Solution Manual for Algebra and Trigonometry, 5th Edition

Chapter 2 More on Functions 6. a) For x-values from 1 to 4, the y-values increase from 2 to 11. Thus the function is increasing on the interval (1, 4). Exercise Set 2.1 1. a) For x-values from −5 to 1, the y-values increase from −3 to 3. Thus the function is increasing on the interval (−5, 1). b) For x-values from 3 to 5, the y-values decrease from 3 to 1. Thus the function is decreasing on the interval (3, 5). c) For x-values from 1 to 3, y is 3. Thus the function is constant on (1, 3). 2. a) For x-values from 1 to 3, the y-values increase from 1 to 2. Thus, the function is increasing on the interval (1, 3). b) For x-values from −5 to 1, the y-values decrease from 4 to 1. Thus the function is decreasing on the interval (−5, 1). c) For x-values from 3 to 5, y is 2. Thus the function is constant on (3, 5). 3. a) For x-values from −3 to −1, the y-values increase from −4 to 4. Also, for x-values from 3 to 5, the y-values increase from 2 to 6. Thus the function is increasing on (−3, −1) and on (3, 5). b) For x-values from −1 to 1, the y-values decrease from 6 to 2. Also, for x-values from 4 to ∞, the yvalues decrease from 11 to −∞. Thus the function is decreasing on (−1, 1) and on (4, ∞). c) For x-values from −∞ to −1, y is 3. Thus the function is constant on (−∞, −1). 7. The x-values extend from −5 to 5, so the domain is [−5, 5]. The y-values extend from −3 to 3, so the range is [−3, 3]. 8. Domain: [−5, 5]; range: [1, 4] 9. The x-values extend from −5 to −1 and from 1 to 5, so the domain is [−5, −1] ∪ [1, 5]. The y-values extend from −4 to 6, so the range is [−4, 6]. 10. Domain: [−5, 5]; range: [1, 3] 11. The x-values extend from −∞ to ∞, so the domain is (−∞, ∞). The y-values extend from −∞ to 3, so the range is (−∞, 3]. 12. Domain: (−∞, ∞); range: (−∞, 11] b) For x-values from 1 to 3, the y-values decrease from 3 to 2. Thus the function is decreasing on the interval (1, 3). 13. From the graph we see that a relative maximum value of the function is 3.25. It occurs at x = 2.5. There is no relative minimum value. c) For x-values from −5 to −3, y is 1. Thus the function is constant on (−5, −3). The graph starts rising, or increasing, from the left and stops increasing at the relative maximum. From this point, the graph decreases. Thus the function is increasing on (−∞, 2.5) and is decreasing on (2.5, ∞). 4. a) For x-values from 1 to 2, the y-values increase from 1 to 2. Thus the function is increasing on the interval (1, 2). b) For x-values from −5 to −2, the y-values decrease from 3 to 1. For x-values from −2 to 1, the y-values decrease from 3 to 1. And for x-values from 3 to 5, the y-values decrease from 2 to 1. Thus the function is decreasing on (−5, −2), on (−2, 1), and on (3, 5). c) For x-values from 2 to 3, y is 2. Thus the function is constant on (2, 3). 5. a) For x-values from −∞ to −8, the y-values increase from −∞ to 2. Also, for x-values from −3 to −2, the y-values increase from −2 to 3. Thus the function is increasing on (−∞, −8) and on (−3, −2). b) For x-values from −8 to −6, the y-values decrease from 2 to −2. Thus the function is decreasing on the interval (−8, −6). c) For x-values from −6 to −3, y is −2. Also, for xvalues from −2 to ∞, y is 3. Thus the function is constant on (−6, −3) and on (−2, ∞). 14. From the graph we see that a relative minimum value of 2 occurs at x = 1. There is no relative maximum value. The graph starts falling, or decreasing, from the left and stops decreasing at the relative minimum. From this point, the graph increases. Thus the function is increasing on (1, ∞) and is decreasing on (−∞, 1). 15. From the graph we see that a relative maximum value of the function is 2.370. It occurs at x = −0.667. We also see that a relative minimum value of 0 occurs at x = 2. The graph starts rising, or increasing, from the left and stops increasing at the relative maximum. From this point it decreases to the relative minimum and then increases again. Thus the function is increasing on (−∞, −0.667) and on (2, ∞). It is decreasing on (−0.667, 2). 16. From the graph we see that a relative maximum value of 2.921 occurs at x = 3.601. A relative minimum value of 0.995 occurs at x = 0.103. c 2016 Pearson Education, Inc. Copyright 58 Chapter 2: More on Functions The graph starts decreasing from the left and stops decreasing at the relative minimum. From this point it increases to the relative maximum and then decreases again. Thus the function is increasing on (0.103, 3.601) and is decreasing on (−∞, 0.103) and on (3.601, ∞). 17. 20. y 5 4 3 2 1 y –5 –4 –3 –2 –1 –1 5 –2 4 –3 3 –4 f (x ) = x 2 2 1 2 3 4 5 x f (x ) = | x + 3 | — 5 –5 1 –5 –4 –3 –2 –1 –1 1 2 3 4 5 Increasing: (−3, ∞) x Decreasing: (−∞, −3) –2 –3 –4 Maxima: none –5 Minimum: −5 at x = −3 The function is increasing on (0, ∞) and decreasing on (−∞, 0). We estimate that the minimum is 0 at x = 0. There are no maxima. 21. y 5 4 3 18. y 2 1 5 –5 –4 –3 –2 –1 f (x ) = 4 — x 2 4 3 –2 2 –1 1 2 3 4 5 x f (x ) = x 2 — 6x + 10 –3 1 –5 –4 –3 –2 –1 –1 –4 1 2 3 4 5 x –5 –2 The function is decreasing on (−∞, 3) and increasing on (3, ∞). We estimate that the minimum is 1 at x = 3. There are no maxima. –3 –4 –5 Increasing: (−∞, 0) 22. Decreasing: (0, ∞) y 10 2 f (x ) = — x — 8x — 9 8 Maximum: 4 at x = 0 6 4 Minima: none 2 19. y –10 –8 –6 –4 –2 5 2 4 6 8 10 x –4 f (x ) = 5 — | x | 4 –2 –6 3 –8 2 –10 1 –5 –4 –3 –2 –1 –1 1 2 3 4 5 Increasing: (−∞, −4) x Decreasing: (−4, ∞) –2 –3 Maximum: 7 at x = −4 –4 –5 Minima: none The function is increasing on (−∞, 0) and decreasing on (0, ∞). We estimate that the maximum is 5 at x = 0. There are no minima. 23. If x = the length of the rectangle, in meters, then the 480 − 2x , or 240 − x. We use the formula Area = width is 2 length × width: A(x) = x(240 − x), or A(x) = 240x − x2 c 2016 Pearson Education, Inc. Copyright Exercise Set 2.1 59 24. Let h = the height of the scarf, in inches. Then the length of the base = 2h − 7. 1 A(h) = (2h − 7)(h) 2 7 A(h) = h2 − h 2 25. We use the Pythagorean theorem. [h(d)]2 + 35002 = d2 [h(d)]2 = d2 − 35002 √ h(d) = d2 − 35002 We considered only the positive square root since distance must be nonnegative. 26. After t minutes, the balloon has risen 120t ft. We use the Pythagorean theorem. [d(t)]2 = (120t)2 + 4002 d(t) = (120t)2 + 4002 We considered only the positive square root since distance must be nonnegative. 27. Let w = the width of the rectangle. Then the 40 − 2w , or 20 − w. Divide the rectangle into length = 2 quadrants as shown below. 1 3d 2 = s 7 12 1 s · 3d = 7 · 12 2 7 sd = 4 2 4 7 d = · , so s 2 14 d(s) = . s 30. The volume of the tank is the sum of the volume of a sphere with radius r and a right circular cylinder with radius r and height 6 ft. 4 V (r) = πr3 + 6πr2 3 31. a) After 4 pieces of float line, each of length x ft, are used for the sides perpendicular to the beach, there remains (240−4x) ft of float line for the side parallel to the beach. Thus we have a rectangle with length 240 − 4x and width x. Then the total area of the three swimming areas is A(x) = (240 − 4x)x, or 240x − 4×2 . b) The length of the sides labeled x must be positive and their total length must be less than 240 ft, so 4x < 240, or x < 60. Thus the domain is {x|0 < x < 60}, or (0, 60). c) We see from the graph that the maximum value of the area function on the interval (0, 60) appears to be 3600 when x = 30. Thus the dimensions that yield the maximum area are 30 ft by 240 − 4 · 30, or 240 − 120, or 120 ft. 20 – w w In each quadrant there are two congruent triangles. One triangle is part of the rhombus and both are part of the rectangle. Thus, in each quadrant the area of the rhombus is one-half the area of the rectangle. Then, in total, the area of the rhombus is one-half the area of the rectangle. 1 A(w) = (20 − w)(w) 2 w2 A(w) = 10w − 2 28. Let w = the width, in feet. Then the length = or 23 − w. A(w) = (23 − w)w 46 − 2w , 2 A(w) = 23w − w2 29. We will in use similar triangles, expressing all distances s 1 ft, and d yd = 3d ft We feet. 6 in. = ft, s in. = 2 12 have 32. a) If the length = x feet, then the width = 24 − x feet. A(x) = x(24 − x) A(x) = 24x − x2 b) The length of the rectangle must be positive and less than 24 ft, so the domain of the function is {x|0 < x < 24}, or (0, 24). c) We see from the graph that the maximum value of the area function on the interval (0, 24) appears to be 144 when x = 12. Then the dimensions that yield the maximum area are length = 12 ft and width = 24 − 12, or 12 ft. 33. a) When a square with sides of length x is cut from each corner, the length of each of the remaining sides of the piece of cardboard is 12 − 2x. Then the dimensions of the box are x by 12 − 2x by 12 − 2x. We use the formula Volume = length × width × height to find the volume of the box: V (x) = (12 − 2x)(12 − 2x)(x) V (x) = (144 − 48x + 4×2 )(x) V (x) = 144x − 48×2 + 4×3 This can also be expressed as V (x) = 4x(x − 6)2 , or V (x) = 4x(6 − x)2 . c 2016 Pearson Education, Inc. Copyright 60 Chapter 2: More on Functions b) The length of the sides of the square corners that are cut out must be positive and less than half the length of a side of the piece of cardboard. Thus, the domain of the function is {x|0 < x < 6}, or (0, 6). c) We see from the graph that the maximum value of the area function on the interval (0, 6) appears to be 128 when x = 2. When x = 2, then 12 − 2x = 12 − 2 · 2 = 8, so the dimensions that yield the maximum volume are 8 cm by 8 cm by 2 cm. 1  x, for x < 0, 39. f (x) = 2  x + 3, for x ≥ 0 1 x for 2 inputs x less than 0. Then graph f (x) = x + 3 for inputs x greater than or equal to 0. We create the graph in two parts. Graph f (x) = y 4 34. a) V (x) = 8x(14 − 2x), or 112x − 16×2 14 b) The domain is x0 < x < , or 2 {x|0 < x 1 Since −4 ≤ 1, g(−4) = −4 + 4 = 0. x ⫺4  1  − x + 2, for x ≤ 0, 3 40. f (x) =  x − 5, for x > 0 Since 0 ≤ 1, g(0) = 0 + 4 = 4. y Since 1 ≤ 1, g(1) = 1 + 4 = 5. 4 Since 3 > 1, g(3) = 8 − 3 = 5. 2    3, 36. f (x) = ⫺4 ⫺2 for x ≤ −2, 1   x + 6, for x > −2 2 4 x ⫺4  3  − x + 2, for x < 4, 4 41. f (x) =  −1, for x ≥ 4 We create the graph in two parts. Graph 3 f (x) = − x + 2 for inputs x less than 4. Then graph 4 f (x) = −1 for inputs x greater than or equal to 4. f (−5) = 3 f (−2) = 3 1 f (0) = · 0 + 6 = 6 2 1 f (2) = · 2 + 6 = 7 2 37. h(x) = 2 ⫺2 −3x − 18, for x < −5, 1, for −5 ≤ x < 1, x + 2, for x ≥ 1 y 4 Since −5 is in the interval [−5, 1), h(−5) = 1. Since 0 is in the interval [−5, 1), h(0) = 1. ⫺4 ⫺2 Since 1 ≥ 1, h(1) = 1 + 2 = 3. 2 4 ⫺2 Since 4 ≥ 1, h(4) = 4 + 2 = 6. x ⫺4  −5x − 8, for x 4 42. h(x) = 2x − 1, for x < 2 2 − x, for x ≥ 2 y Since −4 4, f (6) = 10 − 2 · 6 = −2. 4 2 ⫺4 ⫺2 2 ⫺2 ⫺4 c 2016 Pearson Education, Inc. Copyright 4 x Exercise Set 2.1 61  2   x − 9 , for x = −3, x+3 46. f (x) =   5, for x = −3   x + 1, for x ≤ −3,       for −3 < x < 4 43. f (x) = −1,     1   x, for x ≥ 4  2 y We create the graph in three parts. Graph f (x) = x + 1 for inputs x less than or equal to −3. Graph f (x) = −1 for inputs greater than −3 and less than 4. Then graph 1 f (x) = x for inputs greater than or equal to 4. 2 4 2 ⫺4 ⫺2 y ⫺6 4  2,   ⫺4 ⫺2 2 4 ⫺2 47. f (x) = x for x = 5, x − 25   , for x = 5 x−5 When x = 5, the denominator of (x2 − 25)/(x − 5) is nonzero so we can simplify: x ⫺4 2 (x + 5)(x − 5) x2 − 25 = = x + 5. x−5 x−5 Thus, f (x) = x + 5, for x = 5. 