t = 2. √ 95. Use the function you found in the previous exercise to find the total area burned after 5 minutes. 97. The number of bacteria in a refrigerated food product is given by N(T) = 23T 2 − 56T + 1, 3 < T < 33, where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T(t) = 5t + 1.5, where t is the time in hours. a. Find the composite function N(T(t)). b. Find the time (round to two decimal places) when the bacteria count reaches 6,752. 64 CHAPTER 1 Functions leARnInG OBjeCTIVeS In this section, you will:
• • • • • Combine transformations. Graph functions using vertical and horizontal shifts. Graph functions using reflections about the x-axis and the y-axis. Determine whether a function is even, odd, or neither from its graph. Graph functions using compressions and stretches. 1. 5 TRAnSFORMATIOn OF FUnCTIOnS Figure 1 (credit: "Misko"/Flickr) We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations. Graphing Functions Using Vertical and Horizontal Shifts Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve. Identifying Vertical Shifts One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function g (x) = f (x) + k, the function f (x) is shifted vertically k units. See Figure 2 for an example. SECTION 1.5 transFormation oF Functions 65 f(x) f(x)+1 f(x) 3 4 5 21 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 2 Vertical shift by k = 1 of the cube root function f ( x) = 3 √ — x. To help you visualize the concept of a vertical shift, consider that y = f (x). Therefore, f (x) + k is equivalent to y + k. Every unit of y is replaced
by y + k, so the y-value increases or decreases depending on the value of k. The result is a shift upward or downward. vertical shift Given a function f (x), a new function g(x) = f (x) + k, where k is a constant, is a vertical shift of the function f (x). All the output values change by k units. If k is positive, the graph will shift up. If k is negative, the graph will shift down. Example 1 Adding a Constant to a Function To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 3 shows the area of open vents V (in square feet) throughout the day in hours after midnight, t. During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function. V 300 250 200 150 100 50 – –4 50– 4 8 12 16 20 24 28 t Figure 3 Solution We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically up, as shown in Figure 4. V 300 250 240 200 150 100 50 20 – –4 50– Up 20 4 8 12 16 20 24 28 t Figure 4 66 CHAPTER 1 Functions Notice that in Figure 4, for each input value, the output value has increased by 20, so if we call the new function S(t), we could write S(t) = V(t) + 20 This notation tells us that, for any value of t, S(t) can be found by evaluating the function V at the same input and then adding 20 to the result. This defines S as a transformation of the function V, in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change. See Table 1. t V(t) S(t) 0 0 20 8 0 20 17 220 240 19 0 20 24 0 20 10 220 240 Table 1 How To… Given a tabular function, create a new row to represent a vertical shift. 1. Identify the output row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down. Example 2 Sh
ifting a Tabular Function Vertically A function f (x) is given in Table 2. Create a table for the function g(x) = f (x) − 3. x f (x) 2 1 4 3 Table 2 6 7 8 11 Solution The formula g (x) = f (x) − 3 tells us that we can find the output values of g by subtracting 3 from the output values of f. For example: Given Given transformation f (2) = 1 g(x) = f (x) − 3 g(2) = f (2) − 3 = 1 − 3 = −2 Subtracting 3 from each f (x) value, we can complete a table of values for g(x) as shown in Table 3. x f (x) g(x) 2 1 −2 4 3 0 Table 3 6 7 4 8 11 8 Analysis As with the earlier vertical shift, notice the input values stay the same and only the output values change. Try It #1 The function h(t) = –4.9t2 + 30t gives the height h of a ball (in meters) thrown upward from the ground after t seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function b(t) to h(t), and then find a formula for b(t). SECTION 1.5 transFormation oF Functions 67 Identifying Horizontal Shifts We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in Figure 5. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 f(x+1) f(x) 21 3 4 5 x Figure 5 Horizontal shift of the function f (x) = 3 √ — x. note that h = +1 shifts the graph to the left, that is, towards negative values of x. For example, if f (x) = x2, then g(x) = (x − 2)2 is a new function. Each input is reduced by 2 prior to squaring the function. The result is that
the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in f. horizontal shift Given a function f, a new function g(x) = f (x − h), where h is a constant, is a horizontal shift of the function f. If h is positive, the graph will shift right. If h is negative, the graph will shift left. Example 3 Adding a Constant to an Input Returning to our building airflow example from Figure 3, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function. Solution We can set V(t) to be the original program and F (t) to be the revised program. V(t) = the original venting plan F(t) = starting 2 hrs sooner In the new graph, at each time, the airflow is the same as the original function V was 2 hours later. For example, in the original function V, the airflow starts to change at 8 a.m., whereas for the function F, the airflow starts to change at 6 a.m. The comparable function values are V(8) = F(6). See Figure 6. Notice also that the vents first opened to 220 ft2 at 10 a.m. under the original plan, while under the new plan the vents reach 220 ft2 at 8 a.m., so V(10) = F(8). In both cases, we see that, because F (t) starts 2 hours sooner, h = −2. That means that the same output values are reached when F (t) = V(t − (−2)) = V(t + 2). V 300 250 240 200 150 100 50 –2 0 50– – –4 Left 2 4 8 12 16 20 24 28 t Figure 6 68 CHAPTER 1 Functions Analysis Note that V(t + 2) has the effect of shifting the graph to the left. Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function F(t) uses the same outputs as V(t), but matches those outputs to inputs 2 hours earlier than those of V(t). Said another way, we must add 2 hours to the input of V to find the corresponding output for F : F(t
) = V(t + 2). How To… Given a tabular function, create a new row to represent a horizontal shift. 1. Identify the input row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each input cell. Example 4 Shifting a Tabular Function Horizontally A function f (x) is given in Table 4. Create a table for the function g (x) = f (x − 3). x f (x) 2 1 4 3 Table 4 6 7 8 11 Solution The formula g(x) = f (x − 3) tells us that the output values of g are the same as the output value of f when the input value is 3 less than the original value. For example, we know that f (2) = 1. To get the same output from the function g, we will need an input value that is 3 larger. We input a value that is 3 larger for g (x) because the function takes 3 away before evaluating the function f. We continue with the other values to create Table 5. g (5) = f (5 − 3) = f (2) = 1 x x − 3 f (x − 3) g(x) 5 2 1 1 7 4 3 3 Table 5 9 6 7 7 11 8 11 11 The result is that the function g (x) has been shifted to the right by 3. Notice the output values for g (x) remain the same as the output values for f (x), but the corresponding input values, x, have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11. Analysis Figure 7 represents both of the functions. We can see the horizontal shift in each point. y 12 10 8 6 4 2 –2–2 –4 –6 –8 –10 –12 –12 –10 –8 –6 –4 42 6 8 10 12 x f (x) g (x) = f(x – 3) Figure 7 SECTION 1.5 transFormation oF Functions 69 Example 5 Identifying a Horizontal Shift of a Toolkit Function Figure 8 represents a transformation of the toolkit function f (x) = x2. Relate this new function g (x) to f (x), and then find a formula for g (x). f (x) 6 5 4 3 2 1 – –1 1–
1 2 3 4 5 6 7 x Figure 8 Solution Notice that the graph is identical in shape to the f (x) = x2 function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so g (x) = f (x − 2) Notice how we must input the value x = 2 to get the output value y = 0; the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the f (x) function to write a formula for g (x) by evaluating f (x − 2). f (x) = x2 g (x) = f (x − 2) g (x) = f (x − 2) = (x − 2)2 Analysis To determine whether the shift is +2 or −2, consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, f (0) = 0. In our shifted function, g (2) = 0. To obtain the output value of 0 from the function f, we need to decide whether a plus or a minus sign will work to satisfy g (2) = f (x − 2) = f (0) = 0. For this to work, we will need to subtract 2 units from our input values. Example 6 Interpreting Horizontal versus Vertical Shifts The function G (m) gives the number of gallons of gas required to drive m miles. Interpret G (m) + 10 and G (m + 10). Solution G (m) + 10 can be interpreted as adding 10 to the output, gallons. This is the gas required to drive m miles, plus another 10 gallons of gas. The graph would indicate a vertical shift. G (m + 10) can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than m miles. The graph would indicate a horizontal shift. Try It #2 Given the function f (x) = √ Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units? — x, graph the original function f (x) and the transformation g (x) = f (x
+ 2) on the same axes. Combining Vertical and Horizontal Shifts Now that we have two transformations, we can combine them together. Vertical shifts are outside changes that affect the output ( y-) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and right or left. 7 0 CHAPTER 1 Functions How To… Given a function and both a vertical and a horizontal shift, sketch the graph. 1. Identify the vertical and horizontal shifts from the formula. 2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant. 3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant. 4. Apply the shifts to the graph in either order. Example 7 Graphing Combined Vertical and Horizontal Shifts Given f (x) = ∣ x ∣, sketch a graph of h (x) = f (x + 1) − 3. Solution The function f is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: f (x + 1) is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in f (x + 1) − 3 is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 9. Let us follow one point of the graph of f (x) = ∣ x ∣. • The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) → (−1, 0) • The point (−1, 0) is transformed next by shifting down 3 units: (−1, 0) → (−1, −3) y = \|x + 1\| y = \|x\| y = \|x + 1\| – 3 21 1 –1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 10 shows the graph of h. Figure 9 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2
y = \|x + 1\| – 3 21 3 4 5 x Figure 10 Try It #3 Given f (x) = ∣ x ∣, sketch a graph of h(x) = f (x − 2) + 4. SECTION 1.5 transFormation oF Functions 71 Example 8 Identifying Combined Vertical and Horizontal Shifts Write a formula for the graph shown in Figure 11, which is a transformation of the toolkit square root function. h(x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 11 Solution The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as Using the formula for the square root function, we can write h(x(x) = f (x − 1) + 2 Analysis Note that this transformation has changed the domain and range of the function. This new graph has domain [1, ∞) and range [2, ∞). Try It #4 1 __ Write a formula for a transformation of the toolkit reciprocal function f (x) = that shifts the function’s graph one x unit to the right and one unit up. Graphing Functions Using Reflections about the Axes Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 12. y Horizontal reflection f(x) Original function f(–x) x Vertical reflection –f(x) Figure 12 Vertical and horizontal reflections of a function. Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis. 7 2 CHAPTER 1 Functions reflections Given a function f (x), a new function g(x) = −f (x) is a vertical reflection of the function f (x), sometimes called a reflection about (or over, or through) the x-axis. Given a function f (x), a new function g(x) = f (−x) is a horizontal reflection of the function f (x), sometimes called a reflection about the y-axis. How To… Given a function,
