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, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reflection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 l... |
integer function, 67 set of, 2 intercept definition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 definition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse... |
��nition, 567 determinant definition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 definition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 6... |
a quadratic function, 188 latus rectum, 507 reflective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Tria... |
rantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function definition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for ... |
8 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common difference, 654 definition of, 654 formula for nth term, 656 sum of first n terms, 666 definition of, 652 geometric common ratio, 654 definiti... |
, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coefficient matrix, 590 consistent, 553 constant matrix, 590 definition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Ga... |
20 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1... |
4) comprise the Cartesian coordinates9 of the point P. In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of ‘the point (2, −4).’ We can think of (2, −4) as instructions on how to 5So named in honor of Ren´e Descartes. 6The labels can vary depending on the context o... |
: A(5, 8), B − 5 F (0, 5), G(−7, 0), H(0, −9), O(0, 0).10 2, 3, C(−5.8, −3), D(4.5, −1), E(5, 0), Solution. To plot these points, we start at the origin and move to the right if the x-coordinate is positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it is negative. If th... |
quadrants. One of the most important concepts in all of Mathematics is symmetry.11 There are many types of symmetry in Mathematics, but three of them can be discussed easily using Cartesian Coordinates. Definition 1.3. Two points (a, b) and (c, d) in the plane are said to be symmetric about the x-axis if a = c and b = ... |
. If we start with our point (−2, 3) and pretend that the x-axis is a mirror, then the reflection of (−2, 3) across the x-axis would lie at (−2, −3). If we pretend that the y-axis is a mirror, the reflection of (−2, 3) across that axis would be (2, 3). If we reflect across the x-axis and then the y-axis, we would go from ... |
d2 (x1 − x0)2 + (y1 − y0)2 = d2 (Do you remember why we can replace the absolute value notation with parentheses?) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get Equation 1.1. The Distance Formula: The distance d between the points P (x0, y... |
root We obtain two answers: (1, 2 + 2 why there are two answers. √ 3) and (1, 2 − 2 √ 3). The reader is encouraged to think about Related to finding the distance between two points is the problem of finding the midpoint of the line segment connecting two points. Given two points, P (x0, y0) and Q (x1, y1), the midpoint ... |
4 Exercises 1. Fill in the chart below: Set of Real Numbers Interval Notation Region on the Real Number Line {x | − 1 ≤ x < 5} {x | − 5 < x ≤ 0} {x | x ≤ 3} {x | x ≥ −3} [0, 3) (−3, 3) (−∞, 9) 7 7 2 5 4 In Exercises 2 - 7, find the indicated intersection or union and simplify if possible. Express your answers in interva... |
/on which the point lies. Find the point symmetric to the given point about the x-axis. Find the point symmetric to the given point about the y-axis. Find the point symmetric to the given point about the origin. 16 Relations and Functions In Exercises 22 - 29, find the distance d between the points and the midpoint M of... |