4, for x ≤ −2, x + 1, for −2 < x < 3 −x, for x ≥ 3 The graph of this part of the function consists of a line with a “hole” at the point (5, 10), indicated by an open dot. At x = 5, we have f (5) = 2, so the point (5, 2) is plotted below the open dot. y 4 2 ⫺4 ⫺2 4 ⫺4 2 44. f (x) = 2 ⫺2 2 4 ⫺2 y x 8 ⫺4 2 ⫺8 ⫺4 1   x − 1, for x 1 4 8 x ⫺4 ⫺8 1 x−1 2 for inputs less than 0. Graph g(x) = 3 for inputs greater than or equal to 0 and less than or equal to 1. Then graph g(x) = −2x for inputs greater than 1. We create the graph in three parts. Graph g(x) =  2   x + 3x + 2 , for x = −1, x+1 48. f (x) =   7, for x = −1 y 8 y 6 4 4 2 2 ⫺4 ⫺2 2 4 x ⫺4 ⫺2 ⫺2 ⫺4 c 2016 Pearson Education, Inc. Copyright 2 4 x 62 Chapter 2: More on Functions 49. f (x) = [[x]] 1 [[x]] − 2 2 This function can be defined by a piecewise function with an infinite number of statements:   .   .     .      −2 12 , for −1 ≤ x < 0,    −2, for 0 ≤ x < 1, f (x) = −1 1 , for 1 ≤ x < 2,  2   for 2 ≤ x < 3,  −1,    .     .     .  52. f (x) = See Example 9. 50. f (x) = 2[[x]] This function can be defined by a piecewise function with an infinite  number of statements:   .     .     .     −4, for −2 ≤ x < −1,    −2, for −1 ≤ x < 0, f (x) = 0, for 0 ≤ x < 1,    2, for 1 ≤ x < 2,     .     .     . 6 2 2 4 2 ⫺2 x ⫺4 53. From the graph we see that the domain is (−∞, ∞) and the range is (−∞, 0) ∪ [3, ∞). 57. From the graph we see that the domain is (−∞, ∞) and the range is {y|y ≤ −2 or y = −1 or y ≥ 2}. 58. Domain: (−∞, ∞); range: (−∞, −3] ∪ (−1, 4] 51. f (x) = 1 + [[x]] This function can be defined by a piecewise function with an infinite  number of statements:  .     .     .      −1, for −2 ≤ x < −1,    0, for −1 ≤ x < 0, f (x) = 1, for 0 ≤ x < 1,    for 1 ≤ x < 2,  2,     .     .     . y 59. From the graph we see that the domain is (−∞, ∞) and the range is {−5, −2, 4}. An equation for the function is: f (x) = −2, for x 2 60. Domain: (−∞, ∞); range: {y|y = −3 or y ≥ 0} −3, for x < 0, g(x) = x, for x ≥ 0 61. From the graph we see that the domain is (−∞, ∞) and the range is (−∞, −1] ∪ [2, ∞). Finding the slope of each segment and using the slope-intercept or point-slope formula, we find that an equation for the function is: x, for x ≤ −1, 2, for −1 2 This can also be expressed as follows: g(x) = 4 2 2 ⫺4 ⫺4 ⫺2 56. Domain: (∞, ∞); range: (−∞, 3) x f(x) ⫽ 2冚x军 ⫺2 h(x) ⫽ q 冚x军 ⫺ 2 2 55. From the graph we see that the domain is (−∞, ∞) and the range is [−1, ∞). 4 ⫺4 4 54. Domain: (−∞, ∞); range: (−5, ∞) y ⫺4 y 4 x g(x) ⫽ 1 ⫹ 冀x冁 g(x) = x, for x ≤ −1, 2, for −1 < x < 2, x, for x ≥ 2 c 2016 Pearson Education, Inc. Copyright Exercise Set 2.1 63 62. Domain: (−∞, ∞); range: {y|y = −2 or y ≥ 0}. An equation for the function is: |x|, for x < 3, h(x) = −2, for x ≥ 3 This can also be expressed as follows: −x, for x ≤ 0, x, for 0 < x < 3, −2, for x ≥ 3 It can also be expressed as follows: h(x) = h(x) = 68. −x, for x < 0, x, for 0 ≤ x < 3, −2, for x ≥ 3 63. From the graph we see that the domain is [−5, 3] and the range is (−3, 5). Finding the slope of each segment and using the slope-intercept or point-slope formula, we find that an equation for the function is: h(x) = x + 8, for −5 ≤ x < −3, 3, for −3 ≤ x ≤ 1, 3x − 6, for 1 < x ≤ 3 64. Domain: [−4, ∞); range: [−2, 4] −2x − 4, for −4 ≤ x ≤ −1, x − 1, for −1 < x < 2, 2, for x ≥ 2 This can also be expressed as: f (x) = f (x) = y − y1 = m(x − x1 ) 1 y − 1 = − [x − (−1)] 8 1 y − 1 = − (x + 1) 8 1 1 y−1 = − x− 8 8 1 7 y = − x+ 8 8 2x − 9y + 1 = 0 2x + 1 = 9y 1 2 x+ = y 9 9 1 2 Slope: ; y-intercept: 0, 9 9 69. a) The function C(t) can be defined piecewise.  3, for 0 < t < 1,     6, for 1 ≤ t < 2,     9, for 2 ≤ t < 3, C(t) = .    .     . We graph this function. −2x − 4, for −4 ≤ x < −1, x − 1, for −1 ≤ x 0. a) f (2) = 4 · 2 − 5 · 2 = 4 · 8 − 5 · 2 = 32 − 10 = 22 3 b) f (−2) = 4(−2) − 5(−2) = 4(−8) − 5(−2) = −32 + 10 = −22 3 c) f (a) = 4a3 − 5a d) f (−a) = 4(−a)3 − 5(−a) = 4(−a3 ) − 5(−a) = −4a3 + 5a 67. First find the slope of the given line. 70. If [[x + 2]] = −3, then −3 ≤ x + 2 < −2, or −5 ≤ x < −4. The possible inputs for x are {x| − 5 ≤ x < −4}. 71. If [[x]]2 = 25, then [[x]] = −5 or [[x]] = 5. For −5 ≤ x < −4, [[x]] = −5. For 5 ≤ x < 6, [[x]] = 5. Thus, the possible inputs for x are {x| − 5 ≤ x < −4 or 5 ≤ x 0 for all values of x in its domain, the domain of F/G is [2, 9]. 44. (F + G)(x) = F (x) + G(x) y 10 37. The domain of G/F is the set of numbers in the domains of both F and G (See Exercise 35.), excluding those for which F = 0. Since F (3) = 0, the domain of G/F is [2, 3) ∪ (3, 9]. 8 38. 2 FG 6 4 y 2 4 6 8 10 6 8 10 x 8 6 45. 4 F G 2 2 39. 4 6 y 6 8 10 4 x 2 2 y ⫺2 G⫺F 2 4 x 6 46. G F 4 2 y 4 2 4 6 8 10 x FG 2 2 2 4 4 6 8 2 10 x 4 40. y 6 47. a) P (x) = R(x) − C(x) = 60x − 0.4×2 − (3x + 13) = F G 4 60x − 0.4×2 − 3x − 13 = −0.4×2 + 57x − 13 2 b) R(100) = 60·100−0.4(100)2 = 6000−0.4(10, 000) = 2 4 6 8 10 x 6000 − 4000 = 2000 2 C(100) = 3 · 100 + 13 = 300 + 13 = 313 4 P (100) = R(100) − C(100) = 2000 − 313 = 1687 41. From the graph, we see that the domain of F is [0, 9] and the domain of G is [3, 10]. The domain of F + G is the set of numbers in the domains of both F and G. This is [3, 9]. 42. The domain of F − G and F G is the set of numbers in the domains of both F and G. (See Exercise 41.) This is [3, 9]. The domain of F/G is the set of numbers in the domains of both F and G, excluding those for which G = 0. Since G > 0 for all values of x in its domain, the domain of F/G is [3, 9]. 43. The domain of G/F is the set of numbers in the domains of both F and G (See Exercise 41.), excluding those for which F = 0. Since F (6) = 0 and F (8) = 0, the domain of G/F is [3, 6) ∪ (6, 8) ∪ (8, 9]. 48. a) P (x) = 200x − x2 − (5000 + 8x) = 200x − x2 − 5000 − 8x = −x2 + 192x − 5000 b) R(175) = 200(175) − 1752 = 4375 C(175) = 5000 + 8 · 175 = 6400 P (175) = R(175) − C(175) = 4375 − 6400 = −2025 (We could also use the function found in part (a) to find P (175).) 49. f (x) = 3x − 5 f (x + h) = 3(x + h) − 5 = 3x + 3h − 5 3x + 3h − 5 − (3x − 5) f (x + h) − f (x) = h h 3x + 3h − 5 − 3x + 5 = h 3h = =3 h c 2016 Pearson Education, Inc. Copyright 70 50. Chapter 2: More on Functions f (x) = 4x − 1 f (x + h) − f (x) 4(x + h) − 1 − (4x − 1) = = h h 4h 4x + 4h − 1 − 4x + 1 = =4 h h 51. f (x) = 6x + 2 f (x + h) = 6(x + h) + 2 = 6x + 6h + 2 6x + 6h + 2 − (6x + 2) f (x + h) − f (x) = h h 6x + 6h + 2 − 6x − 2 = h 6h =6 = h 55. f (x) = 1 3x f (x + h) = 1 3(x + h) 1 1 − f (x + h) − f (x) 3(x + h) 3x = h h x 1 x+h 1 · − · 3(x + h) x 3x x + h = h x+h x − 3x(x + h) 3x(x + h) = h x−x−h x − (x + h) 3x(x + h) 3x(x + h) = = h h −h −h 1 3x(x + h) = · = h 3x(x + h) h 52. f (x) = 5x + 3 5(x + h) + 3 − (5x + 3) f (x + h) − f (x) = = h h 5h 5x + 5h + 3 − 5x − 3 = =5 h h 1 x+1 3 1 1 1 f (x + h) = (x + h) + 1 = x + h + 1 3 3 3 1 1 1 x+ h+1− x+1 f (x + h) − f (x) 3 3 3 = h h 1 1 1 x+ h+1− x−1 3 3 3 = h 1 h 1 = 3 = h 3 −1 · h/ −h = 3x(x + h) · h 3x(x + h) · h/ 1 −1 , or − = 3x(x + h) 3x(x + h) = 53. f (x) = 1 54. f (x) = − x + 7 2 1 1 − (x + h) + 7 − − x + 7 f (x + h) − f (x) 2 = 2 = h h 1 1 1 1 − h − x− h+7+ −7 1 2 2 2 2 = =− h h 2 56. f (x) = 1 2x 1 1 1 x 1 x+h − · − · f (x+h)−f (x) 2(x+h) 2x 2(x+h) x 2x x+h = = = h h h x+h x−x−h −h x − 2x(x + h) 2x(x + h) 2x(x + h) 2x(x + h) = = = h h h 1 −1 1 −h · = , or − 2x(x + h) h 2x(x + h) 2x(x + h) 57. f (x) = − 1 4x f (x + h) = − 1 4(x + h) f (x + h) − f (x) = h − 1 − 4(x + h) h − 1 4x x 1 x+h 1 · − − · − 4(x + h) x 4x x+h = h x+h x + − 4x(x + h) 4x(x + h) = h h −x + x + h 4x(x + h) 4x(x + h) = = h h = c 2016 Pearson Education, Inc. Copyright 1 h/ ·1 1 h · = = 4x(x+h) h 4x(x+h)·h/ 4x(x+h) Exercise Set 2.2 58. f (x) = − 71 1 x 63. f (x) = 3×2 − 2x + 1 1 1 − − − f (x + h) − f (x) x+h x = = h h 1 x+h 1 x x x+h · − − · − − + x+h x x x+h x(x+h) x(x+h) = = h h h −x + x + h h 1 1 x(x + h) x(x + h) = = · = h h x(x + h) h x(x + h) f (x + h) = 3(x + h)2 − 2(x + h) + 1 = 3(x2 + 2xh + h2 ) − 2(x + h) + 1 = 3×2 + 6xh + 3h2 − 2x − 2h + 1 f (x) = 3×2 − 2x + 1 f (x + h) − f (x) = h (3×2 + 6xh + 3h2 −2x−2h + 1)−(3×2 − 2x + 1) = h 3×2 + 6xh + 3h2 − 2x − 2h + 1 − 3×2 + 2x − 1 = h h(6x + 3h − 2) 6xh + 3h2 − 2h = = h h·1 h 6x + 3h − 2 · = 6x + 3h − 2 h 1 59. f (x) = x2 + 1 f (x + h) = (x + h)2 + 1 = x2 + 2xh + h2 + 1 x2 + 2xh + h2 + 1 − (x2 + 1) f (x + h) − f (x) = h h = = x2 + 2xh + h2 + 1 − x2 − 1 h 2xh + h h 64. f (x) = 5×2 + 4x f (x+h)−f (x) (5×2+10xh+5h2+4x+4h)−(5×2+4x) = = h h 10xh + 5h2 + 4h = 10x + 5h + 4 h 2 h(2x + h) h h 2x + h = · h 1 = 2x + h = 65. f (x) = 4 + 5|x| f (x + h) = 4 + 5|x + h| 4 + 5|x + h| − (4 + 5|x|) f (x + h) − f (x) = h h 60. f (x) = x2 − 3 (x + h)2 − 3 − (x2 − 3) f (x + h) − f (x) = = h h x + 2xh + h − 3 − x + 3 2xh + h h(2x + h) = = = h h h 2x + h 2 2 2 2 61. f (x) = 4 − x2 f (x + h) = 4 − (x + h)2 = 4 − (x2 + 2xh + h2 ) = 4 − x2 − 2xh − h2 f (x + h) − f (x) 4 − x2 − 2xh − h2 − (4 − x2 ) = h h 4 − x2 − 2xh − h2 − 4 + x2 = h h/(−2x − h) −2xh − h2 = = h h/ = −2x − h −2xh − h2 2 − x2 − 2xh − h2 − 2 + x2 = = h h h(−2x − h) = −2x − h h 4 + 5|x + h| − 4 − 5|x| h = 5|x + h| − 5|x| h 66. f (x) = 2|x| + 3x f (x+h)−f (x) (2|x+h|+3x+3h)−(2|x|+3x) = = h h 2|x + h| − 2|x| + 3h h 67. f (x) = x3 f (x + h) = (x + h)3 = x3 + 3×2 h + 3xh2 + h3 f (x) = x3 x3 + 3×2 h + 3xh2 + h3 − x3 f (x + h) − f (x) = = h h 2 2 3 2 2 3x h + 3xh + h h(3x + 3xh + h ) = = h h·1 2 2 h 3x + 3xh + h · = 3×2 + 3xh + h2 h 1 62. f (x) = 2 − x2 2 − (x + h)2 − (2 − x2 ) f (x + h) − f (x) = = h h = 68. f (x) = x3 − 2x f (x+h)−f (x) (x+h)3 −2(x+h)−(x3 −2x) = = h h x3 + 3×2 h + 3xh2 + h3 −2x − 2h−x3 +2x = h 3×2 h+3xh2 + h3 −2h h(3×2 +3xh + h2 −2) = = h h 3×2 + 3xh + h2 − 2 c 2016 Pearson Education, Inc. Copyright 72 Chapter 2: More on Functions 69. f (x) = x−4 x+3 72. y 4 x+h−4 x−4 − f (x + h) − f (x) x = +h+3 x+3 = h h 2x y 4 2 4 2 2 4 x 2 x+h−4 x−4 − x + h + 3 x + 3 · (x + h + 3)(x + 3) = h (x + h + 3)(x + 3) 4 (x + h − 4)(x + 3) − (x − 4)(x + h + 3) = h(x + h + 3)(x + 3) 73. Graph x − 3y = 3. First we find the x- and y-intercepts. x2+hx−4x+3x+3h−12−(x2+hx+3x−4x−4h−12) = h(x+h+3)(x+3) x−3·0 = 3 x=3 x2 + hx − x + 3h − 12 − x2 − hx + x + 4h + 12 = h(x + h + 3)(x + 3) h 7 7h = · = h(x + h + 3)(x + 3) h (x + h + 3)(x + 3) 7 (x + h + 3)(x + 3) The x-intercept is (3, 0). 0 − 3y = 3 −3y = 3 y = −1 The y-intercept is (0, −1). We find a third point as a check. We let x = −3 and solve for y. x 70. f (x) = 2−x x+h x − f (x + h) − f (x) 2 − (x + h) 2 − x = = h h −3 − 3y = 3 −3y = 6 y = −2 (x + h)(2 − x) − x(2 − x − h) (2 − x − h)(2 − x) = h Another point on the graph is (−3, −2). We plot the points and draw the graph. y 2x − x2 + 2h − hx − 2x + x2 + hx (2 − x − h)(2 − x) = h 4 2 2h (2 − x − h)(2 − x) = h 4 2 x 3y 3 2 4 2 4 x 2 4 1 2 2h · = (2 − x − h)(2 − x) h (2 − x − h)(2 − x) 74. y 71. Graph y = 3x − 1. 4 We find some ordered pairs that are solutions of the equation, plot these points, and draw the graph. When x = −1, y = 3(−1) − 1 = −3 − 1 = −4. When x = 0, y = 3 · 0 − 1 = 0 − 1 = −1. 2 4 2 2 4 x y x2 1 When x = 2, y = 3 · 2 − 1 = 6 − 1 = 5. x y y −1 −4 4 0 −1 2 2 5 4 2 75. Answers may vary; f (x) = 76. The domain of h + f , h − f , and hf consists of all numbers that are in the domain of both h and f , or {−4, 0, 3}. 2 4 2 4 1 1 , g(x) = x+7 x−3 y 3x 1 x The domain of h/f consists of all numbers that are in the domain of both h and f , excluding any for which the value of f is 0, or {−4, 0}. 7 , and the domain of g(x) 77. The domain of h(x) is xx = 3 7 is {x|x = 3}, so and 3 are not in the domain of (h/g)(x). 3 We must also exclude the value of x for which g(x) = 0. c 2016 Pearson Education, Inc. Copyright Exercise Set 2.3 x4 − 1 =0 5x − 15 x4 − 1 = 0 73 17. Multiplying by 5x − 15 x4 = 1 x = ±1 Then the domain of (h/g)(x) is 7 xx = and x = 3 and x = −1 and x = 1 , or 3 7 7 ∪ , 3 ∪ (3, ∞). (−∞, −1) ∪ (−1, 1) ∪ 1, 3 3 Exercise Set 2.3 1. (g ◦ f )(x) = g(f (x)) = g(x + 3) = x + 3 − 3 = x The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 5 4 5 18. (f ◦ g)(x) = f x = · x=x 4 5 4 5 4 4 x = · x=x (g ◦ f )(x) = g 5 4 5 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 19. (f ◦ g)(−1) = f (g(−1)) = f ((−1)2 − 2(−1) − 6) = f (1 + 2 − 6) = f (−3) = 3(−3) + 1 = −9 + 1 = −8 2. (g ◦ f )(−2) = g(f (−2)) = g(3(−2) + 1) = g(−5) = (h ◦ f )(1) = h(f (1)) = h(3 · 1 + 1) = h(3 + 1) = 20. h(4) = 43 = 64 4. 