reflect the graph both vertically and horizontally. 1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis. 2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis. Example 9 Reflecting a Graph Horizontally and Vertically Reflect the graph of s(t) = √ — t a. vertically and b. horizontally. Solution a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in Figure 13. s(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 t –5 –4 –3 –2 v(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 13 Vertical reflection of the square root function Because each output value is the opposite of the original output value, we can write V(t) = −s(t) or V(t) = − √ — t Notice that this is an outside change, or vertical shift, that affects the output s(t) values, so the negative sign belongs outside of the function. b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 14. s(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 t –5 –4 –3 –2 H(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 14 Horizontal reflection of the square root function SECTION 1.5 transFormation oF Functions 73 Because each input value is the opposite of the original input value, we can write H(t) = s(−t) or H(t) = √ — −t Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function. Note that these transformations can affect the domain and range of the functions. While the original square root function has domain [0, ∞) and range [0, ∞), the vertical reflection gives the V(t) function the range (−∞, 0] and the horizontal
reflection gives the H(t) function the domain (−∞, 0]. Try It #5 Reflect the graph of f (x) = \|x − 1\| a. vertically and b. horizontally. Example 10 Reflecting a Tabular Function Horizontally and Vertically A function f (x) is given as Table 6. Create a table for the functions below. a. g(x) = −f (x) b. h(x) = f (−x) x f (x) 2 1 4 3 Table 6 6 7 8 11 Solution a. For g(x), the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See Table 7. x g (x) 2 –1 6 –7 8 –11 4 –3 Table 7 b. For h(x), the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the h(x) values stay the same as the f (x) values. See Table 8. x h(x) –2 1 –4 3 Table 8 –6 7 –8 11 Try It #6 A function f (x) is given as Table 9. Create a table for the functions below. x f(x) −2 5 0 10 Table 9 2 15 4 20 a. g(x) = −f (x) b. h(x) = f (−x) 7 4 CHAPTER 1 Functions Example 11 Applying a Learning Model Equation A common model for learning has an equation similar to k(t) = −2−t + 1, where k is the percentage of mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2t shown in Figure 15. Sketch a graph of k(t). f (t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Solution This equation combines three transformations into one equation. Figure 15 • A horizontal reflection: f (−t) = 2−t • A vertical reflection: −f (−t) = −2−t • A vertical shift: −f (−t) + 1 = −2−t + 1 We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each
of the three transformations. We will choose the points (0, 1) and (1, 2). 1. First, we apply a horizontal reflection: (0, 1) (−1, 2). 2. Then, we apply a vertical reflection: (0, −1) (1, −2). 3. Finally, we apply a vertical shift: (0, 0) (1, 1). This means that the original points, (0,1) and (1,2) become (0,0) and (1,1) after we apply the transformations. In Figure 16, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit. f (t) f (t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (a) 1 2 3 4 5 t –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t –5 –4 –3 –2 (b) Figure 16 k(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (c) 21 3 4 5 t Analysis As a model for learning, this function would be limited to a domain of t ≥ 0, with corresponding range [0, 1). Try It #7 Given the toolkit function f (x) = x 2, graph g(x) = −f (x) and h(x) = f (−x). Take note of any surprising behavior for these functions. SECTION 1.5 transFormation oF Functions 75 Determining even and Odd Functions Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions f (x) = x2 or f (x) = x will result in the original graph. We say that these types of graphs are symmetric about the y-axis. Functions whose graphs are symmetric about the y-axis are called even functions. 1 _ If the graphs of f (x) = x3 or f (x) = x were reflected over both axes, the result would be the original graph, as shown in Figure 17. y y f (x) = x3 f (x) = x3 y y f (−x) f (−xf (−x) −f (−
x5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –2 –3 –4 –5 –1 –2 –3 –4 –5 –5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –2 –3 –4 –5 –1 –2 –3 –4 –5 21 21 3 3 4 4 5 5 x x –5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –2 –3 –4 –5 –1 –2 –3 –4 –5 21 21 3 3 4 4 5 5 x x (a) (a) (b) (b) (c) (c) Figure 17 (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical reflections reproduce the original cubic function. We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function. Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f (x) = 2x is neither even nor odd. Also, the only function that is both even and odd is the constant function f (x) = 0. even and odd functions A function is called an even function if for every input x: f (x) = f (−x) The graph of an even function is symmetric about the y-axis. A function is called an odd function if for every input x: f (x) = −f (−x) The graph of an odd function is symmetric about the origin. How To… Given the formula for a function, determine if the function is even, odd, or neither. 1. Determine whether the function satisfies f (x) = f (−x). If it does, it is even. 2. Determine whether the function satisfies f (x) = −f (−x). If it does, it is odd. 3. If the function does not satisfy either rule, it is neither even nor odd. Example 12 Determining whether a Function Is Even, Odd, or Neither Is the function f (x) = x 3 + 2x even, odd, or neither? Solution Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return
us to the original function. Let’s begin with the rule for even functions. f (−x) = (−x)3 + 2(−x) = −x 3 − 2x This does not return us to the original function, so this function is not even. We can now test the rule for odd functions. Because −f (−x) = f (x), this is an odd function. −f (−x) = −(−x 3 − 2x) = x 3 + 2x 7 6 CHAPTER 1 Functions Analysis Consider the graph of f in Figure 18. Notice that the graph is symmetric about the origin. For every point (x, y) on the graph, the corresponding point (−x, −y) is also on the graph. For example, (1, 3) is on the graph of f, and the corresponding point (−1, −3) is also on the graph. –5 –4 –3 –2 f (x) f 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (−1, −3) (1, 3) 21 3 4 5 x Try It #8 Is the function f (s) = s 4 + 3s 2 + 7 even, odd, or neither? Figure 18 Graphing Functions Using Stretches and Compressions Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity. We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically. Vertical Stretches and Compressions When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. Figure 19 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression. y Vertical stretch Vertical compression 2f(x) f(x) 0.5f(x) x Figure 19 Vertical stretch and compression vertical stretches and compressions Given a function f (x), a new function g(x) = af (x), where a is a
constant, is a vertical stretch or vertical compression of the function f (x). • If a > 1, then the graph will be stretched. • If 0 < a < 1, then the graph will be compressed. • If a < 0, then there will be combination of a vertical stretch or compression with a vertical reflection. SECTION 1.5 transFormation oF Functions 77 How To… Given a function, graph its vertical stretch. 1. Identify the value of a. 2. Multiply all range values by a. 3. If a > 1, the graph is stretched by a factor of a. If 0 < a < 1, the graph is compressed by a factor of a. If a < 0, the graph is either stretched or compressed and also reflected about the x-axis. Example 13 Graphing a Vertical Stretch A function P(t) models the population of fruit flies. The graph is shown in Figure 20. P(t) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 t Figure 20 A scientist is comparing this population to another population, Q, whose growth follows the same pattern, but is twice as large. Sketch a graph of this population. Solution Because the population is always twice as large, the new population’s output values are always twice the original function’s output values. Graphically, this is shown in Figure 21. If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2. The following shows where the new points for the new graph will be located. (0, 1) → (0, 2) (3, 3) → (3, 6) (6, 2) → (6, 4) (7, 0) → (7, 0) Q(t) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 t Figure 21 Symbolically, the relationship is written as Q(t) = 2P(t) This means that for any input t, the value of the function Q is twice the value of the function P. Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, t, stay the same while the output values are twice as large as before. 7 8 CHAPTER 1 Functions How To… Given a tab
ular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression. 1. Determine the value of a. 2. Multiply all of the output values by a. Example 14 Finding a Vertical Compression of a Tabular Function A function f is given as Table 10. Create a table for the function g(x) = 1 __ f (x). 2 x f (x) 2 1 6 7 8 11 4 3 Table 10 Solution The formula g(x) = 1 __ f (x) tells us that the output values of g are half of the output values of f with the same 2 inputs. For example, we know that f (4) = 3. Then We do the same for the other values to produce Table 11. (3) = 3 f (4) = 1 g(4) = 1 __ __ __ 2 2 2 x g(x) 2 1 __ 2 4 3 __ 2 Table 11 6 7 __ 2 8 11 __ 2 1 __ Analysis The result is that the function g(x) has been compressed vertically by. Each output value is divided in half, 2 so the graph is half the original height. Try It #9 A function f is given as Table 12. Create a table for the function g(x) = 3 __ f (x). 4 x f(x) 2 12 4 16 Table 12 6 20 8 0 Example 15 Recognizing a Vertical Stretch The graph in Figure 22 is a transformation of the toolkit function f (x) = x3. Relate this new function g(x) to f (x), and then find a formula for g(x). g(x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 22 SECTION 1.5 transFormation oF Functions 79 Solution When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g(2) = 2. With the basic cubic function at the same input, f (2) = 23 = 8. 1 the outputs of the function f because g(2) = 1 __ __ f (2). From this we Based on that, it appears that the outputs of g are 4 4 can fairly safely conclude that g(x) = 1 __ f (x). 4 We can write a formula for g by
