Michael Sullivan, Kathleen Miranda-Instructor's Solution Manuals to Calculus Early Transcendentals-W

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A Transition to Advanced Mathematics
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Chapter 4 / Exercise 12
A Transition to Advanced Mathematics
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Unformatted text preview: 1 Chapter P: Preparing for Calculus P.1: Functions and Their Graphs Concepts and Vocabulary 1. If f is a function defined by y = f (x), then x is called the independent variable and y is the dependent variable. 2. True . The independent variable is sometimes referred to as the argument of the function. 3. False . If no domain is specified for a function f , then the domain of f is taken to be the largest set of real numbers for which the value f (x) is defined and is a real number. 4. False . The domain of the function f (x) = 3(x2 − 1) is the set {x|x 6= 1}. x−1 5. False . A function can have at most one y-intercept; otherwise, there would be more than one y-value corresponding to x = 0. 6. A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. 7. If the point (5, −3) is on the graph of f , then f ( 5 ) = −3 . 8. Let f (x) = ax2 + 4. For the point (−1, 2) to be on the graph of f , we must have f (−1) = 2. But f (−1) = a(−1)2 + 4 = a + 4, so we must have a + 4 = 2, or a = −2 . 9. A function f is (a) increasing on an interval I if, for any choice of x1 and x2 in I, with x1 < x2 , then f (x1 ) < f (x2 ). 10. A function is (a) even if for every number x in its domain, the number −x is also in the domain and f (−x) = f (x). A function f is (b) odd if for every number x in its domain, the number −x is also in the domain and f (−x) = −f (x). 11. False . Even functions have graphs that are symmetric with respect to the y-axis. Odd functions have graphs that are symmetric with respect to the origin. 12. The average rate of change of f (x) = 2x3 − 3 from 0 to 2 is 2(2)3 − 3 − (2(0)3 − 3) 13 − (−3) 16 f (2) − f (0) = = = = 8. 2−0 2 2 2 Practice Problems 13. Let f (x) = 3x2 + 2x − 4. Then (a) f (0) = 3(0)2 + 2(0) − 4 = −4 . (b) f (−x) = 3(−x)2 + 2(−x) − 4 = 3x2 − 2x − 4 . (c) −f (x) = −(3x2 + 2x − 4) = −3x2 − 2x + 4 . (d) f (x + 1) = 3(x + 1)2 + 2(x + 1) − 4 = 3(x2 + 2x + 1) + 2x + 2 − 4 = 3x2 + 8x + 1 . 2 (e) f (x+h) = 3(x+h)2 +2(x+h)−4 = 3(x2 +2xh+h2 )+2x+2h−4 = 3x2 + 6xh + 3h2 + 2x + 2h − 4 . x . Then x2 + 1 14. Let f (x) = 0 = 0. +1 x −x . (b) f (−x) = = − 2 2 (−x) + 1 x +1 (a) f (0) = 02 (c) −f (x) = − x . x2 + 1 (d) f (x + 1) = x+1 x+1 . = 2 2 (x + 1) + 1 x + 2x + 2 (e) f (x + h) = x+h x+h = 2 . 2 (x + h) + 1 x + 2xh + h2 + 1 15. Let f (x) = |x| + 4. Then (a) f (0) = |0| + 4 = 4 . (b) f (−x) = | − x| + 4 = |x| + 4 . (c) −f (x) = −(|x| + 4) = −|x| − 4 . (d) f (x + 1) = |x + 1| + 4 . (e) f (x + h) = |x + h| + 4 . 16. Let f (x) = √ (c) (d) (e) √ √ 3−0= 3 . p √ f (−x) = 3 − (−x) = 3 + x . √ −f (x) = − 3 − x . p √ f (x + 1) = 3 − (x + 1) = 2 − x . p √ f (x + h) = 3 − (x + h) = 3 − x − h . (a) f (0) = (b) 3 − x. Then 17. Because f (x) = x3 − 1 is defined for any real number x, the domain of f is the set of all real numbers , or in interval notation (−∞, ∞) . x is defined for any real +1 number x. The domain of f is therefore the set of all real numbers , or in interval notation (−∞, ∞) . 