1 1 1 1 3 =g h =g = =g (g ◦ h) 2 2 2 8 1 2 1 1 1 399 −6 = − −6=− −2 8 8 64 4 64 5. (g ◦ f )(5) = g(f (5)) = g(3 · 5 + 1) = g(15 + 1) = 6. g(16) = 162 − 2 · 16 − 6 = 218 1 1 1 1 2 =f g =f −6 = −2 (f ◦ g) 3 3 3 3 1 2 59 59 56 f − − 6 =f − =3 − + 1=− 9 3 9 9 3 7. (f ◦ h)(−3) = f (h(−3)) = f ((−3)3 ) = f (−27) = 3(−27) + 1 = −81 + 1 = −80 8. (h ◦ g)(3) = h(g(3)) = h(32 − 2 · 3 − 6) = h(9 − 6 − 6) = h(−3) = (−3)3 = −27 9. (g ◦ g)(−2) = g(g(−2)) = g((−2)2 − 2(−2) − 6) = g(4 + 4 − 6) = g(2) = 22 − 2 · 2 − 6 = 4 − 4 − 6 = −6 10. (g ◦ g)(3) = g(g(3)) = g(32 − 2 · 3 − 6) = g(9 − 6 − 6) = g(−3) = (−3)2 − 2(−3) − 6 = 9 + 6 − 6 = 9 11. (h ◦ h)(2) = h(h(2)) = h(23 ) = h(8) = 83 = 512 12. (h ◦ h)(−1) = h(h(−1)) = h((−1)3 ) = h(−1) = (−1)3 = −1 13. (f ◦ f )(−4) = f (f (−4)) = f (3(−4) + 1) = f (−12 + 1) = f (−11) = 3(−11) + 1 = −33 + 1 = −32 14. (f ◦ f )(1) = f (f (1)) = f (3 · 1 + 1) = f (3 + 1) = f (4) = 3 · 4 + 1 = 12 + 1 = 13 15. (h ◦ h)(x) = h(h(x)) = h(x3 ) = (x3 )3 = x9 16. (f ◦ f )(x) = f (f (x)) = f (3x + 1) = 3(3x + 1) + 1 = 9x + 3 + 1 = 9x + 4 (f ◦ g)(x) = f (g(x)) = f (3×2 −2x−1) = 3×2 −2x−1+1 = 3×2 − 2x (g ◦ f )(x) = g(f (x)) = g(x+1) = 3(x+1)2 −2(x+1)−1 = 3(x2 +2x+1)−2(x+1)−1 = 3×2 +6x+3−2x−2−1 = 3×2 + 4x The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). (−5)2 − 2(−5) − 6 = 25 + 10 − 6 = 29 3. (f ◦ g)(x) = f (g(x)) = f (x − 3) = x − 3 + 3 = x (f ◦ g)(x) = f (x2 + 5) = 3(x2 + 5) − 2 = 3×2 + 15 − 2 = 3×2 + 13 (g ◦ f )(x) = g(3x−2) = (3x−2)2 +5 = 9×2 −12x+4+5 = 9×2 − 12x + 9 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 21. (f ◦ g)(x) = f (g(x)) = f (4x−3) = (4x−3)2 −3 = 16×2 − 24x + 9 − 3 = 16×2 − 24x + 6 (g ◦ f )(x) = g(f (x)) = g(x2 −3) = 4(x2 −3)−3 = 4×2 − 12 − 3 = 4×2 − 15 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 22. (f ◦ g)(x) = f (2x − 7) = 4(2x − 7)2 − (2x − 7) + 10 = 4(4×2 − 28x + 49) − (2x − 7) + 10 = 16×2 − 112x + 196 − 2x + 7 + 10 = 16×2 − 114x + 213 (g ◦ f )(x) = g(4×2 − x + 10) = 2(4×2 − x + 10) − 7 = 8×2 − 2x + 20 − 7 = 8×2 − 2x + 13 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 1 4 4 = = = 23. (f ◦ g)(x) = f (g(x)) = f 1 5 x 1−5· 1− x x x 4 4x =4· = x−5 x−5 x−5 x 4 1 = = (g ◦ f )(x) = g(f (x)) = g 4 1 − 5x 1 − 5x 1 − 5x 1 − 5x = 1· 4 4 1 The domain of f is xx = and the domain of g is 5 {x|x = 0}. Consider the domain of f ◦ g. Since 0 is not in 1 the domain of g, 0 is not in the domain of f ◦ g. Since 5 1 is not in the domain of f , we know that g(x) cannot be . 5 1 We find the value(s) of x for which g(x) = . 5 c 2016 Pearson Education, Inc. Copyright 74 Chapter 2: More on Functions 1 1 = x 5 5=x Multiplying by 5x Thus 5 is also not in the domain of f ◦ g. Then the domain of f ◦g is {x|x = 0 and x = 5}, or (−∞, 0)∪(0, 5)∪(5, ∞). 1 Now consider the domain of g ◦ f . Recall that is not in 5 the domain of f , so it is not in the domain of g ◦ f . Now 0 is not in the domain of g but f (x) is never 0, so the domain 1 1 1 , or − ∞, ∪ ,∞ . of g ◦ f is xx = 5 5 5 24. 1 = 2x + 1 2x + 1 6 =6· = 1 1 2x + 1 6(2x + 1), or 12x + 6 6 1 1 1 = = = = (g ◦ f )(x) = g 6 12 12 + x x +1 2· +1 x x x x x = 1· 12 + x 12 + x The domain of f is {x|x = 0} and the domain of g 1 is xx = − . Consider the domain of f ◦ g. Since 2 1 1 − is not in the domain of g, − is not in the domain 2 2 of f ◦ g. Now 0 is not in the domain of f but g(x) 1 is never 0, so the domain of f ◦ g is xx = − , or 2 1 1 ∪ − ,∞ . − ∞, − 2 2 Now consider the domain of g ◦ f . Since 0 is not in the domain of f , then 0 is not in the domain of g ◦ f . Also, 1 since − is not in the domain of g, we find the value(s) of 2 1 x for which f (x) = − . 2 1 6 =− x 2 −12 = x Then the domain of g ◦ f is xx = −12 and x = 0 , or (f ◦ g)(x) = f (−∞, −12) ∪ (−12, 0) ∪ (0, ∞). x+7 25. (f ◦ g)(x) = f (g(x)) = f = 3 x+7 −7=x+7−7=x 3 3 (3x − 7) + 7 = (g ◦ f )(x) = g(f (x)) = g(3x − 7) = 3 3x =x 3 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 26. (f ◦ g)(x) = f (1.5x + 1.2) = 4 2 (1.5x + 1.2) − 3 5 4 x + 0.8 − = x 5 4 2 4 2 x− = 1.5 x − + 1.2 = (g ◦ f )(x) = g 3 5 3 5 x − 1.2 + 1.2 = x The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). √ √ 27. (f ◦ g)(x) = f (g(x)) = f ( x) = 2 x + 1 √ (g ◦ f )(x) = g(f (x)) = g(2x + 1) = 2x + 1 The domain of f is (−∞, ∞) and the domain of g is {x|x ≥ 0}. Thus the domain of f ◦ g is {x|x ≥ 0}, or [0, ∞). Now consider the domain of g ◦f . There are no restrictions on the domain of f , but the domain of g is {x|x ≥ 0}. Since 1 1 f (x) ≥ 0 for x ≥ − , the domain of g ◦ f is xx ≥ − , 2 2 1 or − , ∞ . 2 √ 28. (f ◦ g)(x) = f (2 − 3x) = 2 − 3x √ √ (g ◦ f )(x) = g( x) = 2 − 3 x The domain of f is {x|x ≥ 0} and the domain of g is 2 (−∞, ∞). Since g(x) ≥ 0 when x ≤ , the domain of f ◦ g 3 2 is − ∞, . 3 Now consider the domain of g ◦ f . Since the domain of f is {x|x ≥ 0} and the domain of g is (−∞, ∞), the domain of g ◦ f is {x|x ≥ 0}, or [0, ∞). 29. (f ◦ g)(x) = f (g(x)) = f (0.05) = 20 (g ◦ f )(x) = g(f (x)) = g(20) = 0.05 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). √ 30. (f ◦ g)(x) = ( 4 x)4 = x √ 4 (g ◦ f )(x) = x4 = |x| The domain of f is (−∞, ∞) and the domain of g is {x|x ≥ 0}, so the domain of f ◦ g is {x|x ≥ 0}, or [0, ∞). Now consider the domain of g ◦f . There are no restrictions on the domain of f and f (x) ≥ 0 for all values of x, so the domain is (−∞, ∞). 31. (f ◦ g)(x) = f (g(x)) = f (x2 − 5) = √ √ x2 − 5 + 5 = x2 = |x| √ (g ◦ f )(x) = g(f (x)) = g( x + 5) = √ ( x + 5)2 − 5 = x + 5 − 5 = x The domain of f is {x|x ≥ −5} and the domain of g is (−∞, ∞). Since x2 ≥ 0 for all values of x, then x2 −5 ≥ −5 for all values of x and the domain of g ◦ f is (−∞, ∞). Now consider the domain of f ◦g. There are no restrictions on the domain of g, so the domain of f ◦ g is the same as the domain of f , {x|x ≥ −5}, or [−5, ∞). √ 32. (f ◦ g)(x) = ( 5 x + 2)5 − 2 = x + 2 − 2 = x √ √ 5 (g ◦ f )(x) = 5 x5 − 2 + 2 = x5 = x The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). c 2016 Pearson Education, Inc. Copyright Exercise Set 2.3 33. 75 √ √ (f ◦ g)(x) = f (g(x)) = f ( 3 − x) = ( 3 − x)2 + 2 = 1−x = −1 x 1 − x = −x Multiplying by x 3−x+2=5−x (g ◦ f )(x) = g(f (x)) = g(x2 + 2) = √ √ 3 − x2 − 2 = 1 − x2 3 − (x2 + 2) = The domain of f is (−∞, ∞) and the domain of g is {x|x ≤ 3}, so the domain of f ◦ g is {x|x ≤ 3}, or (−∞, 3]. Now consider the domain of g ◦f . There are no restrictions on the domain of f and the domain of g is {x|x ≤ 3}, so we find the values of x for which f (x) ≤ 3. We see that x2 + 2 ≤ 3 for −1 ≤ x ≤ 1, so the domain of g ◦ f is {x| − 1 ≤ x ≤ 1}, or [−1, 1]. √ √ 34. (f ◦ g)(x) = f ( x2 − 25) = 1 − ( x2 − 25)2 = 1 − (x2 − 25) = 1 − x2 + 25 = 26 − x2 1=0 (g ◦ f )(x) = g(1 − x2 ) = (1 − x2 )2 − 25 = √ √ 1 − 2×2 + x4 − 25 = x4 − 2×2 − 24 1 +2 x − 2 = 1 x−2 2x − 3 1 + 2x − 4 x − 2 = x−2 = 1 1 x−2 x−2 2x − 3 x − 2 · = 2x − 3 = x−2 1 The domain of f is {x|x = 2} and the domain of g is {x|x = 0}, so 0 is not in the domain of f ◦ g. We find the value of x for which g(x) = 2. x+2 =2 x x + 2 = 2x 1 (g ◦ f )(x) = g x−2 The domain of f is (−∞, ∞) and the domain of g is {x|x ≤ −5 or x ≥ 5}, so the domain of f ◦ g is {x|x ≤ −5 or x ≥ 5}, or (−∞, −5] ∪ [5, ∞). Now consider the domain of g ◦ f . There are no restrictions on the domain of f and the domain of g is {x|x ≤ −5 or x ≥ 5}, so we find the values of x for 2 which f (x) ≤ √ −5 or f (x) √ ≥ 5. We 2see that 1 − x ≤ −5 when x ≤ − 6 or x ≥ 6 and 1 − x√≥ 5 has no solution, √ so the domain of g ◦ f is {x|x ≤ − 6 or x ≥ 6}, or √ √ (−∞, − 6] ∪ [ 6, ∞). 1 35. (f ◦ g)(x) = f (g(x)) = f = 1+x 1− 1 1+x 1 1+x 1+x−1 = 1+x = 1 1+x 1+ Then the domain of f ◦ g is (−∞, 0) ∪ (0, 2) ∪ (2, ∞). Now consider the domain of g ◦ f . Since the domain of f is {x|x = 2}, we know that 2 is not in the domain of g ◦ f . Since the domain of g is {x|x = 0}, we find the value of x for which f (x) = 0. 1 =0 x−2 1=0 1 1 = = x+1−x 1−x x x 1 x =1· =x 1 1 x The domain of f is {x|x = 0} and the domain of g is {x|x = −1}, so we know that −1 is not in the domain of f ◦ g. Since 0 is not in the domain of f , values of x for which g(x) = 0 are not in the domain of f ◦ g. But g(x) is never 0, so the domain of f ◦ g is {x|x = −1}, or (−∞, −1) ∪ (−1, ∞). Now consider the domain of g ◦ f . Recall that 0 is not in the domain of f . Since −1 is not in the domain of g, we know that g(x) cannot be −1. We find the value(s) of x for which f (x) = −1. 2=x 1+x x · =x 1+x 1 1−x = (g ◦ f )(x) = g(f (x)) = g x False equation We see that there are no values of x for which f (x) = −1, so the domain of g ◦ f is {x|x = 0}, or (−∞, 0) ∪ (0, ∞). 1 36. (f ◦ g)(x) = f x + 2 = x+2 x −2 x 1 1 = = x + 2 − 2x −x + 2 x x x x x = , or = 1· −x + 2 −x + 2 2−x We get a false equation, so there are no such values. Then the domain of g ◦ f is (−∞, 2) ∪ (2, ∞). 37. (f ◦ g)(x) = f (g(x)) = f (x + 1) = (x + 1)3 − 5(x + 1)2 + 3(x + 1) + 7 = x3 + 3×2 + 3x + 1 − 5×2 − 10x − 5 + 3x + 3 + 7 = x3 − 2×2 − 4x + 6 (g ◦ f )(x) = g(f (x)) = g(x3 − 5×2 + 3x + 7) = x3 − 5×2 + 3x + 7 + 1 = x3 − 5×2 + 3x + 8 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). c 2016 Pearson Education, Inc. Copyright 76 38. Chapter 2: More on Functions (g ◦ f )(x) = x3 + 2×2 − 3x − 9 − 1 = b) r = x3 + 2×2 − 3x − 10 (g ◦ f )(x) = (x − 1)3 + 2(x − 1)2 − 3(x − 1) − 9 = x3 − 3×2 + 3x − 1 + 2×2 − 4x + 2 − 3x + 3 − 9 = x3 − x2 − 4x − 5 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). h 2 2 h h h + 2π 2 2 πh2 2 S(h) = πh + 2 3 2 S(h) = πh 2 S(h) = 2π 53. f (x) = (t ◦ s)(x) = t(s(x)) = t(x − 3) = x − 3 + 4 = x + 1 39. h(x) = (4 + 3x)5 This is 4 + 3x to the 5th power. The most obvious answer is f (x) = x5 and g(x) = 4 + 3x. √ 40. f (x) = 3 x, g(x) = x2 − 8 1 (x − 2)4 This is 1 divided by (x − 2) to the 4th power. One obvious 1 answer is f (x) = 4 and g(x) = x − 2. Another possibility x 1 is f (x) = and g(x) = (x − 2)4 . x We have f (x) = x + 1. 54. The manufacturer charges m + 6 per drill. The chain store sells each drill for 150%(m + 6), or 1.5(m + 6), or 1.5m + 9. Thus, we have P (m) = 1.5m + 9. 41. h(x) = 55. Equations (a) − (f ) are in the form y = mx + b, so we can read the y-intercepts directly from the equations. Equa2 tions (g) and (h) can be written in this form as y = x − 2 3 and y = −2x + 3, respectively. We see that only equation (c) has y-intercept (0, 1). 1 42. f (x) = √ , g(x) = 3x + 7 x 56. None (See Exercise 55.) 43. f (x) = 57. If a line slopes down from left to right, its slope is negative. The equations y = mx + b for which m is negative are (b), (d), (f), and (h). (See Exercise 55.) x−1 , g(x) = x3 x+1 58. The equation for which |m| is greatest is the equation with the steepest slant. This is equation (b). (See Exercise 55.) 44. f (x) = |x|, g(x) = 9×2 − 4 2 + x3 2 − x3 √ 4 46. f (x) = x , g(x) = x − 3 45. f (x) = x6 , g(x) = √ 59. The only equation that has (0, 0) as a solution is (a). 60. Equations (c) and (g) have the same slope. (See Exercise 55.) x−5 x+2 √ √ 48. f (x) = 1 + x, g(x) = 1 + x 47. f (x) = x, g(x) = 61. Only equations (c) and (g) have the same slope and different y-intercepts. They represent parallel lines. 62. The only equations for which the product of the slopes is −1 are (a) and (f). 