and Q. 37. Show that the points A, B and C below are the vertices of a right triangle. (a) A(−3, 2), B(−6, 4), and C(1, 8) (b) A(−3, 1), B(4, 0) and C(0, −3) 38. Find a point D(x, y) such that the points A(−3, 1), B(4, 0), C(0, −3) and D are the corners of a square. Justify your answer. 39. Discuss with your classmate... |
�, 0) ∪ [1, 5] = (−∞, 0) ∪ [1, 5] 7. (−∞, 5] ∩ [5, 8) = {5} 8. (−∞, 5) ∪ (5, ∞) 9. (−∞, −1) ∪ (−1, ∞) 10. (−∞, −3) ∪ (−3, 4) ∪ (4, ∞) 11. (−∞, 0) ∪ (0, 2) ∪ (2, ∞) 12. (−∞, −2) ∪ (−2, 2) ∪ (2, ∞) 13. (−∞, −4) ∪ (−4, 0) ∪ (0, 4) ∪ (4, ∞) 18 Relations and Functions 14. (−∞, −1] ∪ [1, ∞) 15. (−∞, ∞) 16. (−∞, −3] ∪ (0, ∞) ... |
y-axis with (3, −7) symmetric about origin with (3, 7) in Quadrant IV symmetric about x-axis with (1.3, 2) symmetric about y-axis with (−1.3, −2) symmetric about origin with (−1.3, 2) (c) The point C(π, √ 10) is (d) The point D(0, 8) is in Quadrant I symmetric about x-axis with (π, − symmetric about y-axis with (−π, s... |
� 3, 0) 34. (−3, −4), 5 miles, (4, −4) 3 29. d = √ 23. d = 4 √ 25. d = 27. d = 3 37 10, M = (1, −4, − x2 + y2, M = x 5 31. (0, 3) √ 33 37. (a) The distance from A to B is |AB| = 52, and the distance from B to C is |BC| =, we are guaranteed by the converse of the Pythagorean Theorem that the triangle is a right triangle... |
4, 2), (−3, 2)} 2. HLS1 = {(x, 3) | − 2 ≤ x ≤ 4} 3. HLS2 = {(x, 3) | − 2 ≤ x < 4} 4. V = {(3, y) | y is a real number} 5. H = {(x, y) | y = −2} 6. R = {(x, y) | 1 < y ≤ 3} 1Carl’s, of course. 1.2 Relations Solution. 21 1. To graph A, we simply plot all of the points which belong to A, as shown below on the left. 2. Don... |
4’. This means that we still get a horizontal line segment which includes (−2, 3) and extends to (4, 3), but we do not include (4, 3) because of the strict inequality x < 4. How do we denote this on our graph? It is a common mistake to make the graph start at (−2, 3) end at (3, 3) as pictured below on the left. The pr... |
1 2 3 4 x −1 −2 −3 −4 The graph of H The graph of V 6. For our last example, we turn to R = {(x, y) | 1 < y ≤ 3}. As in the previous example, x is free to be whatever it likes. The value of y, on the other hand, while not completely free, is permitted to roam between 1 and 3 excluding 1, but including 3. After plottin... |
continue to unite Algebra and Geometry. 1.2.1 Graphs of Equations In this section, we delve more deeply into the connection between Algebra and Geometry by focusing on graphing relations described by equations. The main idea of this section is the following. The Fundamental Graphing Principle The graph of an equation ... |
y). For example, substituting x = −3 into the equation yields 1 − x2 1 − x2 3 y = 3 1 − x2 = 3 1 − (−3)2 = 3√ −8 = −2, so the point (−3, −2) is on the graph. Continuing in this manner, we generate a table of points which are on the graph of the equation. These points are then plotted in the plane as shown below. y y x... |
previous example the graph had two x-intercepts, (−1, 0) and (1, 0), and one y-intercept, (0, 1). The graph of an equation can have any number of intercepts, including none at all! Since x-intercepts lie on the x-axis, we can find them by setting y = 0 in the equation. Similarly, since y-intercepts lie on the y-axis, w... |
into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the y-axis. about the x-axis – substitute (x, −y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the x-axis. about the origin - substitu... |
The only symmetry left to test is symmetry about the x-axis. To that end, we substitute (x, −y) into the equation and simplify (x − 2)2 + y2 = 1? = 1 (x − 2)2 + (−y)2 (x − 2)2 + y2 = 1 Since we have obtained our original equation, we know the graph is symmetric about the x-axis. This means we can cut our ‘plug and plo... |
+ y2 (−1 − 2)2 + 02? = 1 = 1 9 = 1. This proves that (−1, 0) is not on the graph. 6Without the use of a calculator, if you can believe it! 1.2 Relations 1.2.2 Exercises In Exercises 1 - 20, graph the given relation. 1. {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} 2. {(−2, 0), (−1, 1), (−1, −1), (0, 2), ... |