18. Because x2 + 1 is never equal to zero for any real number x, f (x) = x2 19. Because the square root of a negative number is not a real number, the value of t2 − 9 must be nonnegative. The solution of the inequality t2 − 9 ≥ 0 is {t|t ≤ −3} ∪ {t|t ≥ 3}, so the domain of v is the set of real numbers {t|t ≤ −3} ∪ {t|t ≥ 3} , or in interval notation (−∞, −3] ∪ [3, ∞) . 20. Because the expression x − 1 appears under the square root and in the denominator, the value of x − 1 must be positive; that is, x − 1 > 0. The solution of this inequality is {x|x > 1}, so the domain of g is the set of real numbers {x|x > 1} , or in interval notation (1, ∞) . 3 21. Because division by zero is not defined, x3 − 4x = x(x2 − 4) = x(x − 2)(x + 2) cannot be zero; that is, x 6= 0, x 6= 2, and x 6= −2. Thus, the domain of h is the set of real numbers {x|x 6= −2, x 6= 0, x 6= 2} . 22. Because the square root of a negative number is not a real number, the value of t + 1 must be nonnegative. The solution of the inequality t + 1 ≥ 0 is {t|t ≥ −1}. Moreover, because division by zero is not defined, t − 5 cannot be 0; that is, t cannot be equal to 5. The intersection of the sets {t|t ≥ −1} and {t|t 6= 5} is the set {t| − 1 ≤ t < 5} ∪ {t|t > 5}. The domain of s is therefore the set of real numbers {t| − 1 ≤ t < 5} ∪ {t|t > 5} , or in interval notation [−1, 5) ∪ (5, ∞) . 23. Let f (x) = −3x + 1. Then −3(x + h) + 1 − (−3x + 1) −3x − 3h + 1 + 3x − 1 −3h f (x + h) − f (x) = = = = −3 . h h h h 24. Let f (x) = 1 . Then x+3 f (x + h) − f (x) h = = 25. Let f (x) = √ 1 x+h+3 − 1 x+3 h · (x + h + 3)(x + 3) (x + h + 3)(x + 3) −h x + 3 − (x + h + 3) 1 = = − . h(x + h + 3)(x + 3) h(x + h + 3)(x + 3) (x + h + 3)(x + 3) x + 7. Then f (x + h) − f (x) h √ = = 26. Let f (x) = √ x+h+7− h √ x+7 √ √ x+h+7+ x+7 √ ·√ x+h+7+ x+7 h 1 x + h + 7 − (x + 7) √ √ √ √ = √ = √ . h( x + h + 7 + x + 7) h( x + h + 7 + x + 7) x+h+7+ x+7 2 . Then x+7 f (x + h) − f (x) h √ √ x+h+7 x+7 √ ·√ h x+h+7 x+7 √ √ √ √ 2 x+7−2 x+h+7 x+7+ x+h+7 √ √ √ ·√ h x+h+7 x+7 x+7+ x+h+7 2(x + 7) − 2(x + h + 7) √ √ √ √ h x + h + 7 x + 7( x + 7 + x + h + 7) −2h √ √ √ √ h x + h + 7 x + 7( x + 7 + x + h + 7) √ = = = = = 2 x+h+7 −√ − √2 x+7 2 √ √ √ . x + h + 7 x + 7( x + 7 + x + h + 7) 27. Let f (x) = x2 + 2x. Then f (x + h) − f (x) h = = (x + h)2 + 2(x + h) − (x2 + 2x) x2 + 2xh + h2 + 2x + 2h − x2 − 2x = h h 2xh + h2 + 2h h(2x + h + 2) = = 2x + h + 2 . h h 4 28. Let f (x) = (2x + 3)2 . Then f (x + h) − f (x) h = = = (2(x + h) + 3)2 − (2x + 3)2 4(x + h)2 + 12(x + h) + 9 − (4x2 + 12x + 9) = h h 4x2 + 8xh + 4h2 + 12x + 12h + 9 − 4x2 − 12x − 9 h 8xh + 4h2 + 12h h(8x + 4h + 12) = = 8x + 4h + 12 . h h 29. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two points), so it does not represent a function . 30. This graph passes the vertical line test, and so it does represent a function . (a) From the graph, the domain of the function is the set of all real numbers or the interval (−∞, ∞) , and the range of the function is the set of all positive real numbers or the interval (0, ∞) . (b) The graph has no x-intercepts, and the y-intercept is 1. Thus, the only intercept is (0, 1) . (c) The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. 