using the definition of the function f. f (x) = 1 g(x) = 1 __ __ x3 4 4 Try It #10 Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units. Horizontal Stretches and Compressions Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function. Horizontal compression y = x2 y = (0.5x)2 Horizontal stretch y = (2x)2 y 10 5 –4 –3 –2 –1 1 2 3 4 5 x –1 Figure 23 Given a function y = f (x), the form y = f (bx) results in a horizontal stretch or compression. Consider the function y = x2. Observe Figure 23. The graph of y = (0.5x)2 is a horizontal stretch of the graph of the function y = x 2 by a factor of 2. The graph of y = (2x)2 is a horizontal compression of the graph of the function y = x 2 by a factor of 2. horizontal stretches and compressions Given a function f (x), a new function g(x) = f (bx), where b is a constant, is a horizontal stretch or horizontal compression of the function f (x). 1 __ • If b > 1, then the graph will be compressed by. b 1 __ • If 0 < b < 1, then the graph will be stretched by. b • If b < 0, then there will be combination of a horizontal stretch or compression with a horizontal reflection. How To… Given a description of a function, sketch a horizontal compression or stretch. 1. Write a formula to represent the function. 2. Set g(x) = f (bx) where b > 1 for a compression or 0 < b < 1 for a stretch. 80 CHAPTER 1 Functions Example 16 Graphing a Horizontal Compression Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R, will progress in
1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population. Solution Symbolically, we could write R(1) = P(2), R(2) = P(4), and in general, R(t) = P(2t). See Figure 24 for a graphical comparison of the original population and the compressed population. f (x) 6 5 4 3 2 1 – –1 1– Original population, P(t) Transformed population, R(t) f (x1 1a) (b) Figure 24 (a) Original population graph (b) Compressed population graph Analysis Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis. Example 17 Finding a Horizontal Stretch for a Tabular Function 1 __ x . A function f (x) is given as Table 13. Create a table for the function g(x) = f  2 x f (x) 2 1 6 7 8 11 4 3 Table 13 1 __ x  tells us that the output values for g are the same as the output values for Solution The formula g(x) = f  2 the function f at an input half the size. Notice that we do not have enough information to determine g(2) because 1 __ ⋅ 2  = f (1), and we do not have a value for f (1) in our table. Our input values to g will need to be twice as g(2) = f  2 large to get inputs for f that we can evaluate. For example, we can determine g(4). We do the same for the other values to produce Table 14. 1 __ ⋅ 4  = f (2) = 1 g(4) = f  2 x g(x) 4 1 8 3 12 7 16 11 Table 14 SECTION 1.5 transFormation oF Functions 81 Figure 25 shows the graphs of both of these sets of points. –20 –16 –12 –8 y 12 10 8 6 4 2 –4 –2 –4 –6 –8 –10 –12 (a) 84 12 16 20 x –20 –16 –12 –8 Figure 25 y 12 10 8 6 4 2 –4 –2 –4 –6
–8 –10 –12 (b) 84 12 16 20 x Analysis Because each input value has been doubled, the result is that the function g(x) has been stretched horizontally by a factor of 2. Example 18 Recognizing a Horizontal Compression on a Graph Relate the function g(x) to f (x) in Figure 26. f (x) 6 5 4 3 2 1 – –1 1 Figure 26 Solution The graph of g(x) looks like the graph of f (x) horizontally compressed. Because f (x) ends at (6, 4) and g(x) ends at 1 1 __ __  = 2. We might also notice that g(2) = f (6), because 6  (2, 4), we can see that the x-values have been compressed by 3 3 1 __ and g(1) = f (3). Either way, we can describe this relationship as g(x) = f (3x). This is a horizontal compression by. 3 Analysis Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or 1 1 __ __ x . in our function: f  compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 4 4 This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. Try It #11 Write a formula for the toolkit square root function horizontally stretched by a factor of 3. 82 CHAPTER 1 Functions Performing a Sequence of Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first. When we see an expression such as 2f (x) + 3, which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of f (x), we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition. Horizontal transformations are a little trickier to think about. When we write g(x) = f (2x + 3), for
example, we have to think about how the inputs to the function g relate to the inputs to the function f. Suppose we know f (7) = 12. What input to g would produce that output? In other words, what value of x will allow g(x) = f (2x + 3) = 12? We would need 2x + 3 = 7. To solve for x, we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression. This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function. p f (bx + p Let’s work through an example. We can factor out a 2. f (x) = (2x + 4)2 f (x) = (2(x + 2))2 Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally. combining transformations When combining vertical transformations written in the form af (x) + k, first vertically stretch by a and then vertically shift by k. When combining horizontal transformations written in the form f (bx − h), first horizontally shift by h and 1 __ then horizontally stretch by. b 1 _ When combining horizontal transformations written in the form f (b(x − h)), first horizontally stretch by b and then horizontally shift by h. Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first. Example 19 Finding a Triple Transformation of a Tabular Function Given Table 15 for the function f (x), create a table of values for the function g(x) = 2f (3x) + 1. x f (x) 6 10 12 14 Table 15 18 15 24 17 Solution There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal 1 1 __ __. See Table 16., which means we multiply each x-value by transformations, f (3x) is a horizontal compression by 3 3 x f (3x) 2 10 4 14 Table 16 6 15 8 17 SECTION 1.5 transFormation oF Functions 83 Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See Table 17. x 2
f (3x) 2 20 4 28 Table 17 6 30 8 34 Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 18. x g(x) = 2f (3x) + 1 4 29 6 31 8 35 2 21 Table 18 Example 20 Finding a Triple Transformation of a Graph 1 __ Use the graph of f (x) in Figure 27 to sketch a graph of k(x. 2 f (x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 27 Solution To simplify, let’s start by factoring out the inside of the function. 1 1 __ __ (x + 2 __ on the inside of the function. By factoring the inside, we can first horizontally stretch by 2, as indicated by the 2 Remember that twice the size of 0 is still 0, so the point (0, 2) remains at (0, 2) while the point (2, 0) will stretch to (4, 0). See Figure 28. f (x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 28 84 CHAPTER 1 Functions Next, we horizontally shift left by 2 units, as indicated by x + 2. See Figure 29. f (x) –6 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 6 x Figure 29 Last, we vertically shift down by 3 to complete our sketch, as indicated by the −3 on the outside of the function. See Figure 30. f (x) –6 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 6 x Figure 30 Access this online resource for additional instruction and practice with transformation of functions. • Function Transformations (http://openstaxcollege.org/l/functrans) SECTION 1.5 section exercises 85 1.5 SeCTIOn exeRCISeS VeRBAl 1. When examining the formula of a function that is 2. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift? 3. When examining the formula of a function that is the result of
multiple transformations, how can you tell a horizontal compression from a vertical compression? 5. How can you determine whether a function is odd or even from the formula of the function? AlGeBRAIC the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch? 4. When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis? 6. Write a formula for the function obtained when the — x is shifted up 1 unit and to the graph of f (x) = √ left 2 units. 7. Write a formula for the function obtained when the graph of f (x) = \|x\| is shifted down 3 units and to the right 1 unit. 8. Write a formula for the function obtained when the _ x is shifted down 4 units and to the graph of f (x) = 1 right 3 units. 9. Write a formula for the function obtained when the 1 _ x2 is shifted up 2 units and to the graph of f (x) = left 4 units. For the following exercises, describe how the graph of the function is a transformation of the graph of the original function f. 10. y = f (x − 49) 13. y = f (x − 4) 16. y = f (x) − 2 19. y = f (x + 4) − 1 11. y = f (x + 43) 14. y = f (x) + 5 17. y = f (x) − 7 12. y = f (x + 3) 15. y = f (x) + 8 18. y = f (x − 2) + 3 For the following exercises, determine the interval(s) on which the function is increasing and decreasing. 20. f (x) = 4(x + 1)2 − 5 23. k(x) = −3 √ — x − 1 21. g(x) = 5(x + 3)2 − 2 22. a(x) = √ — −x + 4 GRAPHICAl For the following exercises, use the graph of f (x) = 2x shown in Figure 31 to sketch a graph of each transformation of f (x). f 21 3 4 5 x 24. g(x) = 2x + 1 26. w(x) = 2x − 1 25. h(x
) = 2x − 3 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 31 For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. 27. f (t) = (t + 1)2 − 3 28. h(x) = \|x − 1\| + 4 29. k(x) = (x − 2)3 − 1 30. m(t) = 3 + √ — t + 2 86 nUMeRIC CHAPTER 1 Functions 31. Tabular representations for the functions f, g, and h are given below. Write g(x) and h(x) as transformations of f (x). x f (x) x g(x) x h(x) –2 –2 –1 –2 –2 –1 –1 –1 0 –1 –1 0 0 –3 1 –3 0 – 32. Tabular representations for the functions f, g, and h are given below. Write g(x) and h(x) as transformations of f (x). x f (x) x g(x) x h(x) –2 –1 –3 –1 –2 –2 –1 –3 –2 –3 –1 –4 0 4 – For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions. 33. 36. f –5 –4 –3 –2 –5 –4 –3 –2 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –1 –2 –4 –6 –8 –10 34. 21 3 4 5 x –5 –4 –3 –2 37. f 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –1 –2 –4 –6 –8 –10 35. 21 3 4 5 x –5 –4 –3 –2 f f 38. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 f 21 3 4 5 x –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 f 21 3 4 5 x –1 –1 –2 –3
–4 –5 SECTION 1.5 section exercises 87 39. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 40. f 21 3 4 5 6 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 f 21 3 4 5 x For the following exercises, use the graphs of transformations of the square root function to find a formula for each of the functions. 41. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 42. f 21 3 4 5 x –5 –4 –3 –2 f y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions. y y y 43. 44. 45. f 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 f 21 3 4 5 x f –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x f 21 3 4 5 x –5 –4 –3 –2 46. –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 88 CHAPTER 1 Functions For the following exercises, determine whether the function is odd, even, or neither. 47. f (x) = 3x 4 — 48. g(x) = √ x 1 __ + 3x 49. h(x) = x 50. f (x) = (x − 2)2 51. g(x) = 2x4 52. h(x) = 2x − x3 For the following exercises, describe how the graph of each function is a transformation of the graph of the original function f. 53. g(x) = −f (x) 54. g(x) = f (−x) 55. g(x) = 4f (x) 56. g(x) = 6f (x) 57. g(x) = f (5
x) 58. g(x) = f (2x) 61. g(x) = 3f (−x) 62. g(x) = −f (3x) 1 __ x  59. g(x) = f  3 1 __ x  60. g(x) = f  5 For the following exercises, write a formula for the function g that results when the graph of a given toolkit function is transformed as described. 63. The graph of f (x) = ∣ x ∣ is reflected over the y-axis 1 __. and horizontally compressed by a factor of 4 64. The graph of f (x) = √ — x is reflected over the x-axis and horizontally stretched by a factor of 2. 1 __ 65. The graph of f (x) = x2 is vertically compressed by a 1 __ factor of, then shifted to the left 2 units and down 3 3 units. 1 __ 66. The graph of f (x) = is vertically stretched by a x factor of 8, then shifted to the right 4 units and up 2 units. 67. The graph of f (x) = x2 is vertically compressed by a 1 __ factor of, then shifted to the right 5 units and up 2 1 unit. 68. The graph of f (x) = x2 is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. 69. g(x) = 4(x + 1)2 − 5 70. g(x) = 5(x + 3)2 − 2 71. h(x) = − 72. k(x) = −3 √ 73. m(x) = 1 __ x3 2 76. q(x) =  1 75. p(x) =  1 __ __ x  x  4 3 For the following exercises, use the graph in Figure 32 to sketch the given transformations. 77. a(x) = √ 3 + 1 3 − 3 74. n(x) = 1 __ \|x − 2\| 3 — −x + 4 y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 f 42 6