49. f (x) = x3 − 5×2 + 3x − 1, g(x) = x + 2 50. f (x) = 2×5/3 + 5×2/3 , g(x) = x − 1, or f (x) = 2×5 + 5×2 , g(x) = (x − 1)1/3 51. a) Use the distance formula, distance = rate × time. Substitute 3 for the rate and t for time. r(t) = 3t b) Use the formula for the area of a circle. 63. Only the composition (c ◦ p)(a) makes sense. It represents the cost of the grass seed required to seed a lawn with area a. 64. Answers may vary. One example is f (x) = 2x + 5 and x−5 . Other examples are found in Exercises 17, g(x) = 2 18, 25, 26, 32 and 35. A(r) = πr2 c) (A ◦ r)(t) = A(r(t)) = A(3t) = π(3t)2 = 9πt2 Chapter 2 Mid-Chapter Mixed Review This function gives the area of the ripple in terms of time t. 1. The statement is true. See page 100 in the text. 52. a) h = 2r S(r) = 2πr(2r) + 2πr2 2. The statement is false. See page 113 in the text. S(r) = 4πr2 + 2πr2 3. The statement is true. See Example 2 in Section 2.3 in the text, for instance. S(r) = 6πr 2 4. a) For x-values from 2 to 4, the y-values increase from 2 to 4. Thus the function is increasing on the interval (2, 4). c 2016 Pearson Education, Inc. Copyright Chapter 2 Mid-Chapter Mixed Review 77 b) For x-values from −5 to −3, the y-values decrease from 5 to 1. Also, for x-values from 4 to 5, the yvalues decrease from 4 to −3. Thus the function is decreasing on (−5, −3) and on (4, 5). c) For x-values from −3 to −1, y is 3. Thus the function is constant on (−3, −1). 5. From the graph we see that a relative maximum value of 6.30 occurs at x = −1.29. We also see that a relative minimum value of −2.30 occurs at x = 1.29. The graph starts rising, or increasing, from the left and stops increasing at the relative maximum. From this point it decreases to the relative minimum and then increases again. Thus the function is increasing on (−∞, −1.29) and on (1.29, ∞). It is decreasing on (−1.29, 1.29). 6. The x-values extend from −5 to −1 and from 2 to 5, so the domain is [−5, −1] ∪ [2, 5]. The y-values extend from −3 to 5, so the range is [−3, 5]. 1 7. A(h) = (h + 4)h 2 h2 1 + 2h A(h) = h2 + 2h, or 2 2  x − 5, for x ≤ −3,      2x + 3, for −3 0,   2 x, Since −5 ≤ −3, f (−5) = −5 − 5 = −10. Since −3 ≤ −3, f (−3) = −3 − 5 = −8. Since −3 0, f (6) = · 6 = 3. 2 x + 2, for x < −4, 9. g(x) = −x, for x ≥ −4 We create the graph in two parts. Graph g(x) = x + 2 for inputs less than −4. Then graph g(x) = −x for inputs greater than or equal to −4. 12. (g − f )(3) = g(3) − f (3) = (32 + 4) − (3 · 3 − 1) = 9 + 4 − (9 − 1) = 9+4−9+1 =5 13. 1 g 1 3 = (g/f ) 1 3 f 3 2 1 +4 3 = 1 3· −1 3 1 +4 9 = 1−1 37 = 9 0 1 does not exist. Since division by 0 is not defined, (g/f ) 3 14. f (x) = 2x + 5, g(x) = −x − 4 a) The domain of f and of g is the set of all real numbers, or (−∞, ∞). Then the domain of f + g, f − g, f g, and f f is also (−∞, ∞). For f /g we must exclude −4 since g(−4) = 0. Then the domain of f /g is (−∞, −4) ∪ (−4, ∞). 5 5 = 0. For g/f we must exclude − since f − 2 2 Then the domain of g/f is 5 5 ∪ − ,∞ . − ∞, − 2 2 b) (f +g)(x) = f (x)+g(x) = (2x+5)+(−x−4) = x+1 y (f − g)(x) = f (x) − g(x) = (2x + 5) − (−x − 4) = 2x + 5 + x + 4 = 3x + 9 4 (f g)(x) = f (x) · g(x) = (2x + 5)(−x − 4) = −2×2 − 8x − 5x − 20 = −2×2 − 13x − 20 2 ⫺4 ⫺2 2 4 x ⫺2 ⫺4 (f f )(x) = f (x) · f (x) = (2x + 5) · (2x + 5) = 4×2 + 10x + 10x + 25 = 4×2 + 20x + 25 2x + 5 f (x) = (f /g)(x) = g(x) −x − 4 −x − 4 g(x) = f (x) 2x + 5 √ 15. f (x) = x − 1, g(x) = x + 2 (g/f )(x) = 10. (f + g)(−1) = f (−1) + g(−1) = [3(−1) − 1] + [(−1)2 + 4] = −3 − 1 + 1 + 4 =1 11. (f g)(0) = f (0) · g(0) = (3 · 0 − 1) · (02 + 4) = −1 · 4 a) Any number can be an input for f , so the domain of f is the set of all real numbers, or (−∞, ∞). The domain of g consists of all values for which x+2 is nonnegative, so we have x + 2 ≥ 0, or x ≥ −2, or [−2, ∞). Then the domain of f + g, f − g, and f g is [−2, ∞). = −4 c 2016 Pearson Education, Inc. Copyright 78 Chapter 2: More on Functions The domain of f f is (−∞, ∞). Since g(−2) = 0, the domain of f /g is (−2, ∞). Since f (1) = 0, the domain of g/f is [−2, 1)∪(1, ∞). √ x+2 √ (f − g)(x) = f (x) − g(x) = x − 1 − x + 2 √ (f g)(x) = f (x) · g(x) = (x − 1) x + 2 b) (f + g)(x) = f (x) + g(x) = x − 1 + (f f )(x) = f (x) · f (x) = (x − 1)(x − 1) = x2 − x − x + 1 = x2 − 2x + 1 x−1 f (x) =√ (f /g)(x) = g(x) x+2 √ x+2 g(x) = (g/f )(x) = f (x) x−1 16. f (x) = 4x − 3 4(x + h) − 3 − (4x − 3) f (x + h) − f (x) = = h h 4h 4x + 4h − 3 − 4x + 3 = =4 h h 17. f (x) = 6 − x2 6 − (x + h)2 − (6 − x2 ) f (x + h) − f (x) = = h h 6−(x2 +2xh+h2 )−6+x2 6−x2 −2xh−h2 −6 + x2 = = h h h/(−2x − h) −2xh − h2 = = −2x − h h h/ · 1 18. (f ◦ g)(1) = f (g(1)) = f (1 + 1) = f (1 + 1) = f (2) = 5 · 2 − 4 = 10 − 4 = 6 3 19. (g ◦ h)(2) = g(h(2)) = g(22 − 2 · 2 + 3) = g(4 − 4 + 3) = g(3) = 33 + 1 = 27 + 1 = 28 20. (f ◦ f )(0) = f (f (0)) = f (5 · 0 − 4) = f (−4) = 5(−4) − 4 = −20 − 4 = −24 21. (h ◦ f )(−1) = h(f (−1)) = h(5(−1) − 4) = h(−5 − 4) = h(−9) = (−9)2 − 2(−9) + 3 = 81 + 18 + 3 = 102 1 22. (f ◦ g)(x) = f (g(x)) = f (6x + 4) = (6x + 4) = 3x + 2 2 1 1 x = 6 · x + 4 = 3x + 4 (g ◦ f )(x) = g(f (x)) = g 2 2 The domain of f and g is (−∞, ∞), so the domain of f ◦ g and g ◦ f is (−∞, ∞). √ √ 23. (f ◦ g)(x) = f (g(x)) = f ( x) = 3 x + 2 √ (g ◦ f )(x) = g(f (x)) = g(3x + 2) = 3x + 2 The domain of f is (−∞, ∞) and the domain of g is [0, ∞). Consider the domain of f ◦ g. Since any number can be an input for f , the domain of f ◦ g is the same as the domain of g, [0, ∞). Now consider the domain of g ◦ f . Since the inputs of g 2 must be nonnegative, we must have 3x+2 ≥ 0, or x ≥ − . 3 2 Thus the domain of g ◦ f is − , ∞ . 3 24. The graph of y = (h − g)(x) will be the same as the graph of y = h(x) moved down b units. 25. Under the given conditions, (f + g)(x) and (f /g)(x) have different domains if g(x) = 0 for one or more real numbers x. 26. If f and g are linear functions, then any real number can be an input for each function. Thus, the domain of f ◦ g = the domain of g ◦ f = (−∞, ∞). 27. This approach is not valid. Consider Exercise 23 in Exercise Set 2.3 in the text, for example. Since 4x (f ◦ g)(x) = , an examination of only this composed x−5 function would lead to the incorrect conclusion that the domain of f ◦ g is (−∞, 5) ∪ (5, ∞). However, we must also exclude from the domain of f ◦ g those values of x that are not in the domain of g. Thus, the domain of f ◦ g is (−∞, 0) ∪ (0, 5) ∪ (5, ∞). Exercise Set 2.4 1. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would coincide, so the graph is symmetric with respect to the y-axis. If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin. 2. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would coincide, so the graph is symmetric with respect to the y-axis. If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin. 3. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin. 4. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. c 2016 Pearson Education, Inc. Copyright Exercise Set 2.4 79 If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin. The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 8. y 5. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. 10 8 6 If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. 2 —10 —8 —6 —4 —2 —2 If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin. 8 10 x —8 The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. Test algebraically for symmetry with respect to the x-axis: y = |x + 5| Original equation −y = |x + 5| Replacing y by −y y = −|x + 5| Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test algebraically for symmetry with respect to the y-axis: y y = |x + 5| 5 Original equation y = | − x + 5| Replacing x by −x 4 3 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. 2 1 1 2 3 4 5 x Test algebraically for symmetry with respect to the origin: y = |x + 5| —2 —3 y = −| − x + 5| Simplifying —5 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin. Test algebraically for symmetry with respect to the x-axis: y = |x| − 2 Original equation −y = |x| − 2 Replacing y by −y y = −|x| + 2 9. y 5 4 5y = 4x + 5 Simplifying 1 —5 —4 —3 —2 —1 —1 Test algebraically for symmetry with respect to the y-axis: y = |x| − 2 Original equation y = | − x| − 2 Replacing x by −x y = |x| − 2 Simplifying 2 3 4 5 x —4 —5 Test algebraically for symmetry with respect to the origin: Original equation Replacing x by −x and y by −y Simplifying 1 —2 —3 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. −y = | − x| − 2 3 2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. y = |x| − 2 Original equation −y = | − x + 5| Replacing x by −x and y by −y y = |x | – 2 —4 y = −|x| + 2 6 —10 If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin. −y = |x| − 2 4 —6 If the graph were folded on the y-axis, the parts to the left and right of the y-axis would coincide, so the graph is symmetric with respect to the y-axis. —5 —4 —3 —2 —1 —1 2 —4 6. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis. 7. y = |x + 5 | 4 The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. Test algebraically for symmetry with respect to the x-axis: 5y = 4x + 5 Original equation 5(−y) = 4x + 5 Replacing y by −y −5y = 4x + 5 Simplifying 5y = −4x − 5 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. c 2016 Pearson Education, Inc. Copyright 80 Chapter 2: More on Functions Test algebraically for symmetry with respect to the y-axis: 5y = 4x + 5 Original equation 5y = 4(−x) + 5 Replacing x by −x 5y = −4x + 5 Simplifying 11. y 5 4 3 2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. 1 —5 —4 —3 —2 —1 —1 Test algebraically for symmetry with respect to the origin: 5y = 4x + 5 Original equation Replacing x by −x and y by −y 4 x 5 5y = 2 x 2 – 3 —5 The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin. Simplifying Test algebraically for symmetry with respect to the x-axis: The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 5y = 2×2 − 3 Original equation 5(−y) = 2x − 3 Replacing y by −y 2 −5y = 2×2 − 3 Simplifying 5y = −2x + 3 2 y The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. 5 4 3 —4 5y = 4x − 5 10. 2 —3 5(−y) = 4(−x) + 5 −5y = −4x + 5 —2 1 2x – 5 = 3y 3 Test algebraically for symmetry with respect to the y-axis: 2 1 —5 —4 —3 —2 —1 —1 1 2 3 4 5 x —2 5y = 2×2 − 3 Original equation 5y = 2(−x)2 − 3 Replacing x by −x 5y = 2×2 − 3 —3 —4 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. —5 The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. Test algebraically for symmetry with respect to the origin: Test algebraically for symmetry with respect to the x-axis: 5(−y) = 2(−x) − 3 2x − 5 = 3y Original equation 2x − 5 = 3(−y) Replacing y by −y −2x + 5 = 3y 5y = 2×2 − 3 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 12. y 2x − 5 = 3y Original equation 2(−x) − 5 = 3y Replacing x by −x 5 Simplifying 3 4 2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. 