31. This graph passes the vertical line test, and so it does represent a function . (a) From the graph, the domain of the function is the set {x| − π ≤ x ≤ π} or the closed interval [−π, π] , and the range of the function is the set {y| − 1 ≤ y ≤ 1} or the closed interval [−1, 1] .  π  π (b) The graph has x-intercepts of ± and a y-intercept of 1, so the intercepts are ± , 0 and (0, 1) . 2 2 (c) The graph is symmetric with respect to the y-axis but not symmetric with respect to the x-axis or the origin. 32. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two points), so it does not represent a function . 33. (a) f (−1) = −1 + 3 = 2; f (0) = 0 + 3 = 3; f (1) = 5; f (8) = −8 + 2 = −6 (b) The graph of f consists of three pieces corresponding to each equation in the definition. On the interval [−2, 1), the graph is the line y = x + 3, and on the interval (1, ∞), the graph is the line y = −x + 2. The graph also contains the point (1, 5). 5 4 3 2 1 -2 -1 1 -1 2 3 5 (c) The individual components of f have domains of {x| − 2 ≤ x < 1}, {x|x = 1}, and {x|x > 1}. The domain of f is the union of these three sets; that is, {x|x ≥ −2} in set notation or [−2, ∞) in interval notation. Moreover, the individual components of f have ranges of {y|1 ≤ y < 4}, {y|y = 5}, and {y|y < 1}. The range of f is the union of these three sets; that is, {y|y = 5 or y < 4} in set notation or (−∞, 4) ∪ {5} in interval notation. The x-intercept is 2, and the y-intercept is 3, so the intercepts are (2, 0) and (0, 3) . 34. (a) f (−1) = 2(−1) + 5 = 3; f (0) = −3; f (1) = −5(1) = −5; f (8) = −5(8) = −40 (b) The graph of f consists of three pieces corresponding to each equation in the definition. On the interval [−3, 0), the graph is the line y = 2x + 5, and on the interval (0, ∞), the graph is the line y = −5x. The graph also contains the point (0, −3). 5 -3 -2 -1 1 2 3 -5 -10 -15 (c) The individual components of f have domains of {x| − 3 ≤ x < 0}, {x|x = 0}, and {x|x > 0}. The domain of f is the union of these three sets; that is, {x|x ≥ −3} in set notation or [−3, ∞) in interval notation. Moreover, the individual components of f have ranges of {y| − 1 ≤ y < 5}, {y|y = −3}, and {y|y < 0}. The range of f is the union of these three sets; that is, {y|y < 5} in 5 set notation or (−∞, 5) in interval notation. The x-intercept is − , and the y-intercept is −3, 2   5 so the intercepts are − , 0 and (0, −3) . 2 35. (a) f (−1) = 1 + (−1) = 0; f (0) = 02 = 0; f (1) = 12 = 1; f (8) = 82 = 64 (b) The graph of f consists of two pieces corresponding to each equation in the definition. On the interval (−∞, 0), the graph is the line y = 1 + x, and on the interval [0, ∞), the graph is the parabola y = x2 . 6 8 6 4 2 -5 -4 -3 -2 -1 1 2 3 -2 -4 (c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is the union of these two sets; that is, all real numbers or the interval (−∞, ∞) . Moreover, the individual components of f have ranges of {y|y < 1} and {y|y ≥ 0}. The range of f is the union of these two sets; that is, all real numbers or the interval (−∞, ∞) . The x-intercepts are −1 and 0, and the y-intercept is 0, so the intercepts are (−1, 0) and (0, 0) . 