8 10 x Figure 32 78. g(x) = f (x) − 2 79. g(x) = −f (x) 80. g(x) = f (x + 1) 81. g(x) = f (x − 2) SECTION 1.6 aBsolute value Functions 89 leARnInG OBjeCTIVeS In this section you will: • • • Graph an absolute value function. Solve an absolute value equation. Solve an absolute value inequality. 1. 6 ABSOlUTe VAlUe FUnCTIOnS Figure 1 Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: “s58y”/Flickr) Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate absolute value functions. Understanding Absolute Value Recall that in its basic form f (x) = \| x \|, the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. absolute value function The absolute value function can be defined as a piecewise function x { f (x) = ∣ x ∣ = −x if x ≥ 0 if x < 0 Example 1 Determine a Number within a Prescribed Distance Describe all values x within or including a distance of 4 from the number 5. Solution We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as the one in Figure 2, to represent the condition to be satisfied. 4 4 5 Figure 2 The distance from x to 5 can be represented using the absolute value as ∣ x − 5 ∣. We want the values of x that satisfy the condition ∣ x − 5 ∣ ≤ 4. Analysis Note that − So �
� x − 5 ∣ ≤ 4 is equivalent to 1 ≤ x ≤ 9. However, mathematicians generally prefer absolute value notation. 90 CHAPTER 1 Functions Try It #1 Describe all values x within a distance of 3 from the number 2. Example 2 Resistance of a Resistor Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ±5%, or ±10%. Suppose we have a resistor rated at 680 ohms, ±5%. Use the absolute value function to express the range of possible values of the actual resistance. Solution 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R in ohms, \| R − 680 \| ≤ 34 Try It #2 Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation. Graphing an Absolute Value Function The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 3. –5 –4 –3 –2 y = \|x\| 21 1–1 –2 –3 –4 –5 Figure 3 Figure 4 shows the graph of y = 2\| x − 3\| + 4. The graph of y = \| x\| has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at (3, 4) for this transformed function. y = 2\|x − 3\| Vertical stretch y = \|x\| y 10 \|x − 3\| + 4 Up 4 y = \|x − 3\| Right 3 –6 –5 –4 –3 –2 –1–1 1 2 3 4 5 6 x Figure 4 SECTION 1.6 aBsolute value Functions 91 Example 3 Writing an Equation for an Absolute Value Function Write an equation for the function graphed in Figure 5. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure
5 Solution The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 6. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x (3, –2) Figure 6 We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 7. Ratio 2/1 y 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 Ratio 1/1 21 3 4 5 x (3, −2) From this information we can write the equation Figure 7 f (x) = 2\| x − 3 \| − 2, treating the stretch as a vertical stretch, or f (x) = \| 2(x − 3) \| − 2, treating the stretch as a horizontal compression. Analysis Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. 92 CHAPTER 1 Functions Q & A… If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it? Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x and f (x). Now substituting in the point (1, 2) f (x) = a\| 1 − 3 \| − 2 4 = 2a a = 2 Try It #3 Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units. Q & A… Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis? Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero. No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one,
or two points (see Figure 8). y 5 4 3 2 1 –1 –1 –2 –3 –4 (a) –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 (b) 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 (c) 21 3 4 5 x Figure 8 (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points. Solving an Absolute Value equation Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8 = \| 2x − 6\|, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or −8. This leads to two different equations we can solve independently. 2x − 6 = 8 or 2x − 6 = −8 2x = −2 x = −1 2x = 14 x = 7 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example, \| x \| = 4, \| 2x − 1 \| = 3 \| 5x + 2 \| − 4 = 9 SECTION 1.6 aBsolute value Functions 93 solutions to absolute value equations For real numbers A and B, an equation of the form \| A \| = B, with B ≥ 0, will have solutions when A = B or A = −B. If B < 0, the equation \| A \| = B has no solution. How To… Given the formula for an absolute value function, find the horizontal intercepts of its graph. 1. Isolate the absolute value term. 2. Use \| A \| = B to write A = B or −A = B, assuming B > 0. 3. Solve for x. Example 4 Finding the Zeros of an Absolute Value Function For the function f (x) = \| 4x + 1 \| − 7, find the values of x such that f (x) = 0
. Solution 0 = \| 4x + 1 \| − 7 7 = \| 4x + 1 \| 7 = 4x + 1 or −7 = 4x + 1 6 = 4x 6 __ x = = 1.5 4 −8 = 4x −8 ___ 4 x = = −2 Substitute 0 for f (x). Isolate the absolute value on one side of the equation. Break into two separate equations and solve. The function outputs 0 when x = 1.5 or x = −2. See Figure 9. y 10 8 6 4 2 –0.5 –2 –4 –6 –8 –10 (−2, 0) –1 –1.5 –2 –2.5 x where f(x) = 0 f x (1.5, 0) 0.5 1 1.5 2 2.5 x where f(x) = 0 Try It #4 For the function f (x) = \| 2x − 1 \| − 3, find the values of x such that f (x) = 0. Figure 9 Q & A… Should we always expect two answers when solving ∣ A ∣ = B? No. We may find one, two, or even no answers. For example, there is no solution to 2 + \| 3x − 5 \| = 1. How To… Given an absolute value equation, solve it. 1. Isolate the absolute value term. 2. Use \| A \| = B to write A = B or A = −B. 3. Solve for x. 94 CHAPTER 1 Functions Example 5 Solving an Absolute Value Equation Solve 1 = 4\| x − 2\| + 2. Solution Isolating the absolute value on one side of the equation gives the following. 1 = 4\| x − 2 \| + 2 −1 = 4 The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. At this point, we notice that this equation has no solutions. Q & A… In Example 5, if f(x) = 1 and g(x) = 4∣ x − 2 ∣ + 2 were graphed on the same set of axes, would the graphs intersect? No. The graphs of f and g would not intersect, as shown in Figure 10. This confirms, graphically, that the equation 1 = 4\| x − 2 \| + 2 has no solution. y g(x) = 4\|x −
2\| + 2 f (x) = 1 x 21 3 4 5 –5 –4 –3 –2 10 8 6 4 2 –1–2 –4 –6 –8 –10 Figure 10 Try It #5 Find where the graph of the function f (x) = −\| x + 2 \| + 3 intersects the horizontal and vertical axes. Solving an Absolute Value Inequality Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of values. We would use an absolute value inequality to solve such an equation. An absolute value inequality is an equation of the form \| A \| < B, \| A \| ≤ B, \| A \| ≥ B, or \| A \| ≥ B, where an expression A (and possibly but not usually B) depends on a variable x. Solving the inequality means finding the set of all x that satisfy the inequality. Usually this set will be an interval or the union of two intervals. There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two functions. The advantage of the algebraic approach is it yields solutions that may be difficult to read from the graph. For example, we know that all numbers within 200 units of 0 may be expressed as \| x \| < 200 or −200 < x < 200 Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of values x such that the distance between x and 600 is less than 200. We represent the distance between x and 600 as \| x − 600 \|. \| x − 600 \| < 200 or −200 < x − 600 < 200 −200 + 600 < x − 600 + 600 < 200 + 600 400 < x < 800 SECTION 1.6 aBsolute value Functions 95 This means our returns would be between $400 and $800. Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed absolute value function, where we must determine for which values of the input the function’s output will be negative or positive. How To… Given an absolute value inequality of the form \| x − A \| ≤ B for real numbers a and b where b is positive, solve the absolute value inequality algebraically. 1. Find boundary points by solving \| x − A \| = B.
2. Test intervals created by the boundary points to determine where \| x − A \| ≤ B. 3. Write the interval or union of intervals satisfying the inequality in interval, inequality, or set-builder notation. Example 6 Solving an Absolute Value Inequality Solve \| x − 5\| < 4. Solution With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where \| x − 5\| = 4. We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve \| x − 5\| = 4. x − 5 = 4 or x − 5 = −4 x = 9 x = 1 After determining that the absolute value is equal to 4 at x = 1 and x = 9, we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals: x < 1, 1 < x < 9, and x > 9. To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in Table 1. Interval test x f (x 11 < 4 or > 4? \| 0 − 5\| = 5 \| 6 − 5\| = 1 Greater than Less than \| 11 − 5\| = 6 Greater than Table 1 Because 1 ≤ x ≤ 9 is the only interval in which the output at the test value is less than 4, we can conclude that the solution to \| x − 5\| ≤ 4 is 1 ≤ x ≤ 9, or [1, 9]. To use a graph, we can sketch the function f (x) = \| x − 5\|. To help us see where the outputs are 4, the line g(x) = 4 could also be sketched as in Figure 11. f (x) 6 5 4 3 2 1 – –1 1– The graph of f is below or on the graph of 10 x Figure 11 Graph to find the points satisfying an absolute value inequality. 96 CHAPTER 1 Functions We can see the following: • The output values of the absolute value are equal to 4 at x = 1 and x = 9. • The graph of f is below the graph of g on 1 < x < 9. This means the output values of f (x)
are less than the output values of g(x). • The absolute value is less than or equal to 4 between these two points, when 1 ≤ x ≤ 9. In interval notation, this would be the interval [1, 9]. Analysis For absolute value inequalities, \| x − A \| < C, \| x − A \| > C, −C < x − A < C, x − A < −C or x − A > C. The < or > symbol may be replaced by ≤ or ≥. So, for this example, we could use this alternative approach. \| x − 5\| ≤ 4 −4 ≤ x − 5 ≤ 4 Rewrite by removing the absolute value bars. − Isolate the x. 1 ≤ x ≤ 9 Try It #6 Solve \| x + 2\| ≤ 6. How To… Given an absolute value function, solve for the set of inputs where the output is positive (or negative). 1. Set the function equal to zero, and solve for the boundary points of the solution set. 2. Use test points or a graph to determine where the function’s output is positive or negative. Example 7 Using a Graphical Approach to Solve Absolute Value Inequalities 1 __ Given the function f (x) = − \| 4x − 5 \| + 3, determine the x-values for which the function values are negative. 2 Solution We are trying to determine where f (x) < 0, which is when − 1 __ \| 4x − 5\| + 3 < 0. We begin by isolating the 2 absolute value. − 1 \| 4x − 5\| < −3 __ 2 \| 4x − 5\| > 6 Multiply both sides by −2, and reverse the inequality. Next we solve for the equality \| 4x − 5\| = 6. 4x − 5 = 6 or 4x − 5 = −6 4x − 5 = 6 x = 11 __ 4 4x = −1 1 __ x = − 4 Now, we can examine the graph of f to observe where the output is negative. We will observe where the branches are below the x-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that 1 __ and x = it crosses the horizontal axis at x = − 4 11 __ and that the graph has been reflected vertically. See Figure 12. 4 SECTION 1.6 aBsolute value Functions 97 y 5 4 3 2 1 x = −0.