3 2 1 −1 −2 −3 1 2 3 x Relations and Functions 24. y 3 2 1 −3 −2 −1 −1 x −2 −3 −4 Relation C Relation D 25. 27. y 3 2 1 −4 −3 −2 −1 1 2 3 4 x Relation E y 3 2 1 26. 28. y 4 3 2 1 −3 −2 −1 1 2 3 x Relation F y 3 2 1 −3 −2 −1 −1 1 2 3 x −4 −3 −2 −1 −1 1 2 3 x −2 −3 Relation G −2 −3 Relation H 1.2 Relations 29. y 5 4 3 2... |
any exist. Follow the procedure in Example 1.2.3 to create a table of sample points on the graph of the equation. Plot the sample points and create a rough sketch of the graph of the equation. Test for symmetry. If the equation appears to fail any of the symmetry tests, find a point on the graph of the equation whose r... |
2.3 Answers 1. 5. −3 −2 −2 −1 −1 1 2 x −2 −3 −4 y 4 3 2 1 −2 −1 1 2 x 33 2. 4. y 3 2 1 −2 −1 −1 1 2 x −2 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 6 34 7. Relations and Functions y 8. −4 −3 −2 −1 −3 −4 −3 −2 −1 1 2 3 4 x 9. y 101 1 2 x 11. y 4 3 2 1 −3 −2 −1 −1 x −2 −3 5 4 3 2 1 −1 −2 −3 1 2 3 x 12. y 3 2 1 −1 −2 −3 −4 1 2 3... |
3 The line x = −2 y 3 2 1 −3 −2 −1 1 2 3 x The line y = 3 y 3 2 1 32. y 3 2 1 −1 −2 −3 1 2 3 x The line x = 3 34. y −3 −2 −1 −1 1 2 3 x −2 −3 The line y = −2 36. y 3 2 1 −3 −2 −1 −1 1 2 3 x −3 −2 −1 −1 1 2 3 x −2 −3 −2 −3 The line x = 0 is the y-axis The line y = 0 is the x-axis 1.2 Relations 41. y = x2 + 1 37 42. y = ... |
symmetric about the origin (e.g. (−3, 7) is on the graph but (3, −7) is not) 38 43. y = x3 − x x-intercepts: (−1, 0), (0, 0), (1, 0) y-intercept: (0, 0) x y (x, y) −2 −6 (−2, −6) (−1, 0) −1 (0, 0) 0 (1, 0) 1 (2, 62−1 −1 1 2 x −2 −3 −4 −5 −6 The graph is not symmetric about the x-axis. (e.g. (2, 6) is on the graph but ... |
(6, 2) 2 3 (11, 3) x 2 3 6 11 10 11 x The graph is not symmetric about the x-axis (e.g. (3, 1) is on the graph but (3, −1) is not) The graph is not symmetric about the y-axis (e.g. (3, 1) is on the graph but (−3, 1) is not) The graph is not symmetric about the origin (e.g. (3, 1) is on the graph but (−3, −1) is not) x... |
8) −1, − 13 2 (0, −5) 1, − 7 2 (2, −2) y 2 1 −3−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 −7 −8 −9 The graph is not symmetric about the x-axis (e.g. (3, 2) is on the graph but (3, −2) is not) The graph is not symmetric about the y-axis (e.g. (3, 2) is on the graph but (−3, 2) is not) The graph is not symmetric about the origin (... |
−1 −1 1 2 3 x −2 −3 −4 −5 The graph is symmetric about the x-axis The graph is not symmetric about the y-axis (e.g. (−6, 0) is on the graph but (6, 0) is not) The graph is not symmetric about the origin (e.g. (−6, 0) is on the graph but (6, 0) is not) y 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 The graph is symmetric about the x-a... |
.3 Introduction to Functions 43 1.3 Introduction to Functions One of the core concepts in College Algebra is the function. There are many ways to describe a function and we begin by defining a function as a special kind of relation. Definition 1.6. A relation in which each x-coordinate is matched with only one y-coordina... |
on the same vertical line. 1We will have occasion later in the text to concern ourselves with the concept of x being a function of y. In this case, R1 represents x as a function of y; R2 does not. 44 Relations and Functions It is worth taking some time to meditate on the Vertical Line Test; it will check to see how we... |
leaving an ‘open circle’ there. Hence, the vertical line x = 1 crosses the graph of S2 only at the point (1, 2). Indeed, any vertical line will cross the graph at most once, so we have that the graph of S2 passes the Vertical Line Test. Thus it describes y as a function of x1 −1 x −1 1 x −1 S1 and the line x = 1 The g... |