36. (a) f (−1) = √ √ √ 1 = −1; f (0) = 3 0 = 0; f (1) = 3 1 = 1; f (8) = 3 8 = 2 −1 (b) The graph of f consists of two pieces corresponding to each equation in the definition. On the √ interval (−∞, 0), the graph is y = x1 , and on the interval [0, ∞), the graph is y = 3 x. 4 2 -4 -2 2 4 6 8 10 -2 -4 -6 -8 (c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is the union of these two sets; that is, all real numbers or the interval (−∞, ∞) . Moreover, the individual components of f have ranges of {y|y < 0} and {y|y ≥ 0}. The range of f is the union of these two sets; that is, all real numbers or the interval (−∞, ∞) . The x-intercept is 0, and the y-intercept is 0, so the only intercept is (0, 0) . 37. Because the graph of f includes the point (0, 3), f (0) = 3 ; because the graph also includes the point (−6, −3), f (−6) = −3 . 38. Because the graph of f lies above the x-axis at x = 3, f (3) is positive . 7 39. Because the graph of f lies below the x-axis at x = −4, f (−4) is negative . 40. Because the graph of f includes the points (−3, 0), (6, 0), and (10, 0), f (x) x = −3, x = 6, and x = 10 . = 0 for 41. Because the graph of f lies above the x-axis for −3 < x < 6 and for 10 < x ≤ 11, f (x) > 0 for −3 < x < 6 and for 10 < x ≤ 11 . In interval notation, this can be written as (−3, 6) ∪ (10, 11] . 42. The points on the graph of f have x-coordinates between −6 and 11 inclusive. The domain of f is therefore the set of real numbers {x| − 6 ≤ x ≤ 11} or the closed interval [−6, 11] . 43. The points on the graph of f have y-coordinates between −3 and 3 inclusive. The range of f is therefore the set of real numbers {y| − 3 ≤ y ≤ 3} or the closed interval [−3, 3] . 44. Because the graph of f includes the points (−3, 0), (6, 0), and (10, 0), the x-intercepts are −3, 6, and 10 . 45. Because the graph of f includes the point (0, 3), the y-intercept is 3 . 46. The graph of the horizontal line y = 1 will intersect the graph of f three times . 2 47. The graph of the vertical line x = 5 will intersect the graph of f once . 48. Because the graph of f includes the point (x, 3) for 0 ≤ x ≤ 4, f (x) = 3 for 0 ≤ x ≤ 4 . 49. Because the graph of f includes the points (−5, −2) and (8, −2), f (x) = −2 for x = −5 and for x=8. 50. The function f is increasing on the intervals (−6, 0) and (8, 11) . 51. The function f is decreasing on the interval (4, 8) . 52. The function f is constant on the interval (0, 4) . 53. The function f is nonincreasing (that is, the function is constant or decreasing) on the interval (0, 8) . 54. The function f is nondecreasing (that is, the function is constant or increasing) on the intervals (−6, 4) and (8, 11) . 55. Because division by zero is not defined, x − 6 cannot be equal to 0. The domain of g is therefore the set {x|x 6= 6} . 56. Note that g(3) = 3+2 5 = − . Because g(3) 6= 14, the point (3, 14) is not on the graph of g. 3−6 3 57. If x = 4, then g(x) = g(4) = 58. If g(x) = 2, then 4+2 = −3 ; therefore, the point (4, −3) is on the graph of g. 4−6 x+2 = 2. Multiplying both sides of this equation by x − 6 yields x−6 x + 2 = 2(x − 6) = 2x − 12, so that x = 14 . Accordingly, the point (14, 2) is on the graph of g. 8 59. The x-intercepts occur when g(x) = 0. The value of g can only be zero when its numerator is zero; therefore, g(x) = 0 only when x + 2 = 0. Thus, the graph of g has only one x-intercept, and this intercept is −2 . 