25 21 –5 –2 –4 Below x-axis –3 –1–1 –2 –3 –4 –5 Figure 12 x = −2.75 x 3 4 5 Below x-axis 1 11 __ __ We observe that the graph of the function is below the x-axis left of x = − and right of x = 4 4 function values are negative to the left of the first horizontal intercept at x = − 1 __, and negative to the right of the 4 second intercept at x =.This gives us the solution to the inequality. This means the 11 __ 4 1 __ x < − or x > 4 11 __ 4 In interval notation, this would be (−∞, −0.25) ∪ (2.75, ∞). Try It #7 Solve −2\| k − 4\| ≤ −6. Access these online resources for additional instruction and practice with absolute value. • Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue) • Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2) • equations of Absolute Value Function (http://openstaxcollege.org/l/findeqabsval) • equations of Absolute Value Function 2 (http://openstaxcollege.org/l/findeqabsval2) • Solving Absolute Value equations (http://openstaxcollege.org/l/solveabsvalueeq) 98 CHAPTER 1 Functions 1.6 SeCTIOn exeRCISeS VeRBAl 1. How do you solve an absolute value equation? 3. When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function? 5. How do you solve an absolute value inequality algebraically? AlGeBRAIC 2. How can you tell whether an absolute value function has two x-intercepts without graphing the function? 4. How can you use the graph of an absolute value function to determine the x-values for which the function values are negative? 6. Describe all numbers x that are at a distance of 4 from the number 8. Express this using absolute value notation. 1 __ 7. Describe all numbers x that are at a distance of 2 from the number −4. Express this using absolute value notation. 8. Describe the situation in which the distance that point
x is from 10 is at least 15 units. Express this using absolute value notation. 9. Find all function values f (x) such that the distance from f (x) to the value 8 is less than 0.03 units. Express this using absolute value notation. For the following exercises, solve the equations below and express the answer using set notation. 10. \| x + 3 \| = 9 11. \| 6 − x \| = 5 12. \| 5x − 2 \| = 11 13. \| 4x − 2 \| = 11 14. 2\| 4 − x \| = 7 15. 3\| 5 − x \| = 5 16. 3\| x + 1 \| − 4 = 5 17. 5\| x − 4 \| − 7 = 2 18. 0 = −\| x − 3 \| + 2 19. 2\| x − 3 \| + 1 = 2 20. \| 3x − 2 \| = 7 21. \| 3x − 2 \| = −7 22.  1 x − 5  = 11 __ 2 23.  1 x + 5  = 14 __ 3 24. −  1 x + 5  + 14 = 0 __ 3 For the following exercises, find the x- and y-intercepts of the graphs of each function. 25. f (x) = 2\| x +1 \| − 10 26. f (x) = 4\| x − 3 \| + 4 27. f (x) = −3\| x − 2 \| − 1 28. f (x) = −2\| x +1 \| + 6 For the following exercises, solve each inequality and write the solution in interval notation. 29. \| x − 2 \| > 10 30. 2\| v − 7 \| − 4 ≥ 42 31. \| 3x − 4 \| ≤ 8 32. \| x − 4 \| ≥ 8 33. \| 3x − 5 \| ≥ 13 34. \| 3x − 5 \| ≥ −13 35.  3 x − 5  ≥ 7 __ 4 36.  3 x − 5  + 1 ≤ 16 __ 4 SECTION 1.6 section exercises 99 GRAPHICAl For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph. 37. y = \| x − 1 \| 38. y = \| x + 1 \| 39. y = \| x \| + 1 For the following exercises, graph
the given functions by hand. 40. y = \| x \| − 2 41. y = −\| x \| 42. y = −\| x \| − 2 43. y = −\| x − 3 \| − 2 44. f (x) = −\| x − 1 \| − 2 45. f (x) = −\| x + 3 \| + 4 46. f (x) = 2\| x + 3 \| + 1 47. f (x) = 3\| x − 2 \| + 3 48. f (x) = \| 2x − 4 \| − 3 49. f (x) = \| 3x + 9 \| + 2 50. f (x) = −\| x − 1 \| − 3 51. f (x) = −\| x + 4 \| −3 1 __ \| x + 4 \| − 3 52. f (x) = 2 TeCHnOlOGY 53. Use a graphing utility to graph f (x) = 10\| x − 2\| on the viewing window [0, 4]. Identify the corresponding range. Show the graph. 54. Use a graphing utility to graph f (x) = −100\| x\| + 100 on the viewing window [−5, 5]. Identify the corresponding range. Show the graph. For the following exercises, graph each function using a graphing utility. Specify the viewing window. 56. f (x) = 4 × 109 ∣ x − (5 × 109) ∣ + 2 × 109 55. f (x) = −0.1\| 0.1(0.2 − x)\| + 0.3 exTenSIOnS For the following exercises, solve the inequality. 2 __ (x + 1) \| + 3 > −1 57. \| −2x − 3 58. If possible, find all values of a such that there are no x-intercepts for f (x) = 2\| x + 1\| + a. 59. If possible, find all values of a such that there are no y-intercepts for f (x) = 2\| x + 1\| + a. ReAl-WORlD APPlICATIOnS 60. Cities A and B are on the same east-west line. Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and x represents the distance from city B to city A, express this
using absolute value notation. 61. The true proportion p of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation. 62. Students who score within 18 points of the number 82 will pass a particular test. Write this statement using absolute value notation and use the variable x for the score. 63. A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using x as the diameter of the bearing, write this statement using absolute value notation. 64. The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is x inches, express the tolerance using absolute value notation. 100 CHAPTER 1 Functions leARnInG OBjeCTIVeS In this section, you will: • • • • Verify inverse functions. Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one. Find or evaluate the inverse of a function. Use the graph of a one-to-one function to graph its inverse function on the same axes. 1.7 InVeRSe FUnCTIOnS A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating. If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions. x f y? y x Figure 1 Can a function “machine” operate in reverse? Verifying That Two Functions Are Inverse Functions Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula and substitutes 75 for F to calculate 5 __ C = (F − 32
) 9 5 __ (75 − 32) ≈ 24°C. 9 Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure 2 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. Mon Tue Web Thu 26°C \| 19°C 29°C \| 19°C 30°C \| 20°C 26°C \| 18°C Figure 2 SECTION 1.7 inverse Functions 101 At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for F after substituting a value for C. For example, to convert 26 degrees Celsius, she could write 26 = 5 __ (F − 32) 9 9 __ = F − 32 26 ⋅ 5 9 __ + 32 ≈ 79 F = 26 ⋅ 5 After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature. The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function. Given a function f (x), we represent its inverse as f −1(x), read as “ f inverse of x.” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1. In other words, f −1(x) does not mean is the reciprocal of f and not the inverse. The “exponent-like” notation comes from an analogy between function composition and multiplication: just as a−1 a = 1 (1 is the identity element for multiplication) for any nonzero number a, so f −1 ∘ f equals the identity function, that is, 1 ___ f (x) 1 ___ f (x) because (f −1 ∘ f )(x) = f −1( f (x)) = f −1 (y) = x This holds for all x in the domain of f. Informally, this means that inverse functions “undo” each other. However, just as zero does not
have a reciprocal, some functions do not have inverses. Given a function f (x), we can verify whether some other function g (x) is the inverse of f (x) by checking whether either g ( f (x)) = x or f (g (x)) = x is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.) For example, y = 4x and y = 1 __ x are inverse functions. 4 and (f −1 ∘ f )(x) = f −1 (4x) = 1 __ (4x) = x 4 1 1 __ __ (f ∘ f −1)(x few coordinate pairs from the graph of the function y = 4x are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs 1 __ x are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each from the graph of the function y = 4 coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function. inverse function For any one-to-one function f (x) = y, a function f −1 (x) is an inverse function of f if f −1(y) = x. This can also be written as f −1( f (x)) = x for all x in the domain of f. It also follows that f ( f −1(x)) = x for all x in the domain of f −1 if f −1 is the inverse of f. The notation f −1 is read “ f inverse.” Like any other function, we can use any variable name as the input for f −1, so we will often write f −1(x), which we read as “ f inverse of x.” Keep in mind that and not all functions have inverses. f −1(x) ≠ 1 ___ f (x) 102 CHAPTER 1 Functions Example 1 Identifying an Inverse Function for a Given Input-Output Pair If for a particular one-to-one function f (2) = 4 and f (5) = 12, what are the corresponding input and output values for the inverse function? Solution The inverse function reverses the input and output quantities, so if f (2) = 4
, then f −1 (4) = 2; f (5) = 12, then f −1 (12) = 5. Alternatively, if we want to name the inverse function g, then g (4) = 2 and g (12) = 5. Analysis Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 1. (x, f (x)) (2, 4) (5, 12) (x, g (x)) (4, 2) (12, 5) Table 1 Try It #1 Given that h−1(6) = 2, what are the corresponding input and output values of the original function h? How To… Given two functions f (x) and g (x), test whether the functions are inverses of each other. 1. Determine whether f (g(x)) = x or g(f (x)) = x. 2. If both statements are true, then g = f −1 and f = g −1. If either statement is false, then both are false, and g ≠ f −1 and f ≠ g−1. Example 2 Testing Inverse Relationships Algebraically If f (x) = 1 ____ x + 2 and g (x) = 1 __ − 2, is g = f −1? x Solution so 1 _ g (f (x)) = − 1 and f = g −1 This is enough to answer yes to the question, but we can also verify the other formula. f (g (x)) = 1 _ 1 __ − 2 + 2 x 1 _ 1 __ x = x = Analysis Notice the inverse operations are in reverse order of the operations from the original function. Try It #2 If f (x) = x3 − 4 and g (x) = 3 √ — x − 4, is g = f −1? SECTION 1.7 inverse Functions 103 Example 3 Determining Inverse Relationships for Power Functions If f (x) = x3 (the cube function) and g (x) = 1 __ x, is g = f −1? 3 x3 __ Solution 27 f (g (x)) = ≠ x No, the functions are not inverses. Analysis The correct inverse to the cube is, of course, the cube root not a multiplier. 3 √ — x = x1/3 that is, the one-third is an
exponent, Try It #3 If f (x) = (x − 1)3 and g (x) = 3 √ — x + 1, is g = f −1? Finding Domain and Range of Inverse Functions The outputs of the function f are the inputs to f −1, so the range of f is also the domain of f −1. Likewise, because the inputs to f are the outputs of f −1, the domain of f is the range of f −1. We can visualize the situation as in Figure 3. Domain of f Range of f f (x) a b Range of f –1 f –1(x) Domain of f –1 Figure 3 Domain and range of a function and its inverse When a function has no inverse function, it is possible to create a new function where that new function on a limited x is f −1(x) = x2, because a square “undoes” domain does have an inverse function. For example, the inverse of f (x) = √ a square root; but the square is only the inverse of the square root on the domain [0, ∞), since that is the range of f (x) = √ x. — — We can look at this problem from the other side, starting with the square (toolkit quadratic) function f (x) = x2. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function. In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function f (x) = x2
with its domain limited to [0, ∞), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function). If f (x) = (x − 1)2 on [1, ∞), then the inverse function is f −1(x) = √ x + 1. — • The domain of f = range of f −1 = [1, ∞). • The domain of f −1 = range of f = [0, ∞). Q & A… Is it possible for a function to have more than one inverse? No. If two supposedly different functions, say, g and h, both meet the definition of being inverses of another function f, then you can prove that g = h. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse. 104 CHAPTER 1 Functions domain and range of inverse functions The range of a function f (x) is the domain of the inverse function f −1(x). The domain of f (x) is the range of f −1(x). How To… Given a function, find the domain and range of its inverse. 1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. 2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. Example 4 Finding the Inverses of Toolkit Functions Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 2. We restrict the domain in such a fashion that the function assumes all y-values exactly once. Constant f (x) = c Identity f (x) = x Quadratic f (x) = x 2 Cubic f (x) = x 3 Reciprocal f (x) = 1 _ x Reciprocal squared Cube root Square root Absolute value f (x) =
1 _ x2 f (x) = 3 √ — x f (x) = √ — x f (x) = ∣x∣ Table 2 Solution The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse. The absolute value function can be restricted to the domain [0, ∞), where it is equal to the identity function. The reciprocal-squared function can be restricted to the domain (0, ∞). Analysis We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse. f (x) f (x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 (a) 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 (b) 321 4 5 x Figure 4 (a ) Absolute value (b ) Reciprocal squared Try It #4 The domain of function f is (1, ∞) and the range of function f is (−∞, −2). Find the domain and range of the inverse function. SECTION 1.7 inverse Functions 105 Finding and evaluating Inverse Functions Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Inverting Tabular Functions Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range. Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Example 5 A function f (t) is given in Table 3, showing distance in miles that a car has traveled in t minutes. Find and interpret f −1(70). Interpreting the Inverse of a Tabular Function t