1 —5 —4 —3 —2 —1 —1 Test algebraically for symmetry with respect to the origin: 2(−x) − 5 = 3(−y) −2x − 5 = −3y Simplifying 5y = −2×2 + 3 Simplifying Test algebraically for symmetry with respect to the y-axis: 2x − 5 = 3y Replacing x by −x and y by −y −5y = 2×2 − 3 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. −2x − 5 = 3y Original equation 2 —2 1 2 3 4 5 x x 2 + 4 = 3y —3 Original equation —4 Replacing x by −x and y by −y —5 Simplifying The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin. The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. Test algebraically for symmetry with respect to the x-axis: 2x + 5 = 3y x2 + 4 = 3y Original equation x2 + 4 = 3(−y) Replacing y by −y −x − 4 = 3y 2 c 2016 Pearson Education, Inc. Copyright Simplifying Exercise Set 2.4 81 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test algebraically for symmetry with respect to the y-axis: x2 + 4 = 3y Original equation 14. y 5 Replacing x by −x (−x)2 + 4 = 3y 4 2 x + 4 = 3y The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test algebraically for symmetry with respect to the origin: Original equation x2 + 4 = 3y 2 (−x) + 4 = 3(−y) x2 + 4 = −3y 4 y=–— x 3 2 1 —5 —4 —3 —2 —1 —1 1 2 3 4 5 x —2 —3 —4 Replacing x by −x and y by −y —5 Simplifying The graph is not symmetric with respect to the x-axis or the y-axis. It is symmetric with respect to the origin. The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. Test algebraically for symmetry with respect to the x-axis: 4 y=− Original equation x 4 −y = − Replacing y by −y x 4 Simplifying y= x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. −x − 4 = 3y 2 13. y 5 4 1 y=— x 3 2 1 —5 —4 —3 —2 —1 —1 1 2 3 4 5 x —2 —3 —4 —5 The graph is not symmetric with respect to the x-axis or the y-axis. It is symmetric with respect to the origin. Test algebraically for symmetry with respect to the x-axis: 1 Original equation y= x 1 Replacing y by −y −y = x 1 y=− Simplifying x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test algebraically for symmetry with respect to the y-axis: 1 Original equation y= x 1 Replacing x by −x y= −x 1 y=− Simplifying x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test algebraically for symmetry with respect to the origin: 1 Original equation y= x 1 −y = Replacing x by −x and y by −y −x 1 Simplifying y= x Test algebraically for symmetry with respect to the y-axis: 4 Original equation y=− x 4 y=− Replacing x by −x −x 4 Simplifying y= x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test algebraically for symmetry with respect to the origin: 4 Original equation y=− x 4 −y = − Replacing x by −x and y by −y −x 4 y=− Simplifying x The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 15. Test for symmetry with respect to the x-axis: 5x − 5y = 0 Original equation 5x − 5(−y) = 0 Replacing y by −y 5x + 5y = 0 Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 5x − 5y = 0 Original equation 5(−x) − 5y = 0 Replacing x by −x −5x − 5y = 0 5x + 5y = 0 c 2016 Pearson Education, Inc. Copyright Simplifying 82 Chapter 2: More on Functions The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 5x − 5y = 0 5(−x) − 5(−y) = 0 −5x + 5y = 0 Original equation 18. Test for symmetry with respect to the x-axis: 5y = 7×2 − 2x Original equation 5(−y) = 7×2 − 2x Replacing y by −y 5y = −7x + 2x Simplifying 2 Replacing x by −x and y by −y The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Simplifying Test for symmetry with respect to the y-axis: 5x − 5y = 0 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 16. Test for symmetry with respect to the x-axis: 5y = 7×2 − 2x Original equation 5y = 7(−x)2 − 2(−x) Replacing x by −x 2 5y = 7x + 2x Simplifying 6x + 7y = 0 Original equation The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. 6x + 7(−y) = 0 Replacing y by −y Test for symmetry with respect to the origin: 6x − 7y = 0 5y = 7×2 − 2x Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 6x + 7y = 0 Original equation 6(−x) + 7y = 0 Replacing x by −x 6x − 7y = 0 6x + 7y = 0 Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. Simplifying Test for symmetry with respect to the origin: 6(−x) + 7(−y) = 0 −5y = 7×2 + 2x Replacing x by −x and y by −y 5y = −7×2 − 2x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. 6x + 7y = 0 Original equation 5(−y) = 7(−x)2 − 2(−x) Original equation 19. Test for symmetry with respect to the x-axis: y = |2x| Original equation −y = |2x| Replacing y by −y y = −|2x| Simplifying Replacing x by −x and y by −y The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Simplifying Test for symmetry with respect to the y-axis: The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 17. Test for symmetry with respect to the x-axis: 3×2 − 2y 2 = 3 Original equation 3×2 − 2(−y)2 = 3 Replacing y by −y 3×2 − 2y 2 = 3 Simplifying y = |2x| Original equation y = |2(−x)| Replacing x by −x y = | − 2x| Simplifying y = |2x| The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the origin: Test for symmetry with respect to the y-axis: y = |2x| Original equation 3×2 − 2y 2 = 3 Original equation −y = |2(−x)| Replacing x by −x and y by −y −y = | − 2x| Simplifying 3(−x)2 − 2y 2 = 3 Replacing x by −x −y = |2x| 3×2 − 2y 2 = 3 y = −|2x| Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 3×2 − 2y 2 = 3 Original equation 3(−x)2 − 2(−y)2 = 3 Replacing x by −x and y by −y 3×2 − 2y 2 = 3 Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 20. Test for symmetry with respect to the x-axis: y 3 = 2×2 Original equation (−y)3 = 2×2 Replacing y by −y −y = 2x 3 2 Simplifying y 3 = −2×2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. c 2016 Pearson Education, Inc. Copyright Exercise Set 2.4 83 Test for symmetry with respect to the y-axis: y 3 = 2×2 Original equation y 3 = 2(−x)2 Replacing x by −x 23. Test for symmetry with respect to the x-axis: Simplifying y 3 = 2×2 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: Original equation y 3 = 2×2 Replacing x by −x and y by −y (−y)3 = 2(−x)2 −y 3 = 2×2 Simplifying y 3 = −2×2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 21. Test for symmetry with respect to the x-axis: Original equation 2×4 + 3 = y 2 Replacing y by −y 2×4 + 3 = (−y)2 Simplifying 2×4 + 3 = y 2 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 2×4 + 3 = y 2 Original equation Replacing x by −x 2(−x)4 + 3 = y 2 2×4 + 3 = y 2 Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: Original equation 2×4 + 3 = y 2 4 2 2(−x) + 3 = (−y) Replacing x by −x and y by −y Simplifying 2×4 + 3 = y 2 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 22. Test for symmetry with respect to the x-axis: 2y 2 = 5×2 + 12 Original equation 2(−y)2 = 5×2 + 12 Replacing y by −y 2y 2 = 5×2 + 12 Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: Original equation 2y 2 = 5×2 + 12 2y 2 = 5(−x)2 + 12 Replacing x by −x 2y 2 = 5×2 + 12 Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: Original equation 2y 2 = 5×2 + 12 2(−y)2 = 5(−x)2 + 12 2 2 2y = 5x + 12 Replacing x by −x and y by −y Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 3y 3 = 4×3 + 2 Original equation 3(−y)3 = 4×3 + 2 Replacing y by −y −3y = 4x + 2 3 3 Simplifying 3y 3 = −4×3 − 2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 3y 3 = 4×3 + 2 Original equation 3y = 4(−x) + 2 Replacing x by −x 3y 3 = −4×3 + 2 Simplifying 3 3 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 3y 3 = 4×3 + 2 Original equation 3(−y)3 = 4(−x)3 + 2 Replacing x by −x and y by −y −3y 3 = −4×3 + 2 Simplifying 3y = 4x − 2 3 3 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 24. Test for symmetry with respect to the x-axis: 3x = |y| Original equation 3x = | − y| Replacing y by −y 3x = |y| Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 3x = |y| Original equation 3(−x) = |y| Replacing x by −x −3x = |y| Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 3x = |y| Original equation 3(−x) = | − y| Replacing x by −x and y by −y −3x = |y| Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 25. Test for symmetry with respect to the x-axis: xy = 12 Original equation x(−y) = 12 Replacing y by −y −xy = 12 Simplifying xy = −12 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. c 2016 Pearson Education, Inc. Copyright 84 Chapter 2: More on Functions Test for symmetry with respect to the y-axis: xy = 12 Original equation −xy = 12 31. x-axis: Replace y with −y; (0, 4) y-axis: Replace x with −x; (0, −4) Replacing x by −x xy = −12 Origin: Replace x with −x and y with −y; (0, 4) Simplifying 32. x-axis: Replace y with −y; (8, 3) The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: xy = 12 Original equation −x(−y) = 12 Replacing x by −x and y by −y xy = 12 Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 26. Test for symmetry with respect to the x-axis: Original equation xy − x2 = 3 Replacing y by −y x(−y) − x2 = 3 xy + x = −3 2 Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: Original equation xy − x2 = 3 −xy − (−x)2 = 3 xy + x = −3 2 34. The graph is symmetric with respect to the y-axis, so the function is even. 35. The graph is symmetric with respect to the origin, so the function is odd. 36. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd. 37. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd. 38. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd. 39. f (x) = −3×3 + 2x f (−x) = −3(−x)3 + 2(−x) = 3×3 − 2x Simplifying −f (x) = −(−3×3 + 2x) = 3×3 − 2x Test for symmetry with respect to the origin: xy − x2 = 3 Original equation xy − x2 = 3 33. The graph is symmetric with respect to the y-axis, so the function is even. Replacing x by −x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. −x(−y) − (−x)2 = 3 y-axis: Replace x with −x; (−8, −3) Origin: Replace x with −x and y with −y; (−8, 3) f (−x) = −f (x), so f is odd. 40. f (−x) = 7(−x)3 + 4(−x) − 2 = −7×3 − 4x − 2 −f (x) = −(7×3 + 4x − 2) = −7×3 − 4x + 2 Replacing x by −x and y by −y f (x) = f (−x), so f is not even. f (−x) = −f (x), so f is not odd. Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. f (x) = 7×3 + 4x − 2 Thus, f (x) = 7×3 + 4x − 2 is neither even nor odd. 41. 27. x-axis: Replace y with −y; (−5, −6) f (x) = 5×2 + 2×4 − 1 f (−x) = 5(−x)2 + 2(−x)4 − 1 = 5×2 + 2×4 − 1 y-axis: Replace x with −x; (5, 6) f (x) = f (−x), so f is even. Origin: Replace x with −x and y with −y; (5, −6) 7 ,0 28. x-axis: Replace y with −y; 2 7 y-axis: Replace x with −x; − , 0 2 7 Origin: Replace x with −x and y with −y; − , 0 2 29. x-axis: Replace y with −y; (−10, 7) 42. f (x) = x + 1 x 1 1 f (−x) = −x + = −x − −x x 1 1 = −x − −f (x) = − x + x x f (−x) = −f (x), so f is odd. 43. f (x) = x17 f (−x) = (−x)17 = −x17 y-axis: Replace x with −x; (10, −7) Origin: Replace x with −x and y with −y; (10, 7) 3 30. x-axis: Replace y with −y; 1, − 8 3 y-axis: Replace x with −x; − 1, 8 3 Origin: Replace x with −x and y with −y; − 1, − 8 −f (x) = −x17 f (−x) = −f (x), so f is odd. √ f (x) = 3 x 44. √ √ f (−x) = 3 −x = − 3 x √ −f (x) = − 3 x f (−x) = −f (x), so f is odd. c 2016 Pearson Education, Inc. Copyright Exercise Set 2.5 45. 85 f (x) = x − |x| 53. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis. f (−x) = (−x) − |(−x)| = −x − |x| −f (x) = −(x − |x|) = −x + |x| If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. f (x) = f (−x), so f is not even. f (−x) = −f (x), so f is not odd. Thus, f (x) = x − |x| is neither even nor odd. 1 46. f (x) = 2 x 1 1 f (−x) = = 2 (−x)2 x f (x) = f (−x), so f is even. 47. f (−x) = 8 If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin. f (x) = f (−x), so f is even. √ 48. f (x) = x2 + 1 √ f (−x) = (−x)2 + 1 = x2 + 1 55. See the answer section in the text. f (x) = f (−x), so f is even. f (−x) − f (x) f (−x) − f (−(−x)) = , 2 2 f (−x) − f (x) f (x) − f (−x) = . Thus, −O(x) = − 2 2 O(−x) = −O(x) and O is odd. 56. O(−x) = y 4 2 4 2 2 2 4 x 57. a), b) See the answer section in the text. 4 50. Let v = the number of volunteers from the University of Wisconsin – Madison. Then v + 464 = the number of volunteers from the University of California – Berkeley. Solve: v + (v + 464) = 6688 v = 3112, so there were 3112 volunteers from the University of Wisconsin – Madison and 3112 + 464, or 3576 volunteers from the University of California – Berkeley. √ 51. f (x) = x 10 − x2 √ f (−x) = −x 10 − (−x)2 = −x 10 − x2 √ −f (x) = −x 10 − x2 Since f (−x) = −f (x), f is odd. 52. f (x) = f (−x) = 54. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. f (x) = 8 49. If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin. 58. Let f (x) = g(x) = x. Now f and g are odd functions, but (f g)(x) = x2 = (f g)(−x). Thus, the product is even, so the statement is false. 