project up The graph of G The graph of G We see from the figure that if we project the graph of G to the x-axis, we get all real numbers less than 1. Using interval notation, we write the domain of G as (−∞, 1). To determine the range of G, we project the curve to the y-axis as follows: y 4 3 2 1 project left project r... |
� equation describes y as a function of x. 1 − x2 returns only one value of y. Hence, this x2y = 1 − 3y x2y + 3y = 1 y x2 + 3 = 1 y = 1 x2 + 3 factor For each choice of x, there is only one value for y, so this equation describes y as a function of x. We could try to use our graphing calculator to verify our responses ... |
(3, 0)} 4. {(1, 2), (4, 4), (9, 6), (16, 8), (25, 10), (36, 12),...} 5. {(x, y) | x is an odd integer, and y is an even integer} 6. {(x, 1) | x is an irrational number} 7. {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5),... } 8. {..., (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9),... } 9. {(−2, y) | − 3 ... |
−1 1 2 x −2 −1 1 2 x y 4 3 2 1 28. y 4 3 2 1 −2 −1 1 2 x −2 −1 1 2 x y 2 1 30. y 2 1 −2 −1 1 2 x −3 −2 −1 1 2 3 x −1 −2 −1 −2 52 31. y 2 1 Relations and Functions 32. y 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −2 −1 −2 In Exercises 33 - 47, determine whether or not the equation represents y as a function of x. 33. y = x3 − x 36... |
. Not a function domain = {−3, −2, −1, 0, 1, 2,3} range = {0, 1, 4, 9} 3. Function domain = {−7, −3, 3, 4, 5, 6} range = {0, 4, 5, 6, 9} 4. Function domain = {1, 4, 9, 16, 25, 36,...} = {x | x is a perfect square} range = {2, 4, 6, 8, 10, 12,...} = {y | y is a positive even integer} 5. Not a function 6. Function domain... |
2} 31. Not a function 22. Not a function 24. Function domain = [0, 3) ∪ (3, 6] range = (−4, −1] ∪ [0, 4] 26. Function domain = (−∞, ∞) range = (−∞, 4] 28. Function domain = (−∞, ∞) range = (−∞, ∞) 30. Function domain = [−3, 3] range = [−2, 2] 32. Function domain = (−∞, ∞) range = {2} 33. Function 34. Function 35. Funct... |
. multiply by 3 2. add 4 If we choose 5 as our input, in step 1 we multiply by 3 to get (5)(3) = 15. In step 2, we add 4 to our result from step 1 which yields 15 + 4 = 19. Using function notation, we would write f (5) = 19 to indicate that the result of applying the process f to the input 5 gives the output 19. In gen... |
. Solve f (x) = 4. Solution. 1. (a) To find f (−1), we replace every occurrence of x in the expression f (x) with −1 f (−1) = −(−1)2 + 3(−1) + 4 = −(1) + (−3) + 4 = 0 Similarly, f (0) = −(0)2 + 3(0) + 4 = 4, and f (2) = −(2)2 + 3(2) + 4 = −4 + 6 + 4 = 6. (b) To find f (2x), we replace every occurrence of x with the quant... |
) = 4 and f (3) = 4. A few notes about Example 1.4.2 are in order. First note the difference between the answers for f (2x) and 2f (x). For f (2x), we are multiplying the input by 2; for 2f (x), we are multiplying the output by 2. As we see, we get entirely different results. Along these lines, note that f (x + 2), f (x)... |
the following functions. 1. g(x) = √ 4 − 3x 3. f (x) = 2 4x x − 3 1 − 5. r(t) = 4 √ t + 3 6 − Solution. √ 2. h(x) = 5 4 − 3x 4. F (x) = √ 4 2x + 1 x2 − 1 6. I(x) = 3x2 x 1. The potential disaster for g is if the radicand4 is negative. To avoid this, we set 4 − 3x ≥ 0. 3, the expression From this, we get 3x ≤ 4 or x ≤ ... |
∞, −1) ∪ (−1, 3) ∪ (3, ∞). 4. In finding the domain of F, we notice that we have two potentially hazardous issues: not only do we have a denominator, we have a fourth (even-indexed) root. Our strategy is to determine the restrictions imposed by each part and select the real numbers which satisfy both conditions. To sati... |
�nd the domain of I to be all real numbers except 0: (−∞, 0) ∪ (0, ∞). It is worth reiterating the importance of finding the domain of a function before simplifying, as evidenced by the function I in the previous example. Even though the formula I(x) simplifies to 5Do you remember why? Consider squaring both sides to ‘so... |