(minutes) f (t) (miles) 30 20 50 40 70 60 90 70 Table 3 Solution The inverse function takes an output of f and returns an input for f. So in the expression f −1(70), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f, 90 minutes, so f −1(70) = 90. The interpretation of this is that, to drive 70 miles, it took 90 minutes. Alternatively, recall that the definition of the inverse was that if f (a) = b, then f −1(b) = a. By this definition, if we are given f −1(70) = a, then we are looking for a value a so that f (a) = 70. In this case, we are looking for a t so that f (t) = 70, which is when t = 90. Try It #5 Using Table 4, find and interpret a. f (60), and b. f −1(60). t (minutes) f (t) (miles) 30 20 60 50 70 60 90 70 50 40 Table 4 Evaluating the Inverse of a Function, Given a Graph of the Original Function We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. How To… Given the graph of a function, evaluate its inverse at specific points. 1. Find the desired input on the y-axis of the given graph. 2. Read the inverse function’s output from the x-axis of the given graph. 106 CHAPTER 1 Functions Example 6 Evaluating a Function and Its Inverse from a Graph at Specific Points A function g(x) is given in Figure 5. Find g(3) and g−1(3). g(x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure
5 Solution To evaluate g(3), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point (3, 1) tells us that g(3) = 1. To evaluate g −1(3), recall that by definition g −1(3) means the value of x for which g(x) = 3. By looking for the output value 3 on the vertical axis, we find the point (5, 3) on the graph, which means g(5) = 3, so by definition, g −1(3) = 5. See Figure 6. g(x) 6 5 4 3 2 1 – –1 1– (5, 3) (3, 1) 1 2 3 4 5 6 7 x Figure 6 Try It #6 Using the graph in Figure 6, a. find g−1(1), and b. estimate g−1(4). Finding Inverses of Functions Represented by Formulas Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula— for example, y as a function of x— we can often find the inverse function by solving to obtain x as a function of y. How To… Given a function represented by a formula, find the inverse. 1. Make sure f is a one-to-one function. 2. Solve for x. 3. Interchange x and y. Example 7 Inverting the Fahrenheit-to-Celsius Function Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. 5 __ C = (F − 32) 9 SECTION 1.7 inverse Functions 107 Solution 5 __ (F − 32) C = 9 9 __ = F − 32 C ⋅ 5 9 __ C + 32 F = 5 By solving in general, we have uncovered the inverse function. If then 5 __ C = h(F) = (F − 32), 9 9 __ F = h−1(C) = C + 32. 5 In this case, we introduced a function h to represent the conversion because the input and output variables are descriptive, and writing C−1 could get confusing. Try It #7 1 __ (x − 5) Solve for x in terms of y given y = 3 Example 8 Solving to Find an Inverse Function Find the inverse of the function f (x) = 2 ____ x − 3 +
4. Solution y = + 4 Set up an equation. 2 ____ x − 3 2 _____ x − 3 2 ____ y − 4 2 ____ = Subtract 4 from both sides. Multiply both sides by x − 3 and divide by y − 4. x = + 3 Add 3 to both sides. So f −1 (y) = + 3 or f −1 (x) = 2 _____ y − 4 2 ____ x − 4 + 3. Analysis The domain and range of f exclude the values 3 and 4, respectively. f and f −1 are equal at two points but are not the same function, as we can see by creating Table 5. x f (x) 1 3 2 2 Table 5 5 5 f −1(y) y Example 9 Solving to Find an Inverse with Radicals Find the inverse of the function f (x) = 2 + √ — x − 4. Solution y − 2)2 = x − 4 x = (y − 2)2 + 4 So f −1 (x) = (x − 2)2 + 4. The domain of f is [4, ∞). Notice that the range of f is [2, ∞), so this means that the domain of the inverse function f −1 is also [2, ∞). Analysis The formula we found for f −1(x) looks like it would be valid for all real x. However, f −1 itself must have an inverse (namely, f) so we have to restrict the domain of f −1 to [2, ∞) in order to make f −1 a one-to-one function. This domain of f −1 is exactly the range of f. 108 CHAPTER 1 Functions Try It #8 What is the inverse of the function f (x) = 2 − √ — x? State the domains of both the function and the inverse function. Finding Inverse Functions and Their Graphs Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function f (x) = x2 restricted to the domain [0, ∞), on which this function is one-to-one, and graph it as in Figure 7. f (x) 5 4 3 2 1 –1–1 –2 –5 –4 –3 –2 21 3 4 5 x Figure 7 Quadr
atic function with domain restricted to [0, ∞). Restricting the domain to [0, ∞) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain. We already know that the inverse of the toolkit quadratic function is the square root function, that is, f −1(x) = √ What happens if we graph both f and f −1 on the same set of axes, using the x-axis for the input to both f and f −1? We notice a distinct relationship: The graph of f −1(x) is the graph of f (x) reflected about the diagonal line y = x, which we will call the identity line, shown in Figure 8. — x. f (x) y = x f –1(x) 21 1–1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 8 Square and square-root functions on the non-negative domain This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes. SECTION 1.7 inverse Functions 109 Example 10 Given the graph of f (x) in Figure 9, sketch a graph of f −1(x). Finding the Inverse of a Function Using Reflection about the Identity Line y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 9 Solution This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0, ∞) and range of (−∞, ∞), so the inverse will have a domain of (−∞, ∞) and range of (0, ∞). If we reflect this graph over the line y = x, the point (1, 0) reflects to (0, 1) and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph gives Figure 10. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 f –1(x) y = x f (x) 21 3 4 5 x Figure 10 The
function and its inverse, showing reflection about the identity line Try It #9 Draw graphs of the functions f and f −1 from Example 8. Q & A… Is there any function that is equal to its own inverse? Yes. If f = f −1, then f (f (x)) = x, and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because Any function f (x) = c − x, where c is a constant, is also equal to its own inverse. 1 _ = x 1 _ x Access these online resources for additional instruction and practice with inverse functions. • • Inverse Functions (http://openstaxcollege.org/l/inversefunction) Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph) • Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/restrictdomain) 110 CHAPTER 1 Functions 1.7 SeCTIOn exeRCISeS VeRBAl 1. Describe why the horizontal line test is an effective way to determine whether a function is one-to-one? 2. Why do we restrict the domain of the function f (x) = x2 to find the function’s inverse? 3. Can a function be its own inverse? Explain. 4. Are one-to-one functions either always increasing or always decreasing? Why or why not? 5. How do you find the inverse of a function algebraically? AlGeBRAIC 6. Show that the function f (x) = a − x is its own inverse for all real numbers a. For the following exercises, find f −1(x) for each function. 7. f (x) = x + 3 10. f (x) = 3 − x 8. f (x) = x + 5 11. f (x) = x _____ x + 2 9. f (x) = 2 − x 12. f (x) = 2x + 3 ______ 5x + 4 For the following exercises, find a domain on which each function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. 13. f (x) = (x + 7)2 14. f (x) = (x − 6)2 15. f
(x) = x2 − 5 16. Given f (x) = and g(x) = x __ 2 + x 2x _____ : 1 − x a. Find f (g(x)) and g (f (x)). b. What does the answer tell us about the relationship between f (x) and g(x)? For the following exercises, use function composition to verify that f (x) and g(x) are inverse functions. 17. f (x) = 3 √ — x − 1 and g (x) = x3 + 1 18. f (x) = −3x + 5 and g (x) = x − 5 _____ −3 GRAPHICAl For the following exercises, use a graphing utility to determine whether each function is one-to-one. — 19. f (x) = √ 21. f (x) = −5x + 1 x 3 √ — 3x + 1 20. f (x) = 22. f (x) = x3 − 27 For the following exercises, determine whether the graph represents a one-to-one function. 23. y 25 20 15 10 5 –10 –5 –5 –10 –15 –20 –25 –25 –20 –15 f 5 10 15 20 25 x 24. y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 f 42 6 8 10 x SECTION 1.7 section exercises 111 For the following exercises, use the graph of f shown in Figure 11. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 f 21 3 4 5 x Figure 11 25. Find f (0). 26. Solve f (x) = 0. 27. Find f −1(0). 28. Solve f −1(x) = 0. For the following exercises, use the graph of the one-to-one function shown in Figure 12. y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 f 42 6 8 10 x 29. Sketch the graph of f −1. 30. Find f (6) and f −1(2). 31. If the complete graph of f is shown, find the domain of f. 32. If the complete graph of f is shown, find the range
Figure 12 of f. nUMeRIC For the following exercises, evaluate or solve, assuming that the function f is one-to-one. 33. If f (6) = 7, find f −1(7). 34. If f (3) = 2, find f −1(2). 35. If f −1(−4) = −8, find f (−8). 36. If f −1(−2) = −1, find f (−1). For the following exercises, use the values listed in Table 6 to evaluate or solve. x f (x Table 37. Find f (1). 39. Find f −1(0). 38. Solve f (x) = 3. 40. Solve f −1(x) = 7. 41. Use the tabular representation of f in Table 7 to create a table for f −1(x). x f (x) 3 1 6 4 9 7 13 12 14 16 Table 7 112 CHAPTER 1 Functions TeCHnOlOGY For the following exercises, find the inverse function. Then, graph the function and its inverse. 42. f (x) = 3 _____ x − 2 43. f (x) = x 3 − 1 44. Find the inverse function of f (x) = and range in interval notation. 1 _ x − 1. Use a graphing utility to find its domain and range. Write the domain ReAl-WORlD APPlICATIOnS 45. To convert from x degrees Celsius to y degrees 9 __ x + 32. Fahrenheit, we use the formula f (x) = 5 Find the inverse function, if it exists, and explain its meaning. 46. The circumference C of a circle is a function of its radius given by C(r) = 2πr. Express the radius of a circle as a function of its circumference. Call this function r(C). Find r(36π) and interpret its meaning. 47. A car travels at a constant speed of 50 miles per hour. The distance the car travels in miles is a function of time, t, in hours given by d(t) = 50t. Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function t(d). Find t(180) and interpret its meaning. CHAPTER 1 review 113 CHAPTeR 1 ReVIeW Key Terms absolute maximum the greatest value of a function over an
interval absolute minimum the lowest value of a function over an interval absolute value equation an equation of the form ∣A∣ = B, with B ≥ 0; it will have solutions when A = B or A = −B absolute value inequality a relationship in the form ∣A∣ < B, ∣A∣ ≤ B, ∣A∣ > B, or ∣A∣ ≥ B average rate of change the difference in the output values of a function found for two values of the input divided by the difference between the inputs composite function the new function formed by function composition, when the output of one function is used as the input of another decreasing function a function is decreasing in some open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a dependent variable an output variable domain the set of all possible input values for a relation even function a function whose graph is unchanged by horizontal reflection, f (x) = f (−x), and is symmetric about the y-axis function a relation in which each input value yields a unique output value horizontal compression a transformation that compresses a function’s graph horizontally, by multiplying the input by a constant b > 1 horizontal line test a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by −1 horizontal shift a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the input horizontal stretch a transformation that stretches a function’s graph horizontally by multiplying the input by a constant 0 < b < 1 increasing function a function is increasing in some open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a independent variable an input variable input each object or value in a domain that relates to another object or value by a relationship known as a function interval notation a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion inverse function for any one-to-one function f (x), the inverse is a function f −1(x) such that f −1(f (x)) = x for all x in
the domain of f; this also implies that f ( f −1(x)) = x for all x in the domain of f −1 local extrema collectively, all of a function’s local maxima and minima local maximum a value of the input where a function changes from increasing to decreasing as the input value increases. local minimum a value of the input where a function changes from decreasing to increasing as the input value increases. odd function a function whose graph is unchanged by combined horizontal and vertical reflection, f (x) = − f (−x), and is symmetric about the origin one-to-one function a function for which each value of the output is associated with a unique input value output each object or value in the range that is produced when an input value is entered into a function piecewise function a function in which more than one formula is used to define the output range the set of output values that result from the input values in a relation 114 CHAPTER 1 Functions rate of change the change of an output quantity relative to the change of the input quantity relation a set of ordered pairs set-builder notation a method of describing a set by a rule that all of its members obey; it takes the form {x ∣ statement about x} vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant 0 < a < 1 vertical line test a method of testing whether a graph represents a function by determining whether a vertical line intersects the graph no more than once vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by −1 vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output vertical stretch a transformation that stretches a function’s graph vertically by multiplying the output by a constant a > 1 Key equations Constant function f (x) = c, where c is a constant Identity function f (x) = x Absolute value function f (x) = ∣x∣ Reciprocal squared function f (x) = Quadratic function Cubic function Reciprocal function Square root function Cube root function Average rate of change f (x) = x2 f (x) = x3 1 _ f (x) = x 1 _ x2 x f (x) = √ — f (x) = ∆y _ ∆x = — x 3 √ f (x2) − f