59. Let f (x) and g(x) be even functions. Then by definition, f (x) = f (−x) and g(x) = g(−x). Thus, (f + g)(x) = f (x) + g(x) = f (−x) + g(−x) = (f + g)(−x) and f + g is even. The statement is true. 60. Let f (x) be an even function, and let g(x) be an odd function. By definition f (x) = f (−x) and g(−x) = −g(x), or g(x) = −g(−x). Then f g(x) = f (x) · g(x) = f (−x) · [−g(−x)] = −f (−x) · g(−x) = −f g(−x), and f g is odd. The statement is true. Exercise Set 2.5 x2 + 1 x3 + 1 1. Shift the graph of f (x) = x2 right 3 units. x2 + 1 (−x)2 + 1 = 3 (−x) + 1 −x3 + 1 y 4 x2 + 1 −f (x) = − 3 x +1 Since f (x) = f (−x), f is not even. 2 4 2 2 Since f (−x) = −f (x), f is not odd. 2 x2 + 1 is neither even nor odd. Thus, f (x) = 3 x +1 4 c 2016 Pearson Education, Inc. Copyright 4 x f(x) (x 3)2 86 Chapter 2: More on Functions 2. Shift the graph of g(x) = x2 up 1 unit. 2 6. Shift the graph of g(x) = √ x right 1 unit. y y 4 4 2 2 4 2 2 4 2 x 4 2 x 4 2 2 4 4 g(x) √ x 1 1 g(x) x 2 2 1 up 4 units. x 7. Shift the graph of h(x) = 3. Shift the graph of g(x) = x down 3 units. y y 8 4 6 2 4 2 4 2 4 x 1 h(x) x 4 2 2 4 2 g(x) x 3 2 4 x 2 4. Reflect the graph of g(x) = x across the x-axis and then shift it down 2 units. 8. Shift the graph of g(x) = 1 right 2 units. x y y 4 4 2 2 4 2 2 4 2 x 4 2 2 2 g(x) x 2 5. Reflect the graph of h(x) = √ x across the x-axis. y h(x) 3x 3 h(x) √x 2 4 2 x 9. First stretch the graph of h(x) = x vertically by multiplying each y-coordinate by 3. Then reflect it across the x-axis and shift it up 3 units. y 4 4 1 g(x) x 2 4 4 4 2 2 4 x 4 2 2 2 2 4 4 4 x 10. First stretch the graph of f (x) = x vertically by multiplying each y-coordinate by 2. Then shift it up 1 unit. y 4 2 4 2 2 4 2 4 f(x) 2x 1 c 2016 Pearson Education, Inc. Copyright x Exercise Set 2.5 87 11. First shrink the graph of h(x) = |x| vertically by multiply1 ing each y-coordinate by . Then shift it down 2 units. 2 15. Shift the graph of g(x) = x2 left 1 unit and down 1 unit. y y 4 4 2 1 h (x) |x| 2 2 4 2 2 2 4 x 2 4 2 4 x 4 2 g(x) (x 1)2 1 4 12. Reflect the graph of g(x) = |x| across the x-axis and shift it up 2 units. 16. Reflect the graph of h(x) = x2 across the x-axis and down 4 units. y 4 2 y 2 x 4 2 4 4 2 6 4 2 2 x 4 8 2 h(x) x 2 4 4 g(x) x 2 13. Shift the graph of g(x) = x3 right 2 units and reflect it across the x-axis. 17. First shrink the graph of g(x) = x3 vertically by multiply1 ing each y-coordinate by . Then shift it up 2 units. 3 y y 4 4 g (x) (x 2)3 2 2 4 4 2 2 4 2 2 x 2 4 4 x 1 3 g(x) x 2 3 4 18. Reflect the graph of h(x) = x3 across the y-axis. 14. Shift the graph of f (x) = x3 left 1 unit. y y 4 4 2 2 4 2 4 2 2 4 2 x 2 x 4 2 4 4 f(x) (x 1)3 h(x) (x)3 √ 19. Shift the graph of f (x) = y 4 2 4 2 2 4 x 2 4 c 2016 Pearson Education, Inc. Copyright f(x) √ x 2 x left 2 units. 88 Chapter 2: More on Functions 20. First shift the graph of f (x) = √ x right 1 unit. Shrink it 1 vertically by multiplying each y-coordinate by and then 2 reflect it across the x-axis. y 4 29. Think of the graph of f (x) = |x|. Since g(x) = 1 1 x − 4, the graph of g(x) = x − 4 is the graph of f 3 3 2 4 2 2 x 4 2 4 f (x) = |x| stretched horizontally by multiplying each xcoordinate by 3 and then shifted down 4 units. 1 f(x) √ x 1 2  21. Shift the graph of f (x) = √ 3 x down 2 units. y 4 3 f(x) √x 2 2 30. Think of the graph of g(x) = x3 . Since 2 2 f (x) = g(x) − 4, the graph of f (x) = x3 − 4 is the 3 3 graph of g(x) = x3 shrunk vertically by multiplying each 2 y-coordinate by and then shifted down 4 units. 3 1 31. Think of the graph of g(x) = x2 . Since f (x) = − g(x − 5), 4 1 the graph of f (x) = − (x − 5)2 is the graph of g(x) = x2 4 shifted right 5 units, shrunk vertically by multiplying each 1 y-coordinate by , and reflected across the x-axis. 4 2 4 2 1 . Since f (x) = 5 − g(x), or x 1 f (x) = −g(x) + 5, the graph of f (x) = 5 − is the graph x 1 of g(x) = reflected across the x-axis and then shifted up x 5 units. 28. Think of the graph of g(x) = x 2 4 √ 22. Shift the graph of h(x) = 3 x left 1 unit. 32. Think of the graph of g(x) = x3 . Since f (x) = g(−x) − 5, the graph of f (x) = (−x)3 − 5 is the graph of g(x) = x3 reflected across the y-axis and shifted down 5 units. y 4 1 . Since f (x) = x 1 + 2 is the graph g(x + 3) + 2, the graph of f (x) = x+3 1 of g(x) = shifted left 3 units and up 2 units. x √ 34. Think of the graph of√f (x) = x. Since g(x) = f (−x) + √5, the graph of g(x) = −x + 5 is the graph of f (x) = x reflected across the y-axis and shifted up 5 units. 33. Think of the graph of g(x) = 2 4 2 2 4 x 2 4 3 h(x) √ x 1 23. Think of the graph of f (x) = |x|. Since g(x) = f (3x), the graph of g(x) = |3x| is the graph of f (x) = |x| shrunk horizontally by dividing each x 1 . coordinate by 3 or multiplying each x-coordinate by 3 √ 1 24. Think of the graph of g(x) = 3 x. Since f (x) = g(x), the 2 √ 1√ graph of f (x) = 3 x is the graph of g(x) = 3 x shrunk 2 1 vertically by multiplying each y-coordinate by . 2 1 25. Think of the graph of f (x) = . Since h(x) = 2f (x), x 1 2 the graph of h(x) = is the graph of f (x) = stretched x x vertically by multiplying each y-coordinate by 2. 26. Think of the graph of g(x) = |x|. Since f (x) = g(x−3)−4, the graph of f (x) = |x − 3| − 4 is the graph of g(x) = |x| shifted right 3 units and down 4 units. √ 27. Think of the graph of g(x) √ = x. Since f (x) = 3g(x) −√5, the graph of f (x) = 3 x − 5 is the graph of g(x) = x stretched vertically by multiplying each y-coordinate by 3 and then shifted down 5 units. 35. Think of the graph of f (x) = x2 . Since h(x) = −f (x−3)+ 5, the graph of h(x) = −(x − 3)2 + 5 is the graph of f (x) = x2 shifted right 3 units, reflected across the x-axis, and shifted up 5 units. 36. Think of the graph of g(x) = x2 . Since f (x) = 3g(x + 4)− 3, the graph of f (x) = 3(x+4)2 −3 is the graph of g(x) = x2 shifted left 4 units, stretched vertically by multiplying each y-coordinate by 3, and then shifted down 3 units. 37. The graph of y = g(x) is the graph of y = f (x) shrunk 1 vertically by a factor of . Multiply the y-coordinate by 2 1 : (−12, 2). 2 38. The graph of y = g(x) is the graph of y = f (x) shifted right 2 units. Add 2 to the x-coordinate: (−10, 4). 39. The graph of y = g(x) is the graph of y = f (x) reflected across the y-axis, so we reflect the point across the y-axis: (12, 4). c 2016 Pearson Education, Inc. Copyright Exercise Set 2.5 89 40. The graph of y = g(x) is the graph of y = f (x) shrunk 1 the horizontally. The x-coordinates of y = g(x) are 4 corresponding x-coordinates of y = f (x), so we divide the 1 : (−3, 4). x-coordinate by 4 or multiply it by 4 54. Shape: h(x) = x2 Shift h(x) right 6 units: g(x) = h(x − 6) = (x − 6)2 Shift g(x) up 2 units: f (x) = g(x) + 2 = (x − 6)2 + 2 55. Shape: m(x) = x2 41. The graph of y = g(x) is the graph of y = f (x) shifted down 2 units. Subtract 2 from the y-coordinate: (−12, 2). Turn m(x) upside-down (that is, reflect it across the xaxis): h(x) = −m(x) = −x2 42. The graph of y = g(x) is the graph of y = f (x) stretched horizontally. The x-coordinates of y = g(x) are twice the corresponding x-coordinates of y = f (x), so we multiply 1 : (−24, 4). the x-coordinate by 2 or divide it by 2 Shift g(x) up 4 units: f (x) = g(x) + 4 = −(x − 3)2 + 4 43. The graph of y = g(x) is the graph of y = f (x) stretched vertically by a factor of 4. Multiply the y-coordinate by 4: (−12, 16). 44. The graph of y = g(x) is the graph y = f (x) reflected across the x-axis. Reflect the point across the x-axis: (−12, −4). 45. g(x) = x2 + 4 is the function f (x) = x2 + 3 shifted up 1 unit, so g(x) = f (x) + 1. Answer B is correct. 46. If we substitute 3x for x in f , we get 9×2 + 3, so g(x) = f (3x). Answer D is correct. 47. If we substitute x − 2 for x in f , we get (x − 2)3 + 3, so g(x) = f (x − 2). Answer A is correct. 2 2 48. If we multiply x + 3 by 2, we get 2x + 6, so g(x) = 2f (x). Answer C is correct. 49. Shape: h(x) = x2 Turn h(x) upside-down (that is, reflect it across the xaxis): g(x) = −h(x) = −x2 Shift g(x) right 8 units: f (x) = g(x − 8) = −(x − 8)2 √ 50. Shape: h(x) = x √ Shift h(x) left 6 units: g(x) = h(x + 6) = x + 6 √ Shift g(x) down 5 units: f (x) = g(x) − 5 = x + 6 − 5 Shift h(x) right 3 units: g(x) = h(x − 3) = −(x − 3)2 56. Shape: h(x) = |x| Stretch h(x) horizontally by a factor of 2 that is, multiply 1 1 1 : g(x) = h x = x each x-value by 2 2 2 1 Shift g(x) down 5 units: f (x) = g(x) − 5 = x − 5 2 √ 57. Shape: m(x) = x √ Reflect m(x) across the y-axis: h(x) = m(−x) = −x Shift h(x) left 2 units: g(x) = h(x + 2) = −(x + 2) Shift g(x) down 1 unit: f (x) = g(x) − 1 = −(x + 2) − 1 58. Shape: h(x) = Reflect h(x) across the x-axis: g(x) = −h(x) = − Shift g(x) up 1 unit: f (x) = g(x) + 1 = − 1 +1 x y (⫺1, 4) (⫺3, 4) g(x) ⫽ ⫺2f(x) 4 (5, 0) (⫺4, 0) ⫺4 ⫺2 2 4 6 x ⫺2 ⫺4 Shift h(x) left 7 units: g(x) = h(x + 7) = |x + 7| Shift g(x) up 2 units: f (x) = g(x) + 2 = |x + 7| + 2 52. Shape: h(x) = x3 Turn h(x) upside-down (that is, reflect it across the xaxis): g(x) = −h(x) = −x3 ⫺6 (2, ⫺6) 1 60. Each y-coordinate is multiplied by . We plot and connect 2 (−4, 0), (−3, −1), (−1, −1), (2, 1.5), and (5, 0). y Shift g(x) right 5 units: f (x) = g(x − 5) = −(x − 5)3 1 x 1 x 59. Each y-coordinate is multiplied by −2. We plot and connect (−4, 0), (−3, 4), (−1, 4), (2, −6), and (5, 0). 51. Shape: h(x) = |x| 53. Shape: h(x) = 1 x 4 1 Shrink h(x) vertically by a factor of 2 1 : multiply each function value by 2 1 1 1 1 g(x) = h(x) = · , or 2 2 x 2x (4, 0) that is, 2 (2, 1.5) 4 2 2 (3, 1) (1, 1) 4 Shift g(x) down 3 units: f (x) = g(x) − 3 = (5, 0) 4 1 −3 2x c 2016 Pearson Education, Inc. Copyright x g(x) qf(x) 90 Chapter 2: More on Functions 61. The graph is reflected across the y-axis and stretched horizontally by a factor of 2. That is, each x-coordinate is 1 multiplied by −2 or divided by − . We plot and con2 nect (8, 0), (6, −2), (2, −2), (−4, 3), and (−10, 0). y 2 (⫺10, 0) 2 ⫺4 2 4 (2, ⫺2) 6 4 (5, 4) (1, 4) x (4, 4) 12 (1, 13) g(x) 3f(x 1) 4 (8, 0) ⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺2 4 (4, 2) g (x) ⫽ f 冢⫺qx冣 4 (⫺4, 3) y (2, 2) 8 x (6, ⫺2) 65. The graph is reflected across the y-axis so each x-coordinate is replaced by its opposite. y 62. The graph is shrunk horizontally by a factor of 2. That 1 . is, each x-coordinate is divided by 2 or multiplied by 2 We plot and connect (−2, 0), (−1.5, −2), (−0.5, −2), (1, 3), and (2.5, 0). y 4 2 (1.5, 2) 4 4 (5, 0) (4, 0) 6 4 2 2 2 4 6 x (3, 2) (1, 2) 4 y (2.5, 0) (2, 0) 4 2 66. The graph is reflected across the x-axis so each y-coordinate is replaced by its opposite. (1, 3) 2 g(x) f (x ) 4 (2, 3) (1, 2) x (3, 2) g(x) f (x ) 4 2 (0.5, 2) (4, 0) g(x) f (2x) 6 4 2 (5, 0) 2 4 6 x 2 63. The graph is shifted right 1 unit so each x-coordinate is increased by 1. The graph is also reflected across the xaxis, shrunk vertically by a factor of 2, and shifted up 3 1 units. Thus, each y-coordinate is multiplied by − and 2 then increased by 3. We plot and connect (−3, 3), (−2, 4), (0, 4), (3, 1.5), and (6, 3). y 6 (⫺2, 4) (⫺3, 3) (9, 1) ⫺2 ⫺4 4 y 4 (3, 3) (2, 1) 2 (3, 0) 10 8 6 4 2 (3, 1.5) 2 67. The graph is shifted left 2 units so each x-coordinate is decreased by 2. It is also reflected across the x-axis so each y-coordinate is replaced with its opposite. In addition, the graph is shifted up 1 unit, so each y-coordinate is then increased by 1. h(x) g (x 2) 1 (0, 4) (1, 3) (6, 3) 2 ⫺4 ⫺2 (2, 3) 4 (1, 0) 2 4 6 x 2 6 x (7, 3) (4, 3) 4 (0, 3) g (x) ⫽ ⫺qf (x ⫺ 1) ⫹ 3 (5, 5) 6 64. The graph is shifted left 1 unit so each x-coordinate is decreased by 1. The graph is also reflected across the x-axis, stretched vertically by a factor of 3, and shifted down 4 units. Thus, each y-coordinate is multiplied by −3 and then decreased by 4. We plot and connect (−5, −4), (−4, 2), (−2, 2), (1, −13), and (4, −4). 