the amount of grapes purchased (in pounds) and c(g) is the cost (in dollars). For example, c(5) represents the cost, in dollars, to purchase 5 pounds of grapes. In this case, c(5) = 1.5(5) = 7.5, so it would cost $7.50. If, on the other hand, we wanted to find the amount of grapes we can purchase for $5, we would need ... |
. Even with this restriction, our model has its limitations. As we saw above, it is virtually impossible to buy exactly 3.3 pounds of grapes so that our cost is exactly $5. In this case, being sensible shoppers, we would most likely ‘round down’ and purchase 3 pounds of grapes or however close the market scale can read... |
+ 100t, so for these values of t, we solve −5t2 + 100t = 375. Rearranging terms, we get 5t2 − 100t + 375 = 0, and factoring gives 5(t − 5)(t − 15) = 0. Our answers are t = 5 and t = 15, and since both of these values of t lie between 0 and 20, we keep both solutions. For t > 20, h(t) = 0, and in this case, there are n... |
.4 Function Notation 1.4.2 Exercises 63 In Exercises 1 - 10, find an expression for f (x) and state its domain. 1. f is a function that takes a real number x and performs the following three steps in the order given: (1) multiply by 2; (2) add 3; (3) divide by 4. 2. f is a function that takes a real number x and perform... |
) subtract 13. In Exercises 11 - 18, use the given function f to find and simplify the following: f (3) f (4x) f (x − 4) f (−1) 4f (x) f (x) − 4 f 3 2 f (−x) f x2 64 Relations and Functions 11. f (x) = 2x + 1 13. f (x) = 2 − x2 15. f (x) = x x − 1 17. f (x) = 6 12. f (x) = 3 − 4x 14. f (x) = x2 − 3x + 2 16. f (x) = 2 x3... |
if Compute the following function values. (b) f (−3) (e) f (−3.001) (c) f (3) (f) f (2) 1.4 Function Notation 65 36. Let f (x) = √ x2 1 − x2 x x ≤ −1 if if −1 < x ≤ 1 x > 1 if Compute the following function values. (a) f (4) (d) f (0) (b) f (−3) (e) f (−1) (c) f (1) (f) f (−0.999) In Exercises 37 - 62, find the... |
The area A enclosed by a square, in square inches, is a function of the length of one of its sides x, when measured in inches. This relation is expressed by the formula A(x) = x2 for x > 0. Find A(3) and solve A(x) = 36. Interpret your answers to each. Why is x restricted to x > 0? 64. The area A enclosed by a circle,... |
2) and C(5). 70. Using data from the Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Use your calculator to find F (0), F (14) and F (28). Round your a... |
talk 750 minutes per month with this plan? (b) How much does it cost to talk 20 hours a month with this plan? (c) Explain the terms of the plan verbally. 75. In Section 1.1.1 we defined the set of integers as Z = {..., −3, −2, −1, 0, 1, 2, 3,...}.14 The greatest integer of x, denoted by x, is defined to be the largest i... |
. Thus we return to our previous statement, in general, f (a + b) = f (a) + f (b). 1 x √ 1.4 Function Notation 1.4.3 Answers 1. f (x) = 2x+3 4 Domain: (−∞, ∞) 69 2. f (x) = 2(x+3) 4 = x+3 2 Domain: (−∞, ∞) 3. f (x Domain: (−∞, ∞) √ 4. f (x) = 2x + 3 Domain: − 3 2, ∞ 5. f (x) = 2(x + 3) = √ 2x + 6 Domain: [−3, ∞) 7. f (... |
x2 f (x − 4) = −x2 + 8x − 14 f (x) − 4 = −x2 − 2 f x2 = 2 − x4 14. For f (x) = x2 − 3x + 2 f (3) = 2 f (−14x) = 16x2 − 12x + 2 4f (x) = 4x2 − 12x + 8 f (−x) = x2 + 3x + 2 f (x − 4) = x2 − 11x + 30 f (x) − 4 = x2 − 3x − 2 f x2 = x4 − 3x2 + 2 15. For f (x) = x x−1 f (3) = 3 2 f (4x) = 4x 4x−1 f (x − 4) = x−4 x−5 16. For... |
) = −9 f (2a) = 4a − 5 2f (a) = 4a − 10 f (a + 2) = 2a − 1 f (a) + f (2) = 2a − −5a a 20. For f (x) = 5 − 2x f (a) 2 = 2a−5 2 f (a + h) = 2a + 2h − 5 f (2) = 1 f (−2) = 9 f (2a) = 5 − 4a 2f (a) = 10 − 4a f (a + 2) = 1 − 2a f (a) + f (2) = 6 − 2a f 2 a = 5 − 4 a = 5a−4 a 21. For f (x) = 2x2 − 1 f (a) 2 = 5−2a 2 f (a + h... |