(x1) _ x2 − x1 Composite function (f ∘ g)(x) = f (g(x)) Vertical shift g(x) = f (x) + k (up for k > 0) Horizontal shift g(x) = f (x − h) (right for h > 0) Vertical reflection g(x) = −f (x) Horizontal reflection g(x) = f (−x) Vertical stretch g(x) = af (x) (a > 0) Vertical compression g(x) = af (x) (0 < a < 1) Horizontal stretch g(x) = f (bx) (0 < b < 1) Horizontal compression g(x) = f (bx) (b > 1) CHAPTER 1 review 115 Key Concepts 1.1 Functions and Function Notation • A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. See Example 1 and Example 2. • Function notation is a shorthand method for relating the input to the output in the form y = f (x). See Example 3 and Example 4. • In tabular form, a function can be represented by rows or columns that relate to input and output values. See Example 5. • To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function can be evaluated by replacing the input variable with a given value. See Example 6 and Example 7. • To solve for a specific function value, we determine the input values that yield the specific output value. See Example 8. • An algebraic form of a function can be written from an equation. See Example 9 and Example 10. • Input and output values of a function can be identified from a table. See Example 11. • Relating input values to output values on a graph is another way to evaluate a function. See Example 12. • A function is one-to-one if each output value corresponds to only one input value. See Example 13. • A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point. See Example 14. • The graph of a one-to-one function passes the horizontal line test. See Example 15. 1.2 Domain and Range • The domain of a function includes all real input values that would not cause us to attempt
an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number. • The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example 1. • The domain of a function can also be determined by identifying the input values of a function written as an equation. See Example 2, Example 3, and Example 4. • Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See Example 5. • For many functions, the domain and range can be determined from a graph. See Example 6 and Example 7. • An understanding of toolkit functions can be used to find the domain and range of related functions. See Example 8, Example 9, and Example 10. • A piecewise function is described by more than one formula. See Example 11 and Example 12. • A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 13. 1.3 Rates of Change and Behavior of Graphs • A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See Example 1. • Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 2. • Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example 3. • An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example 4 and Example 5. • The average rate of change can sometimes be determined as an expression. See Example 6. • A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example 7. • A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values. 116 CHAPTER 1 Functions • A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values. • Minima and maxima are also called extrema. • We can find local extrema from a graph. See Example 8 and Example 9. • The highest and lowest points on a graph indicate the max
ima and minima. See Example 10. 1.4 Composition of Functions • We can perform algebraic operations on functions. See Example 1. • When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function. • The function produced by combining two functions is a composite function. See Example 2 and Example 3. • The order of function composition must be considered when interpreting the meaning of composite functions. See Example 4. • A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function. • A composite function can be evaluated from a table. See Example 5. • A composite function can be evaluated from a graph. See Example 6. • A composite function can be evaluated from a formula. See Example 7. • The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example 8 and Example 9. • Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions. • Functions can often be decomposed in more than one way. See Example 10. 1.5 Transformation of Functions • A function can be shifted vertically by adding a constant to the output. See Example 1 and Example 2. • A function can be shifted horizontally by adding a constant to the input. See Example 3, Example 4, and Example 5. • Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See Example 6. • Vertical and horizontal shifts are often combined. See Example 7 and Example 8. • A vertical reflection reflects a graph about the x-axis. A graph can be reflected vertically by multiplying the output by –1. • A horizontal reflection reflects a graph about the y-axis. A graph can be reflected horizontally by multiplying the input by –1. • A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not affect the final graph. See Example 9. • A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. See Example 10. • A function presented as an equation can be reflected by applying transformations one at a time. See Example 11. • Even functions are symmetric about the y-axis, whereas odd functions are symmetric about
the origin. • Even functions satisfy the condition f (x) = f (−x). • Odd functions satisfy the condition f (x) = −f (−x). • A function can be odd, even, or neither. See Example 12. • A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 13, Example 14, and Example 15. • A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 16, Example 17, and Example 18. CHAPTER 1 review 117 • The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. See Example 19 and Example 20. 1.6 Absolute Value Functions • The absolute value function is commonly used to measure distances between points. See Example 1. • Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See Example 2. • The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See Example 3. • In an absolute value equation, an unknown variable is the input of an absolute value function. • If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See Example 4. • An absolute value equation may have one solution, two solutions, or no solutions. See Example 5. • An absolute value inequality is similar to an absolute value equation but takes the form ∣A∣ < B, ∣A∣ ≤ B, ∣A∣ > B, or ∣A∣ ≥ B. It can be solved by determining the boundaries of the solution set and then testing which segments are in the set. See Example 6. • Absolute value inequalities can also be solved graphically. See Example 7. 1.7 Inverse Functions • If g(x) is the inverse of f (x), then g(f (x)) = f (g(x)) = x. See Example 1, Example 2, and Example 3. • Each of the toolkit functions has an inverse. See Example 4. • For a function to have an inverse, it must be one-to-one (pass the horizontal line test). • A function that is not one-to-one over its entire domain may be one-to-one on part of
its domain. • For a tabular function, exchange the input and output rows to obtain the inverse. See Example 5. • The inverse of a function can be determined at specific points on its graph. See Example 6. • To find the inverse of a formula, solve the equation y = f (x) for x as a function of y. Then exchange the labels x and y. See Example 7, Example 8, and Example 9. • The graph of an inverse function is the reflection of the graph of the original function across the line y = x. See Example 10. 118 CHAPTER 1 Functions CHAPTeR 1 ReVIeW exeRCISeS FUnCTIOnS AnD FUnCTIOn nOTATIOn For the following exercises, determine whether the relation is a function. 1. {(a, b), (c, d), (e, d)} 2. {(5, 2), (6, 1), (6, 2), (4, 8)} 3. y 2 + 4 = x, for x the independent variable and y the dependent variable 4. Is the graph in Figure 1 a function? –25 –20 –15 –10 Figure 1 y 25 20 15 10 5 –5 –5 –10 –15 –20 –25 f 5 1 0 15 20 25 x For the following exercises, evaluate the function at the indicated values: f (−3); f (2); f (−a); −f (a); f (a + h). 5. f (x) = −2x 2 + 3x 6. f ( x) = 2∣3 x − 1∣ For the following exercises, determine whether the functions are one-to-one. 7. f (x) = −3x + 5 8. f (x) = ∣x − 3∣ For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function. 9. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 10. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 11. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises,
graph the functions. 12. f (x) = ∣x + 1∣ 13. f (x) = x 2 − 2 CHAPTER 1 review 119 For the following exercises, use Figure 2 to approximate the values. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 14. f (2) 15. f (−2) 21 3 4 5 x 16. If f (x) = −2, then solve for x. 17. If f (x) = 1, then solve for x. Figure 2 For the following exercises, use the function h(t) = −16t 2 + 80t to find the values. 18. h(2) − h(1) __ 2 − 1 19. h(a) − h(1) __ a − 1 DOMAIn AnD RAnGe For the following exercises, find the domain of each function, expressing answers using interval notation. 20. f (x) = 2 _ 3x + 2 21. f (x) = x − 3 ___________ x 2 − 4x − 12 23. Graph this piecewise function: f (x) = { x + 1 −2x − 3 x < −2 x ≥ −2 — x − 6 22. f (x) = √ _ x − 4 √ — RATeS OF CHAnGe AnD BeHAVIOR OF GRAPHS For the following exercises, find the average rate of change of the functions from x = 1 to x = 2. 24. f (x) = 4x − 3 25. f (x) = 10x 2 + x 26. f (x) = − 2 _ x 2 For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant. 27. y 28. –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 29. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 30. Find the local minimum of the function graphed in Exercise 27. 31. Find the local extrema for the
function graphed in Exercise 28. 120 CHAPTER 1 Functions 32. For the graph in Figure 3, the domain of the function is [−3, 3]. The range is [−10, 10]. Find the absolute minimum of the function on this interval. 33. Find the absolute maximum of the function graphed in Figure 3. –5 –4 –3 –2 y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x Figure 3 COMPOSITIOn OF FUnCTIOnS For the following exercises, find (f ∘ g)(x) and (g ∘ f)(x) for each pair of functions. 34. f (x) = 4 − x, g(x) = −4x 35. f (x) = 3x + 2, g(x) = 5 − 6x 36. f (x) = x 2 + 2x, g(x) = 5x + 1 37. f (x(x) = _ x 38. f (x) =, g(x _____ 2 For the following exercises, find (f ∘ g) and the domain for (f ∘ g)(x) for each pair of functions. 39. f (x(x) = 42. f (x) = 1 _ x2 − 1, g(x) = √ — x + 1 40. f (x) =, g(x(x) = √ 41. f (x) = — x For the following exercises, express each function H as a composition of two functions f and g where H(x) = (f ∘ g)(x). 43. H(x) = √ _______ 2x − 1 ______ 3x + 4 44. H(x) = 1 _ (3x 2 − 4)−3 TRAnSFORMATIOn OF FUnCTIOnS For the following exercises, sketch a graph of the given function. 45. f (x) = (x − 3)2 46. f (x) = (x + 4)3 48. f (x) = −x 3 49. f (x) = 3 √ — −x 51. f (x) = 4[∣x − 2∣ − 6] 52. f (x) = −(x + 2)2 − 1 47. f (x) = √ — x
+ 5 50. f (x) = 5 √ — −x − 4 For the following exercises, sketch the graph of the function g if the graph of the function f is shown in Figure 4. y 5 4 3 2 1 –1 –1 –2 –5 –4 –3 –2 21 3 4 5 x Figure 4 53. g(x) = f (x − 1) 54. g(x) = 3f (x) CHAPTER 1 review 121 For the following exercises, write the equation for the standard function represented by each of the graphs below. 55. y 56. y –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, determine whether each function below is even, odd, or neither. 57. f (x) = 3x 4 58. g(x) = √ — x 1 _ x + 3x 59. h(x) = For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither. 60. y 61. y 62. y –25 –20 –15 –10 25 20 15 10 5 –5 –5 –10 –15 –20 –25 5 10 15 20 25 x –10 –8 –6 –4 25 20 15 10 5 –2 –5 –10 –15 –20 –25 42 6 8 10 x –10 –8 –6 –4 ABSOlUTe VAlUe FUnCTIOnS For the following exercises, write an equation for the transformation of f (x) = \| x \|. 63. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 64. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 65. 21 3 4 5 x –5 –4 –3 –2 10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 42 6 8 10 x 21 3 4 5 x For the following exercises, graph the absolute value function. 66. f (x)