68. The graph is reflected across the y-axis so each x-coordinate is replaced with its opposite. It is also shrunk 1 vertically by a factor of , so each y-coordinate is multi2 1 plied by (or divided by 2). 2 y (7, 3) 4 (0, 0) (2, 2) 2 8 6 4 2 (5, 1) 2 1 (1, ) 2 4 c 2016 Pearson Education, Inc. Copyright (2, 2) (5, 2) (7, 0) 2 4 6 8 x 1 (1, ) 2 1 h(x) g(x) 2 Exercise Set 2.5 91 69. The graph is shrunk horizontally. The x-coordinates of y = h(x) are one-half the corresponding x-coordinates of y = g(x). y 4 7 2 (, 2 0) 7 6 5 (, 2 4) by a factor of 3 that is, each y-coordinate is multiplied 1 and then shifted down 3 units. This is graph (e). by 3 1 77. g(x) = f (x + 2) 3 The graph of g(x) is the graph of f (x) shrunk vertically h(x) g (2x ) 8 (1, 4) (1, 4) (, 2 6) 1 (, 2 1) 4 2 1 2 (, 2 1) 4 2 4 by a factor of 3 that is, each y-coordinate is multiplied 1 and then shifted left 2 units. This is graph (c). by 3 78. g(x) = −f (x + 2) x 5 (, 2 2) (0, 0) 70. The graph is shifted right 1 unit, so each x-coordinate is increased by 1. It is also stretched vertically by a factor of 2, so each y-coordinate is multiplied by 2 or divided 1 . In addition, the graph is shifted down 3 units, so by 2 each y-coordinate is decreased by 3. y 10 (8, 9) 6 (4, 5) (0, 1) 2 (6, 3) 4 6 8 2 4 (1, 3) 6 8 x (2, 1) 1 1 (−x)4 + (−x)3 − 81(−x)2 − 17 = 4 5 83. The graph of f (x) = x3 − 3×2 is shifted left 1 unit. A formula for the transformed function is k(x) = f (x + 1), or k(x) = (x + 1)3 − 3(x + 1)2 . (6, 7) 71. g(x) = f (−x) + 3 The graph of g(x) is the graph of f (x) reflected across the y-axis and shifted up 3 units. This is graph (f). 72. g(x) = f (x) + 3 The graph of g(x) is the graph of f (x) shifted up 3 units. This is graph (h). 73. g(x) = −f (x) + 3 The graph of g(x) is the graph of f (x) reflected across the x-axis and shifted up 3 units. This is graph (f). 74. g(x) = −f (−x) The graph of g(x) is the graph of f (x) reflected across the x-axis and the y-axis. This is graph (a). 1 f (x − 2) 3 The graph of g(x) is the graph of f (x) shrunk vertically by a factor of 3 that is, each y-coordinate is multiplied 1 and then shifted right 2 units. This is graph (d). by 3 75. g(x) = 80. f (−x) = 82. Each y-coordinate of the graph of f (x) = x3 − 3×2 is mul1 tiplied by . A formula for the transformed function is 2 1 1 h(x) = f (x), or h(x) = (x3 − 3×2 ). 2 2 2 6 4 2 79. f (−x) = 2(−x)4 − 35(−x)3 + 3(−x) − 5 = 2×4 + 35×3 − 3x − 5 = g(x) 81. The graph of f (x) = x3 − 3×2 is shifted up 2 units. A formula for the transformed function is g(x) = f (x) + 2, or g(x) = x3 − 3×2 + 2. (3, 5) 4 The graph of g(x) is the graph f (x) reflected across the x-axis and shifted left 2 units. This is graph (b). 1 4 1 3 x − x − 81×2 − 17 = g(x) 4 5 h(x) 2g (x 1) 3 8 (1, 5) 1 f (x) − 3 3 The graph of g(x) is the graph of f (x) shrunk vertically 76. g(x) = 84. The graph of f (x) = x3 − 3×2 is shifted right 2 units and up 1 unit. A formula for the transformed function is t(x) = f (x − 2) + 1, or t(x) = (x − 2)3 − 3(x − 2)2 + 1. 85. Test for symmetry with respect to the x-axis. y = 3×4 − 3 Original equation −y = 3×4 − 3 Replacing y by −y y = −3x + 3 Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. 4 Test for symmetry with respect to the y-axis. Original equation y = 3×4 − 3 y = 3(−x)4 − 3 Replacing x by −x Simplifying y = 3×4 − 3 The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y = 3×4 − 3 −y = 3(−x)4 − 3 Replacing x by −x and y by −y −y = 3×4 − 3 y = −3×4 + 3 c 2016 Pearson Education, Inc. Copyright Simplifying 92 Chapter 2: More on Functions The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 86. Test for symmetry with respect to the x-axis. y 2 = x Original equation (−y)2 = x Replacing y by −y 1828 y 2 = x Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: y2 = x Original equation 2· = n − 418 Solve. 1828 = 2n − 418 2246 = 2n 1123 = n Check. 2 · 1123 = 2246 and 2246 − 418 = 1828. This is the number of guns found in 2013, so the answer checks. y 2 = −x Replacing x by −x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y2 = x Translate. Number of number of guns guns found was twice less 418. found in 2010 in 2013       Original equation (−y)2 = −x Replacing x by −x and y by −y y 2 = −x Simplifying State. In 2010, 1123 guns were found with airline travelers. 90. Let a = the total number of acres of pumpkins harvested in Michigan, Ohio, and Illinois in 2012. Solve: 0.545a = 16, 200 a ≈ 29, 700 acres The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 91. Each point for which f (x) < 0 is reflected across the x-axis. y 87. Test for symmetry with respect to the x-axis: 4 2x − 5y = 0 Original equation (1, 2) (1, 2) 2 (4, 0) 2x − 5(−y) = 0 Replacing y by −y 4 2 2x + 5y = 0 Simplifying 2 (4, 2) 4 x 2 The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. |f(x)| 4 Test for symmetry with respect to the y-axis: 2x − 5y = 0 Original equation 92. The graph of y = f (|x|) consists of the points of y = f (x) for which x ≥ 0 along with their reflections across the y-axis. 2(−x) − 5y = 0 Replacing x by −x −2x − 5y = 0 Simplifying The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 2x − 5y = 0 Original equation 2(−x) − 5(−y) = 0 Replacing x by −x and y by −y −2x + 5y = 0 y 4 (1, 2) (4, 2) 2 4 2 (1, 2) 2 (4, 2) 4 x 2 f(|x|) 4 2x − 5y = 0 Simplifying The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 93. The graph of y = g(|x|) consists of the points of y = g(x) for which x ≥ 0 along with their reflections across the y-axis. 88. Let p = the number of pages of federal tax rules in 1995. Then the number of pages in 2014 was p + 84.2% of p, or 1.842p. Solve: 1.842p = 74, 608 p = 40, 504 pages y 4 (2, 1) 2 (4, 0) (2, 1) 4 2 2 89. Familiarize. Let n = the number of guns that were found with airline travelers in 2010. c 2016 Pearson Education, Inc. Copyright (4, 0) x 4 2 4 g(|x|) Exercise Set 2.6 93 94. Each point for which g(x) < 0 is reflected across the x-axis. 5. 1 = k· y (2, 1) 2 (4, 0) (2, 1) 4 2 2 The variation constant is 4. The equation of variation is y = 4x. (4, 0) x 4 2 4 (The graph of y = f (x − 3) is the graph of y = f (x) shifted right 3 units, so the point (−1, 5) on y = f (x) is transformed to the point (−1 + 3, 5), or (2, 5) on y = f (x − 3).) 96. Call the transformed function g(x). g(5) = 4 − f (−3) = 4 − f (5 − 8), g(8) = 4 − f (0) = 4 − f (8 − 8), and g(11) = 4 − f (3) = 4 − f (11 − 8). Thus g(x) = 4 − f (x − 8), or g(x) = 4 − |x − 8|. Exercise Set 2.6 y = kx 54 = k · 12 9 54 = k, or k = 12 2 9 The variation constant is , or 4.5. The equation of vari2 9 ation is y = x, or y = 4.5x. 2 2. y = kx 0.1 = k(0.2) 1 = k Variation constant 2 1 Equation of variation: y = x, or y = 0.5x. 2 3. k x k 3= 12 36 = k y= The variation constant is 36. The equation of variation is 36 . y= x 4. k x k 12 = 5 60 = k Variation constant y= Equation of variation: y = k x k 0.1 = 0.5 0.05 = k Variation constant 0.05 Equation of variation: y = x k 7. y= x k 32 = 1 8 1 · 32 = k 8 4=k 6. |g(x)| 95. f (2 − 3) = f (−1) = 5, so b = 5. 1. 1 4 4=k 4 Then y = kx 60 x y= The variation constant is 4. The equation of variation is 4 y= . x 8. y = kx 3 = k · 33 1 = k Variation constant 11 1 x Equation of variation: y = 11 9. y = kx 3 = k·2 4 1 3 · =k 2 4 3 =k 8 3 The variation constant is . The equation of variation is 8 3 y = x. 8 k 10. y = x 1 k = 5 35 7 = k Variation constant 7 Equation of variation: y = x k 11. y= x k 1.8 = 0.3 0.54 = k The variation constant is 0.54. The equation of variation 0.54 . is y = x c 2016 Pearson Education, Inc. Copyright 94 12. Chapter 2: More on Functions 17. Let F = the number of grams of fat and w = the weight. y = kx F = kw 0.9 = k(0.4) 9 = k Variation constant 4 9 Equation of variation: y = x, or y = 2.25x 4 F varies directly as w. 60 = k · 120 60 = k, or 120 1 =k 2 13. Let W = the weekly allowance and a = the child’s age. W = ka 1 w 2 1 F = · 180 2 5.50 = k · 6 11 =k 12 F = 11 x 12 11 ·9 W = 12 W = $8.25 F = 90 Substituting Solving for k Variation constant Equation of variation Substituting W = The maximum daily fat intake for a person weighing 180 lb is 90 g. 18. 53 = k · 38, 040, 000 Substituting 53 =k Variation constant 38, 040, 000 53 P N= 38, 040, 000 53 · 26, 060, 000 Substituting N = 38, 040, 000 N ≈ 36 14. Let S = the sales tax and p = the purchase price. S = kp S varies directly as p. 7.14 = k · 119 Substituting 0.06 = k Variation constant S = 0.06p Equation of variation S = 0.06(21) Substituting S ≈ 1.26 The sales tax is $1.26. 15. 16. k r k 5= 80 400 = r 400 t= r 400 t= 70 5 40 , or 5 hr t= 7 7 k W = W varies inversely as L. L k Substituting 1200 = 8 9600 = k Variation constant t= 9600 L 9600 W = 14 W ≈ 686 W = Texas has 36 representatives. k 19. T = T varies inversely as P . P k Substituting 5= 7 35 = k Variation constant 35 P 35 T = 10 T = 3.5 T = Equation of variation Substituting It will take 10 bricklayers 3.5 hr to complete the job. 20. Equation of variation Substituting A 14-m beam can support about 686 kg. N = kP 21. k r k 45 = 600 27, 000 = k 27, 000 t= r 27, 000 t= 1000 t = 27 min t= d = km d varies directly as m. 40 = k · 3 Substituting 40 =k Variation constant 3 c 2016 Pearson Education, Inc. Copyright Exercise Set 2.6 95 40 m Equation of variation 3 200 40 ·5= Substituting d= 3 3 2 d = 66 3 26. d= A 5-kg mass will stretch the spring 66 22. 2 cm. 3 f = kF y= 0.15 = k(0.1)2 f = 0.042F f = 0.042(80) f = 3.36 k P varies inversely as W . P = W k Substituting 330 = 3.2 The equation of variation is y = 15×2 . 28. 550W = 1056 y= Equation of variation 29. y = kxz Dividing by 550 W = 1.92 Simplifying 1=k The equation of variation is y = xz. kx 30. y = z k · 12 4= 15 5=k M varies directly as E. 35.9 = k · 95 Substituting 35.9 =k Variation constant 95 35.9 E Equation of variation M = 95 35.9 M = · 100 Substituting 95 M ≈ 37.8 A 100-lb person would weigh about 37.8 lb on Mars. k 25. y= 2 x k 0.15 = Substituting (0.1)2 k 0.15 = 0.01 0.15(0.01) = k The equation of variation is y = y= 5x z y = kxz 2 31. 105 = k · 14 · 52 Substituting 105 = 350k 105 =k 350 3 =k 10 3 2 The equation of variation is y = xz . 10 xz 32. y = k · w 3 2·3 = k· 2 4 1=k y= 0.0015 = k Substituting 56 = 56k Multiplying by W 1056 W = 550 M = kE 2 2 x 3 56 = k · 7 · 8 Substituting A tone with a pitch of 550 vibrations per second has a wavelength of 1.92 ft. 24. y = kx2 6 = k · 32 2 =k 3 Variation constant 1056 P = W 1056 550 = W Substituting 0.15 = 0.01k 0.15 =k 0.01 15 = k 0.042 = k 1056 = k 54 x2 y = kx2 27. 6.3 = k · 150 23. k x2 k 6= 2 3 54 = k y= xz w 0.0015 . x2 c 2016 Pearson Education, Inc. Copyright 96 33. Chapter 2: More on Functions xz wp 3 3 · 10 =k 28 7·8 30 3 = k· 28 56 3 56 · =k 28 30 1 =k 5 y=k The equation of variation is d = 34. Substitute 72 for d and find r. 1 2 r 72 = 18 1296 = r2 Substituting The equation of variation is y = 36 = r xz 1 xz , or . 5 wp 5wp xz w2 12 16 · 3 = k· 2 5 5 5 =k 4 y = k· A car can travel 36 mph and still stop in 72 ft. k 38. W = 2 d k 220 = (3978)2 3, 481, 386, 480 = k 3, 481, 386, 480 d2 3, 481, 386, 480 W = (3978 + 200)2 W ≈ 199 lb W = 5xz 5 xz , or 4 w2 4w2 k I = 2 d k 90 = 2 Substituting 5 k 90 = 25 2250 = k y= 35. The equation of variation is I = kR I We first find k. 39. E = k · 89 3.75 = 213.1 213.1 =k 3.75 89 9≈k 2250 . d2 d = 7.5 The distance from 5 m to 7.5 m is 7.5 − 5, or 2.5 m, so it is 2.5 m further to a point where the intensity is 40 W/m2 . 222 = k · 37.8 · 40 37 =k 252 37 Av 252 37 · 51v 430 = 252 v ≈ 57.4 mph D= 37. 