= 2 √ 2a + 1 f (a + 2) = √ 2a + 5 f (a)+f (2) = √ 2a + 1+ √ 5 f (a) 2 = √ 2a+1 2 f (a+h) = √ 2a + 2h + +4 a 24. For f (x) = 117 f (2) = 117 f (−2) = 117 f (2a) = 117 2f (a) = 234 f (a + 2) = 117 f (a) + f (2) = 234 f 2 a = 117 25. For f (x) = x 2 f (2) = 1 2f (aa) 2 = 117 2 f (a + h) = 117 f (−2) = −1 f (2a) = a f (a ... |
= 0 when x = 1 2 33. For f (x) = 3 4−x, f (0) = 3 4 and f (x) is never equal to 0 34. For f (x) = 3x2−12x 4−x2, f (0) = 0 and f (x) = 0 when x = 0 or x = 4 35. (a) f (−4) = 1 (b) f (−3) = 2 (d) f (3.001) = 1.999 (e) f (−3.001) = 1.999 36. (a) f (4) = 4 (d) f (0) = 1 37. (−∞, ∞) 39. (−∞, −1) ∪ (−1, ∞) 41. (−∞, ∞) (b) f... |
�, −, 1 2 ∪ 1 2, ∞ 62. [0, 25) ∪ (25, ∞) 63. A(3) = 9, so the area enclosed by a square with a side of length 3 inches is 9 square inches. The solutions to A(x) = 36 are x = ±6. Since x is restricted to x > 0, we only keep x = 6. This means for the area enclosed by the square to be 36 square inches, the length of the s... |
ground. 68. T (0) = 3, so at 6 AM (0 hours after 6 AM), it is 3◦ Fahrenheit. T (6) = 33, so at noon (6 hours after 6 AM), the temperature is 33◦ Fahrenheit. T (12) = 27, so at 6 PM (12 hours after 6 AM), it is 27◦ Fahrenheit. 1.4 Function Notation 75 69. C(0) = 27, so to make 0 pens, it costs15 $2700. C(2) = 11, so to... |
25 for up to 1000 minutes and 10 cents per minute for each minute over 1000 minutes. 75. (a) 0.785 = 0, 117 = 117, −2.001 = −3, and π + 6 = 9 15This is called the ‘fixed’ or ‘start-up’ cost. We’ll revisit this concept on page 82. 76 Relations and Functions 1.5 Function Arithmetic In the previous section we used the newl... |
the real number outputs from f and g. Example 1.5.1. Let f (x) = 6x2 − 2x and g(x) = 3 − 1 x. 1. Find (f + g)(−1) 2. Find (f g)(2) 3. Find the domain of g − f then find and simplify a formula for (g − f )(x). 1.5 Function Arithmetic 77 4. Find the domain of g f then find and simplify a formula for g f (x). Solution. 1. ... |
). In this case, we get common denominators and attempt to reduce the resulting fraction. Doing so, we get (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x = 3 − 1 x − 6x2 + 2x get common denominators = = = + − − 1 x 6x3 x 2x2 3x x x 3x − 1 − 6x3 − 2x2 x −6x3 − 2x2 + 3x − 1 x 4. As in the previous example, we have two w... |
· x x x 3 − 6x2 − 2x 1 x (6x2 − 2x) x 3x − 1 (6x2 − 2x) x 3x − 1 2x2(3x − 1) 1 (3x − 1) 2x2 (3x − 1) 1 2x2 simplify compound fractions factor cancel Please note the importance of finding the domain of a function before simplifying its expression. In number 4 in Example 1.5.1 above, had we waited to find the domain of g ... |
difference quotient is f (x + h) − f (x) h = = = = = x2 + 2xh + h2 − x − h − 2 − x2 − x − 2 h x2 + 2xh + h2 − x − h − 2 − x2 + x + 2 h 2xh + h2 − h h h (2x + h − 1) h h (2x + h − 1) h factor cancel = 2x + h − 1. 80 Relations and Functions 2. To find g(x + h), we replace every occurrence of x in the formula g(x) = 3 2x+1... |
by the conjugate. Difference of Squares. √ x + h2 )2 x (x + h Since we have removed the original ‘h’ from the denominator, we are done. As mentioned before, we will revisit difference quotients in Section 2.1 where we will explain them geometrically. For now, we want to move on to some classic applications of function a... |
, p(x) returns the price charged per item when x items are produced and sold. Our next function to consider is the revenue function, R(x). The function R(x) computes the amount of money collected as a result of selling x items. Since p(x) is the price charged per item, we have R(x) = xp(x). Finally, the profit function,... |
. Solve P (x) = 0 and interpret the result. 5See Example 5.2.4 in Section 5.2. 6Pronounced ‘dopeys’... 1.5 Function Arithmetic 83 Solution. 1. We substitute x = 0 into the formula for C(x) and get C(0) = 100(0) + 2000 = 2000. This means to produce 0 dOpis, it costs $2000. In other words, the fixed (or start-up) costs ar... |