= \| x − 5 \| 67. f (x) = −\| x − 3 \| 68. f (x) = \| 2x − 4 \| 122 CHAPTER 1 Functions For the following exercises, solve the absolute value equation. 69. \| x + 4 \| = 18 x + 5  =  3 70.  1 x − 2  __ __ 4 3 For the following exercises, solve the inequality and express the solution using interval notation. 71. \| 3x − 72.  __ 3 InVeRSe FUnCTIOnS For the following exercises, find f −1(x) for each function. 73. f (x) = 9 + 10x 74. f (x) = x _ x + 2 For the following exercise, find a domain on which the function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. 75. f (x) = x 2 + 1 76. Given f (x) = x 3 − 5 and g(x) = a. Find f (g(x)) and g(f (x)). b. What does the answer tell us about the relationship between f (x) and g(x)? x + 5 : — 3 √ For the following exercises, use a graphing utility to determine whether each function is one-to-one. 1 _ 77. f (x) = x 80. If f (1) = 4, find f −1(4). 78. f (x) = −3x 2 + x 79. If f (5) = 2, find f −1(2). CHAPTER 1 practice test 123 CHAPTeR 1 PRACTICe TeST For the following exercises, determine whether each of the following relations is a function. 1. y = 2x + 8 2. {(2, 1), (3, 2), (−1, 1), (0, −2)} For the following exercises, evaluate the function f (x) = −3x 2 + 2x at the given input. 3. f (−2) 4. f (a) 5. Show that the function f (x) = −2(x − 1)2 + 3 is not one-to-one. 6. Write the domain of the function f (x) = √ — 3 − x in interval
notation. 7. Given f (x) = 2x 2 − 5x, find f (a + 1) − f (1). 8. Graph the function f (x) = { x + 1 if −2 < x < 3 x ≥ 3 −x if 9. Find the average rate of change of the function f (x) = 3 − 2x 2 + x by finding f (b) − f (a) _ b − a. For the following exercises, use the functions f (x) = 3 − 2x 2 + x and g(x) = √ — x to find the composite functions. 10. (g ∘ f )(x) 11. (g ∘ f )(1) 12. Express H(x) = 3 √ — 5x 2 − 3x as a composition of two functions, f and g, where (f ∘ g)(x) = H(x). For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function. 13. f (x) = √ — x + 6 − 1 14. f (x) = 1 _ x + 2 − 1 For the following exercises, determine whether the functions are even, odd, or neither. 5 _ 15. f (x) = − x2 + 9x 6 1 _ 17. f (x) = x 19. Solve \| 2x − 3 \| = 17. 5 _ 16. f (x) = − x 3 + 9x 5 18. Graph the absolute value function f (x) = −2\| x − 1 \| + 3. 1 x − 3  ≥ 17. Express the solution in 20. Solve −  _ 3 interval notation. For the following exercises, find the inverse of the function. 21. f (x) = 3x − 5 22. f (x) = 4 _ x + 7 124 CHAPTER 1 Functions For the following exercises, use the graph of g shown in Figure 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 23. On what intervals is the function increasing? 24. On what intervals is the function decreasing? 21 3 4 5 x 25. Approximate the local minimum of the function. Express the answer as an ordered pair. 26. Approximate the local maximum of the function. Express the answer as
an ordered pair. Figure 1 For the following exercises, use the graph of the piecewise function shown in Figure 2. 27. Find f (2). 28. Find f (−2). 29. Write an equation for the piecewise function. y 5 4 3 2 1 f 21 3 4 5 x –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 Figure 2 For the following exercises, use the values listed in Table 1. x F (x 11 6 13 7 15 8 17 Table 1 30. Find F (6). 31. Solve the equation F (x) = 5. 32. Is the graph increasing or decreasing on its domain? 33. Is the function represented by the graph one-to-one? 34. Find F −1(15). 35. Given f (x) = −2x + 11, find f −1(x). Linear Functions 2 Figure 1 A bamboo forest in China (credit: “JFXie”/Flickr) CHAPTeR OUTlIne 2.1 linear Functions 2.2 Graphs of linear Functions 2.3 Modeling with linear Functions 2.4 Fitting linear Models to Data Introduction Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour.[6] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function. Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data. 6 http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/ 125 126 CHAPTER 2 linear Functions leARnInG OBjeCTIVeS In this section, you will: • • Represent a linear function. Determine whether a linear function is increasing, decreasing,
or constant. • • • Calculate and interpret slope. Write the point-slope form of an equation. Write and interpret a linear function. 2.1 lIneAR FUnCTIOnS Figure 1 Shanghai Maglev Train (credit: “kanegen”/Flickr) Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure 1). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[7] Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time. Representing linear Functions The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method. Representing a Linear Function in Word Form Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship. • The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed. The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station. 7 http://www.chinahighlights.com/shanghai/transportation
/maglev-train.htm SECTION 2.1 linear Functions 127 Representing a Linear Function in Function Notation Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slope-intercept form of a line, where x is the input value, m is the rate of change, and b is the initial value of the dependent variable. Equation form y = mx + b Equation notation f (x) = mx + b In the example of the train, we might use the notation D(t) in which the total distance D is a function of the time t. The rate, m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train. Representing a Linear Function in Tabular Form D(t) = 83t + 250 A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time. 1 second 1 second 1 second t D(t) 0 250 1 333 2 416 3 499 83 meters 83 meters 83 meters Figure 2 Tabular representation of the function D showing selected input and output values Q & A… Can the input in the previous example be any real number? No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers. Representing a Linear Function in Graphical Form Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D(t) = 83t + 250, to draw a graph, represented in Figure 3. Notice the graph is a line. When we plot a linear function, the graph is always a line. The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph in Figure 3 that the y-intercept in the train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a constant speed 500 400 300
200 100 0 1 2 3 4 Time (s) 5 Figure 3 The graph of D(t) = 83t + 250. Graphs of linear functions are lines because the rate of change is constant. Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f (x) = 2x + 1. Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product. 128 CHAPTER 2 linear Functions linear function A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line f (x) = mx + b where b is the initial or starting value of the function (when input, x = 0), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b). Example 1 Using a Linear Function to Find the Pressure on a Diver The pressure, P, in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d, in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate this function in words. Figure 4 (credit: Ilse Reijs and jan-noud Hutten) Solution To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths. Analysis The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases. Determining Whether a linear Function Is Increasing, Decreasing, or Constant The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a
positive slope slants upward from left to right as in Figure 5(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5(c). Increasing function Decreasing function Constant function f (x) f (x) f (x) f f f x x x (a) (b) Figure 5 (c) SECTION 2.1 linear Functions 129 increasing and decreasing functions The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function. • f (x) = mx + b is an increasing function if m > 0. • f (x) = mx + b is an decreasing function if m < 0. • f (x) = mx + b is a constant function if m = 0. Example 2 Deciding Whether a Function Is Increasing, Decreasing, or Constant Some recent studies suggest that a teenager sends an average of 60 texts per day.[8] For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant. a. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent. b. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month. c. A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month. Solution Analyze each function. a. The function can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day. b. The function can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the slope is negative so the function is decreasing
. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x days. c. The cost function can be represented as f (x) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant. Calculating and Interpreting Slope In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Given two values for the input, x1 and x2, and two corresponding values for the output, y1 and y2 —which can be represented by a set of points, (x1, y1) and (x2, y2) —we can calculate the slope m, as follows m = change in output (rise) __ = change in input (run) Δy _ Δx = y2 − y1 _ x2 − x1 where ∆y is the vertical displacement and ∆x is the horizontal displacement. Note in function notation two corresponding values for the output y1 and y2 for the function f, y1 = f (x1) and y2 = f (x2), so we could equivalently write Figure 6 indicates how the slope of the line between the points, (x1, y1) and (x2, y2), is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is. m = f (x2) − f (x1) _ x2 − x1 8 http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/ 130 CHAPTER 2 linear Functions x2 – x1 y2 – y1 y 10 x2, y2) (x1, y1) m = ∆y ∆x = y2 – y1 x2 – x1 0 –1 321 4 x Figure 6 The slope of a function is calculated by the change in y divided by the change in x. It does not matter which coordinate is used as the (x2, y2) and which is the (x1, y1), as long as each calculation is started with the elements from the same coordinate pair. Q &
A… Are the units for slope always units for the output __? units for the input Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input. calculate slope The slope, or rate of change, of a function m can be calculated according to the following: m = change in output (rise) __ = change in input (run) Δy _ Δx = y2 − y1 _ x2 − x1 where x1 and x2 are input values, y1 and y2 are output values. How To… Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. Example 3 Finding the Slope of a Linear Function If f (x) is a linear function, and (3, −2) and (8, 1) are points on the line, find the slope. Is this function increasing or decreasing? Solution The coordinate pairs are (3, −2) and (8, 1). To find the rate of change, we divide the change in output by the change in input. m = change in output __ = change in input 1 − (−2) ________ 8 − 3 3 __ = 5 We could also write the slope as m = 0.6. The function is increasing because m > 0. Analysis As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used. Try It #1 If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function increasing or decreasing? SECTION 2.1 linear Functions 131 Example 4 Finding the Population Change from a Linear Function The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012. Solution The rate of change relates the change in population to the change in time. The population increased by
27,800 − 23,400 = 4,400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years. 4,400 people __ = 4 years 1,100 people __ year So the population increased by 1,100 people per year. Analysis Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable. Try It #2 The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012. Writing the Point-Slope Form of a linear equation Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form. The point-slope form is derived from the slope formula. y − y1 = m(x − x1) m = y − y1 _ x − x1 y − y1 _ x − x1 m(x − x1) = y − y1 m(x − x1) = Assuming x ≠ x1 Simplify. (x − x1) Multiply both sides by (x − x1). y − y1 = m(x − x1) Rearrange. Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope 1 __ (x − 6). form, y − 4 = − 2 We can convert it to the slope-intercept form as shown. 1 __ (x − 6) y − 4 = − 2 1 __ __ x + 7 y = − 2 1 __ Distribute the −. 2 Add 4 to each side. 1 __ Therefore, the same line can be described in slope-intercept form as y = − x + 7. 2 point-slope form of a linear equation The point-slope form of a linear equation takes the form y − y1 = m(x − x1) where m is the slope, x1 and y1 are the x- and y-coordinates of a specific point through which the line passes. 132 CHAPTER 2
linear Functions Writing the Equation of a Line Using a Point and the Slope The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told hat a line has a slope of 2 and passes through the point (4, 1). We know that m = 2 and that x1 = 4 and y1 = 1. We can substitute these values into the general point-slope equation. y − y1 = m(x − x1) y − 1 = 2(x − 4) If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques. y − 1 = 2(x − 4) y − 1 = 2x − 8 y = 2x − 7 Distribute the 2. Add 1 to each side. Both equations, y − 1 = 2(x − 4) and y = 2x − 7, describe the same line. See Figure 7. y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 642 8 10 x Figure 7 Example 5 Writing Linear Equations Using a Point and the Slope Write the point-slope form of an equation of a line with a slope of 3 that passes through the point (6, −1). Then rewrite it in the slope-intercept form. Solution Let’s figure out what we know from the given information. The slope is 3, so m = 3. We also know one point, so we know x1 = 6 and y1 = −1. Now we can substitute these values into the general point-slope equation. y − y1 = m(x − x1) y − (−1) = 3(x − 6) Substitute known values. y + 1 = 3(x − 6) Distribute −1 to find point-slope form. Then we use algebra to find the slope-intercept form. y + 1 = 3(x − 6) y + 1 = 3x − 18 y = 3x − 19 Distribute 3. Simplify to slope-intercept form. Try It #3 Write the point-slope form of an equation of a line with a slope of −2 that passes through the point (−2, 2). Then rewrite it in the slope-intercept form. Writing the Equation of a Line Using Two Points The point-slope form of

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.