200 = k · 602 200 = 3600k 200 =k 3600 1 =k 18 40. kT P k · 42 231 = 20 110 = k V = 110T P 110 · 30 V = 15 V = 220 cm3 Substituting 213.1 89 John Lester would have given up about 98 earned runs if he had pitched 235 innings. V = d = kr2 Multiplying by 9R . I Substitute 3.75 for E and 235 for I and solve for R. 9R 3.75 = 235 235 235 3.75 =R Multiplying by 9 9 98 ≈ R d2 = 56.25 D = kAv Substituting The equation of variation is E = Substitute 40 for I and find d. 2250 40 = 2 d 40d2 = 2250 36. 1 2 r . 18 41. parallel 42. zero 43. relative minimum 44. odd function 45. inverse variation c 2016 Pearson Education, Inc. Copyright Chapter 2 Review Exercises 97 49. We are told A = kd2 , and we know A = πr2 so we have: 46. a) 7xy = 14 2 y= x Inversely kd2 = πr2 kd2 = π b) x − 2y = 12 x y = −6 2 Neither d 2 d 2 r= 2 πd2 4 π k= 4 kd2 = c) −2x + y = 0 Variation constant Chapter 2 Review Exercises y = 2x Directly 1. This statement is true by the definition of the greatest integer function. 3 d) x = y 4 4 y= x 3 Directly 2. This statement is false. See Example 3 in Section 2.3 in the text. 3. The graph of y = f (x − d) is the graph of y = f (x) shifted right d units, so the statement is true. e) x = 2 y 1 y= x 2 Directly 4. The graph of y = −f (x) is the reflection of the graph of y = f (x) across the x-axis, so the statement is true. 47. Let V represent the volume and p represent the price of a jar of peanut butter. V = kp π 5. a) For x-values from −4 to −2, the y-values increase from 1 to 4. Thus the function is increasing on the interval (−4, −2). b) For x-values from 2 to 5, the y-values decrease from 4 to 3. Thus the function is decreasing on the interval (2, 5). V varies directly as p. 2 3 (5) = k(2.89) Substituting 2 3.89π ≈ k Variation constant V = 3.89πp 2 π(1.625) (5.25) = 3.89πp Equation of variation Substituting 3.56 ≈ p If cost is directly proportional to volume, the larger jar should cost $3.56. c) For x-values from −2 to 2, y is 4. Thus the function is constant on the interval (−2, 2). 6. a) For x-values from −1 to 0, the y-values increase from 3 to 4. Also, for x-values from 2 to ∞, the y-values increase from 0 to ∞. Thus the function is increasing on the intervals (−1, 0), and (2, ∞). b) For x-values from 0 to 2, the y-values decrease from 4 to 0. Thus, the function is decreasing on the interval (0, 2). Now let W represent the weight and p represent the price of a jar of peanut butter. c) For x-values from −∞ to −1, y is 3. Thus the function is constant on the interval (−∞, −1). W = kp 18 = k(2.89) Substituting 7. 6.23 ≈ k Variation constant W = 6.23p Equation of variation 5 22 = 6.23p Substituting 3 y 4 2 3.53 = p 1 If cost is directly proportional to weight, the larger jar should cost $3.53. kp2 48. Q = 3 q Q varies directly as the square of p and inversely as the cube of q. –5 –4 –3 –2 –1 –1 –2 1 2 3 4 5 x f (x ) = x 2 – 1 –3 –4 –5 The function is increasing on (0, ∞) and decreasing on (−∞, 0). We estimate that the minimum value is −1 at x = 0. There are no maxima. c 2016 Pearson Education, Inc. Copyright 98 Chapter 2: More on Functions 8. y y 4 5 4 2 f (x ) = 2 – | x | 3 4 2 2 1 –5 –4 –3 –2 –1 –1 2 4 x 2 1 2 3 4 5 x 4 –2 –3 –4 –5 The function is increasing on (−∞, 0) and decreasing on (0, ∞). We estimate that the maximum value is 2 at x = 0. There are no minima. 9. If two sides of the patio are each x feet, then the remaining side will be (48 − 2x) ft. We use the formula Area = length × width.  3 for x 2 We create the graph in three parts. Graph f (x) = x3 for inputs less than −2. Then graph f (x) = |x| for inputs greater than or equal √to −2 and less than or equal to 2. Finally graph f (x) = x − 1 for inputs greater than 2. y A(x) = x(48 − 2x), or 48x − 2×2 2 10. The length of the rectangle is 2x. The width is the second coordinate of the point (x, y) on the circle. The circle has center (0,√0) and radius 2, so its equation is x2 + y 2 = 4 and y = 4 −√x2 . Thus the area of the rectangle is given by A(x) = 2x 4 − x2 . 4 2 108 = x2 h 108 =h x2 Now find the surface area. S = x2 + 4 · x · h 108 S(x) = x2 + 4 · x · 2 x 432 S(x) = x2 + x b) x must be positive, so the domain is (0, ∞). c) From the graph, we see that the minimum value of the function occurs when x = 6 in. For this value of x, 108 108 108 = 3 in. h= 2 = 2 = x 6 36  for x ≤ −4,  −x, 12. f (x) = 1 x + 1, for x > −4 2 4 x 4 6 8 11. a) Let h = the height of the box. Since the volume is 108 in3 , we have: 108 = x · x · h 2 2 10 12  2  x − 1 , for x = −1, 14. f (x) = x+1  3, for x = −1 x2 − 1 x+1 for all inputs except −1. Then graph f (x) = 3 for x = −1. We create the graph in two parts. Graph f (x) = y 4 2 4 2 2 4 x 4 15. f (x) = [[x]]. See Example 9 in Section 2.1 of the text. We create the graph in two parts. Graph f (x) = −x for 1 inputs less than or equal to −4. Then graph f (x) = x + 1 2 for inputs greater than −4. 16. f (x) = [[x − 3]] This function could be defined by a piecewise function with an infinite number of statements. c 2016 Pearson Education, Inc. Copyright Chapter 2 Review Exercises    .     .     .     −4,    −3, f (x) = −2,     −1,     .     .    .  99 4 , g(x) = 3 − 2x x2 a) Division by zero is undefined, so the domain of f is {x|x = 0}, or (−∞, 0) ∪ (0, ∞). The domain of g is the set of all real numbers, or (−∞, ∞). 22. f (x) = for −1 ≤ x < 0, for 0 ≤ x < 1, for 1 ≤ x < 2, for 2 ≤ x < 3,  3 for x 2 Since −1 is in the interval [−2, 2], f (−1) = | − 1| = 1. √ √ Since 5 > 2, f (5) = 5 − 1 = 4 = 2. Since −2 is in the interval [−2, 2], f (−2) = | − 2| = 2. Since −3 0, the graph of y = f (x)+b is the graph of y = f (x) shifted up b units. Answer C is correct. 1 74. The graph of g(x) = − f (x) + 1 is the graph of y = f (x) 2 1 shrunk vertically by a factor of , then reflected across the 2 x-axis, and shifted up 1 unit. The correct graph is B. c) For x-values from −2 to 2, y is 2. Thus the function is constant on the interval (−2, 2). 2. y 5 4 f (x ) = 2 – x 2 3 75. Let f (x) and g(x) be odd functions. Then by definition, f (−x) = −f (x), or f (x) = −f (−x), and g(−x) = −g(x), or g(x) = −g(−x). Thus (f + g)(x) = f (x) + g(x) = −f (−x) + [−g(−x)] = −[f (−x) + g(−x)] = −(f + g)(−x) and f + g is odd. 2 1 –5 –4 –3 –2 –1 –1 1 2 3 4 5 x –2 –3 –4 76. Reflect the graph of y = f (x) across the x-axis and then across the y-axis. 77. f (x) = 4×3 − 2x + 7 a) f (x) + 2 = 4x − 2x + 7 + 2 = 4x − 2x + 9 3 3 b) f (x + 2) = 4(x + 2)3 − 2(x + 2) + 7 = 4(x3 + 6×2 + 12x + 8) − 2(x + 2) + 7 = 4×3 + 24×2 + 48x + 32 − 2x − 4 + 7 = 4×3 + 24×2 + 46x + 35 c) f (x) + f (2) = 4×3 − 2x + 7 + 4 · 23 − 2 · 2 + 7 = 4×3 − 2x + 7 + 32 − 4 + 7 –5 The function is increasing on (−∞, 0) and decreasing on (0, ∞). The relative maximum is 2 at x = 0. There are no minima. 3. If b = the length of the base, in inches, then the height = 4b − 6. We use the formula for the area of a triangle, 1 A = bh. 2 1 A(b) = b(4b − 6), or 2 A(b) = 2b2 − 3b = 4×3 − 2x + 42 f (x) + 2 adds 2 to each function value; f (x + 2) adds 2 to each input before the function value is found; f (x) + f (2) adds the output for 2 to the output for x. c 2016 Pearson Education, Inc. Copyright Chapter 2 Test 105  2 for x 1 1 x+4 2 1 1 1 f (x + h) = (x + h) + 4 = x + h + 4 2 2 2 1 1 1 x+ h+4− x+4 f (x + h) − f (x) 2 2 2 = h h 1 1 1 x+ h+4− x−4 2 2 2 = h 1 h 1 1 1 h 1 = 2 = h· = · = h 2 h 2 h 2 20. f (x) = y 4 2 4 2 2 4 x 2 4 7 7 7 7 ≤ 1, f − = − = . 8 8 8 8 √ √ Since 5 > 1, f (5) = 5 − 1 = 4 = 2. 5. Since −1 ≤ − 21. f (x) = 2×2 − x + 3 f (x+h) = 2(x+h)2 −(x+h)+3 = 2(x2 +2xh+h2 )−x − h+3 = 2×2 + 4xh + 2h2 − x − h + 3 Since −4 < −1, f (−4) = (−4)2 = 16. 6. 7. 8. 2×2+4xh+2h2−x−h + 3−(2×2 −x + 3) f (x+h)−f (x) = h h (f + g)(−6) = f (−6) + g(−6) = (−6)2 − 4(−6) + 3 + 3 − (−6) = √ √ 36 + 24 + 3 + 3 + 6 = 63 + 9 = 63 + 3 = 66 (f − g)(−1) = f (−1) − g(−1) = (−1)2 − 4(−1) + 3 − 3 − (−1) = √ √ 1+4+3− 3+1=8− 4=8−2=6 √ (f g)(2) = f (2) · g(2) = (22 − 4 · 2 + 3)( 3 − 2) = √ (4 − 8 + 3)( 1) = −1 · 1 = −1 = 4xh + 2h2 − h h h/(4x + 2h − 1) h/ = 4x + 2h − 1 22. 1 −4·1+3 f (1) 1−4+3 0 √ √ = = = √ =0 g(1) 3−1 2 2 10. Any real number can be an input for f (x) = x2 , so the domain is the set of real numbers, or (−∞, ∞). √ 11. The domain of g(x) = x − 3 is the set of real numbers for which x − 3 ≥ 0, or x ≥ 3. Thus the domain is {x|x ≥ 3}, or [3, ∞). 2×2 +4xh+2h2−x−h+3−2×2 +x−3 h = 2 9. (f /g)(1) = = (g ◦ h)(2) = g(h(2)) = g(3 · 22 + 2 · 2 + 4) = g(3 · 4 + 4 + 4) = g(12 + 4 + 4) = g(20) = 4 · 20 + 3 = 80 + 3 = 83 23. (f ◦ g)(−1) = f (g(−1)) = f (4(−1) + 3) = f (−4 + 3) = f (−1) = (−1)2 − 1 = 1 − 1 = 0 24. (h ◦ f )(1) = h(f (1)) = h(12 − 1) = h(1 − 1) = h(0) = 3 · 02 + 2 · 0 + 4 = 0 + 0 + 4 = 4 (g ◦ g)(x) = g(g(x)) = g(4x + 3) = 4(4x + 3) + 3 = 12. The domain of f + g is the intersection of the domains of f and g. This is {x|x ≥ 3}, or [3, ∞). 25. 13. The domain of f − g is the intersection of the domains of f and g. This is {x|x ≥ 3}, or [3, ∞). √ 26. (f ◦ g)(x) = f (g(x)) = f (x2 + 1) = x2 + 1 − 5 = √ x2 − 4 √ √ (g ◦ f )(x) = g(f (x)) = g( x − 5) = ( x − 5)2 + 1 = 14. The domain of f g is the intersection of the domains of f and g. This is {x|x ≥ 3}, or [3, ∞). 15. The domain of f /g is the intersection of the domains of f and g, excluding those x-values for which g(x) = 0. Since x − 3 = 0 when x = 3, the domain is (3, ∞). √ 16. (f + g)(x) = f (x) + g(x) = x2 + x − 3 √ 17. (f − g)(x) = f (x) − g(x) = x2 − x − 3 √ 18. (f g)(x) = f (x) · g(x) = x2 x − 3 x2 f (x) =√ 19. (f /g)(x) = g(x) x−3 16x + 12 + 3 = 16x + 15 x−5+1=x−4 27. The inputs for f (x) must be such that x − 5 ≥ 0, or x ≥ 5. Then for (f ◦g)(x) we must have g(x) ≥ 5, or x2 +1 ≥ 5, or x2 ≥ 4. Then the domain of (f ◦g)(x) is (−∞, −2]∪[2, ∞). Since we can substitute any real number for x in g, the domain of (g ◦ f )(x) is the same as the domain of f (x), [5, ∞). 28. Answers may vary. f (x) = x4 , g(x) = 2x − 7 c 2016 Pearson Education, Inc. Copyright 106 Chapter 2: More on Functions 29. y = x4 − 2×2 34. Replace y with −y to test for symmetry with respect to the x-axis. −y = x4 − 2×2 y = −x4 + 2×2 The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Variation constant Equation of variation: y = 35. 30 x y = kx 60 = k · 12 Replace x with −x to test for symmetry with respect to the y-axis. 5=k y = (−x)4 − 2(−x)2 Variation constant Equation of variation: y = 5x y = x − 2x 4 k x k 5= 6 30 = k y= 2 The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. 36. Replace x with −x and y with −y to test for symmetry with respect to the origin. −y = (−x)4 − 2(−x)2 kxz 2 w k(0.1)(10)2 100 = 5 100 = 2k y= 50 = k −y = x4 − 2×2 y= y = −x4 + 2×2 The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 2x x2 + 1 2x 2(−x) =− 2 f (−x) = (−x)2 + 1 x +1 f (x) = f (−x), so f is not even. 2x −f (x) = − 2 x +1 f (−x) = −f (x), so f is odd. 30. f (x) = 37. 50xz w Variation constant 2 Equation of variation d = kr2 200 = k · 602 1 =k 18 1 2 r d= 18 1 · 302 d= 18 d = 50 ft Variation constant Equation of variation 38. The graph of g(x) = 2f (x) − 1 is the graph of y = f (x) stretched vertically by a factor of 2 and shifted down 1 unit. The correct graph is C. 31. Shape: h(x) = x2 Shift h(x) right 2 units: g(x) = h(x − 2) = (x − 2)2 Shift g(x) down 1 unit: f (x) = (x − 2) − 1 2 32. Shape: h(x) = x2 39. Each x-coordinate on the graph of y = f (x) is divided by −3 , 1 , or 3 on the graph of y = f (3x). Thus the point 3 (−1, 1) is on the graph of f (3x). Shift h(x) left 2 units: g(x) = h(x + 2) = (x + 2)2 Shift g(x) down 3 units: f (x) = (x + 2)2 − 3 1 33. Each y-coordinate is multiplied by − . We plot and con2 nect (−5, 1), (−3, −2), (1, 2) and (4, −1). c 2016 Pearson Education, Inc. Copyright