before, the validity of this formula is for 0 ≤ x ≤ 30 only. 6. We find R(0) = 0 which means if no dOpis are sold, we have no revenue, which makes sense. Turning to profit, P (0) = −2000 since P (x) = R(x)−C(x) and P (0) = R(0)−C(0) = −2000. This means that if no dOpis are sold, more money ($2000 to be exact!) was put i... |
that... in Section 2.3. 7Imagine that! Giving something away for free and hardly anyone taking advantage of it... 8We’ve seen this sort of thing before in Section 1.4.1. 84 1.5.1 Exercises Relations and Functions In Exercises 1 - 10, use the pair of functions f and g to find the following values if they exist. (f + g)(... |
+ x + 6 and g(x) = x2 − 9 17. f (x) = x 2 and g(x) = 19. f (x) = x and g(x) = 2 x √ x + 1 18. f (x) = x − 1 and g(x) = 1 x − 1 20. f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 In Exercises 21 - 45, find and simplify the difference quotient f (x + h) − f (x) h for the given function. 21. f (x) = 2x − 5 23. f (x) = 6 25. f ... |
and interpret C(10). Find and interpret p(5) Find and simplify R(x). Find and simplify P (x). Solve P (x) = 0 and interpret. 46. The cost, in dollars, to produce x “I’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x ≥ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 − 2x, 0 ≤ x ≤ 15. 47. The c... |
g)(−3) 54. (g + f )(1) 57. 60. f g g f (−2) (−1) 52. (f − g)(2) 55. (g − f )(3) 58. 61. f g g f (−1) (3) 53. (f g)(−1) 56. (gf )(−3) 59. 62. f g g f (2) (−3) 1.5 Function Arithmetic 87 1.5.2 Answers 1. For f (x) = 3x + 1 and g(x) = 4 − x (f + g)(2) = 9 (f g) 1 2 = 35 4 (f − g)(−1) = −7 (g − f )(1) = −1 f g (0) = 1 4 g... |
) = √ 4 − x and g(x) = √ (f + g)(2f − g)(−1) = −1 + √ 5 (f g) 1 2 = √ 35 2 f g √ 2 (0) = (g − f )(1) = 0 g f (−2) = 0 88 Relations and Functions 7. For f (x) = 2x and g(x) = 1 2x+1 (f + g)(2) = 21 5 (f g) 1 2 = 1 2 (f − g)(−1) = −1 f g (0) = 0 (g − f )(1) = − 5 3 g f (−2) = 1 12 8. For f (x) = x2 and g(x) = 3 2x−3 (f +... |
− 1 Domain: (−∞, ∞) (f − g)(x) = x + 3 Domain: (−∞, ∞) f g (x) = 2x+1 x−2 Domain: (−∞, 2) ∪ (2, ∞) (f − g)(x) = 2 − 6x Domain: (−∞, ∞) f (x) = 1−4x g 2x−1 Domain: −∞, 1 2 ∪ 1 2, ∞ 1.5 Function Arithmetic 89 13. For f (x) = x2 and g(x) = 3x − 1 (f + g)(x) = x2 + 3x − 1 Domain: (−∞, ∞) (f g)(x) = 3x3 − x2 Domain: (−∞, ∞... |
∞, ∞) (f g)(x) = −x4 + x3 + 15x2 − 9x − 54 Domain: (−∞, ∞) (f − g)(x) = −2x2 + x + 15 Domain: (−∞, ∞) f g (x) = − x+2 x+3 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 17. For f (x) = x 2 and g(x) = 2 x (f + g)(x) = x2+4 2x Domain: (−∞, 0) ∪ (0, ∞) (f g)(x) = 1 Domain: (−∞, 0) ∪ (0, ∞) (f − g)(x) = x2−4 2x Domain: (−∞, 0) ∪ (0, ... |
(f + g)(x) = 2 Domain: [5, ∞) x − 5 (f g)(x) = x − 5 Domain: [5, ∞) 21. 2 23. 0 25. −2x − h + 2 27. −2x − h + 1 29. m 31. 33. 35. −2 x(x + h) −(2x + h) x2(x + h)2 −4 (4x − 3)(4x + 4h − 3) (f − g)(x) = 0 Domain: [5, ∞) f g (x) = 1 Domain: (5, ∞) 22. −3 24. 6x + 3h − 1 26. 8x + 4h 28. 3x2 + 3xh + h2 30. 2ax + ah + b 32.... |
1 and x = 13. These are the ‘break even’ points, so selling 1 shirt or 13 shirts will guarantee the revenue earned exactly recoups the cost of production. C(0) = 100, so the fixed costs are $100. C(10) = 20, so when 10 bottles of tonic are produced, the cost per bottle is $20. p(5) = 30, so to sell 5 bottles of tonic, ... |
6 pies or 12 pies a day will guarantee the revenue earned exactly recoups the cost of production. 50. C(0) = 1000, so the monthly fixed costs are 1000 hundred dollars, or $100,000. C(10) = 120, so when 10 scooters are made, the cost per scooter is 120 hundred dollars, or $12,000. p(5) = 130, so to sell 5 scooters a mon... |
points which satisfy the equation y = f (x). That is, the point (x, y) is on the graph of f if and only if y = f (x). Example 1.6.1. Graph f (x) = x2 − x − 6. Solution. To graph f, we graph the equation y = f (x). To this end, we use the techniques outlined in Section 1.2.1. Specifically, we check for intercepts, test ... |
proceed as before – finding intercepts, testing for symmetry and then plotting additional points as needed. To find the x-intercepts, as before, we set f (x) = 0. The twist is that we have two formulas for f (x). For x < 1, we use the formula f (x) = 4 − x2. Setting f (x) = 0 gives 0 = 4 − x2, so that x = ±2. However, o... |
a table as before, we see that as the x values sneak up to x = 1 in this fashion, the f (x) values inch closer and closer1 to 4 − 12 = 3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get (x, f (x)) x f (x) (0.9, 3.1... |
the original equation y = f (x) we need −f (−x) = f (x), or, equivalently, f (−x) = −f (x). These results are summarized below. Testing the Graph of a Function for Symmetry The graph of a function f is symmetric about the y-axis if and only if f (−x) = f (x) for all x in the domain of f. about the origin if and only i... |
this, we check with an x value. After some trial and error, we see that g(1) = 5 whereas g(−1) = −5. This proves that g is not even, but it doesn’t rule out the possibility that g is odd. (Why not?) To check if g is odd, we compare g(−x) with −g(x) −g(x) = − 5x 2 − x2 −5x 2 − x2 = −g(x) = g(−x) Hence, g is odd. Graphi... |
i(−x) the numerator is −5x. To re-write i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x) = = = 5x 2x − x3 (−1)5x (−1) (2x − x3) −5x −2x + x3 Hence, i(x) = i(−x), so i is even. The calculator s... |
0 which means to find p(−x), we need to use the other formula for p(x), namely p(x) = −x+3. Hence, for x < 0, p(−x) = −(−x)+3 = x + 3 = p(x). For x ≥ 0, p(x) = −x + 3 and we have two cases. If x > 0, then −x < 0 so p(−x) = (−x) + 3 = −x + 3 = p(x). If x = 0, then p(0) = 3 = p(−0). Hence, in all cases, p(−x) = p(x), so ... |
5.5) (−2, 4.54 −3 −2 −1 −1 1 2 3 4 5 6 7 x (−4, −3) −2 −3 −4 −5 −6 −7 −8 −9 (4, −6) (3, −8) (5, −6) The graph of y = f (x) Reading from left to right, the graph ‘starts’ at the point (−4, −3) and ‘ends’ at the point (6, 5.5). If we imagine walking from left to right on the graph, between (−4, −3) and (−2, 4.5), we are... |
= −2 is contained in an interval on which f was increasing and one on which it is decreasing. (There’s more on this issue – and many others – in the Exercises.) We’re now ready for the more formal algebraic definitions of what it means for a function to be increasing, decreasing or constant. Definition 1.10. Suppose f i... |
, it is not a precise enough way to define the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional definitions and thoroughly vet the pathologies they induce in the Exercises. We have one last observation to make before we proceed to the algebraic definitions and look at a fa... |
all x in the domain of f. It’s important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function’s graph would have to look like?) We shall see examples of functions in the Exercises which ha... |
from −4 to 4, inclusive. Hence, the domain is [−4, 4]. 2. To find the range, we project the graph to the y-axis. We see that the y values from −3 to 3, inclusive, constitute the range of f. Hence, our answer is [−3, 3]. 3. The x-intercepts are the points on the graph with y-coordinate 0, namely (−2, 0) and (2, 0). 4. T... |
be symmetric about the y-axis. This suggests14 that f is even. 11. As we move from left to right, the graph rises from (−4, −3) to (0, 3). This means f is increasing on the interval [−4, 0]. (Remember, the answer here is an interval on the x-axis.) 12. As we move from left to right, the graph falls from (0, 3) to (4, ... |
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