60. Because g(0) = 0+2 1 1 = − , the y-intercept is − . 0−6 3 3 61. The domain of h is {x|x 6= ±1}, so for every number x in the domain of h, the number −x is also in the domain. Moreover, −x x h(−x) = =− 2 = −h(x), (−x)2 − 1 x −1 so that the function h is odd . Because h is an odd function, its graph is symmetric with respect to the origin . 62. The domain of f is all real numbers, so for every number x in the domain of f , the number −x is also in the domain. Moreover, p p 3 f (−x) = 3 3(−x)2 + 1 = 3x2 + 1 = f (x), so that the function f is even . Because f is an even function, its graph is symmetric with respect to the y-axis . 63. The domain of G is {x|x ≥ 0}. The number x = 1 is in the domain of G, but x = −1 is not; therefore, the function G is neither even nor odd . Because G is neither an even nor an odd function, its graph is not symmetric with respect to either the y-axis or the origin . 64. The domain of F is {x|x 6= 0}, so for every number x in the domain of F , the number −x is also in the domain. Moreover, 2(−x) 2x F (−x) = =− = −F (x), | − x| |x| so that the function F is odd . Because F is an odd function, its graph is symmetric with respect to the origin . 65. Let f (x) = −2x2 + 4. (a) The average rate of change of f from 1 to 2 is f (2) − f (1) −4 − 2 −6 = = = −6 . 2−1 2−1 1 (b) The average rate of change of f from 1 to 3 is f (3) − f (1) −14 − 2 −16 = = = −8 . 3−1 3−1 2 (c) The average rate of change of f from 1 to 4 is f (4) − f (1) −28 − 2 −30 = = = −10 . 4−1 4−1 3 (d) The average rate of change of f from 1 to x for x 6= 1 is f (x) − f (1) −2x2 + 4 − 2 2(1 − x2 ) 2(1 − x)(1 + x) = = = = −2(1 + x) . x−1 x−1 x−1 x−1 9 66. Let s(t) = 20 − 0.8t2 . (a) The average rate of change of s from 1 to 4 is s(4) − s(1) 7.2 − 19.2 −12 = = = −4 . 4−1 4−1 3 (b) The average rate of change of s from 1 to 3 is s(3) − s(1) 12.8 − 19.2 −6.4 = = = −3.2 . 3−1 3−1 2 (c) The average rate of change of s from 1 to 2 is s(2) − f (1) 16.8 − 19.2 −2.4 = = = −2.4 . 2−1 2−1 1 (d) The average rate of change of s from 1 to t for t 6= 1 is s(t) − s(1) 20 − 0.8t2 − 19.2 0.8(1 − t2 ) 0.8(1 − t)(1 + t) = = = = −0.8(1 + t) . t−1 t−1 t−1 t−1 67. For −1 ≤ x < 0, the graph is a line through the points (−1, 1) and (0, 0). The slope of this line is m = −1, and the y-intercept is 0; therefore, the equation for this component of the function is y = −x. For 0 ≤ x ≤ 2, the graph is a line through the points (0, 0) and (2, 1). The slope of this line is m = 21 , and the y-intercept is 0; therefore, the equation for this component of the function is y = 12 x. Combining these two equations, the definition for this piecewise function is  f (x) = −x, 1 2 x, if − 1 ≤ x < 0 if 0 ≤ x ≤ 2. The domain of f is the set {x| − 1 ≤ x ≤ 2} or the closed interval [−1, 2] , and the range is the set {y|0 ≤ y ≤ 1} or the closed interval [0, 1] . 68. For x ≤ 0, the graph is a line with slope 1 through the origin, and for 0 < x ≤ 2, the graph is a horizontal line at height y = 1. The definition for this piecewise function is therefore  f (x) = x, 1, if x ≤ 0 if 0 < x ≤ 2. The domain of f is {x|x ≤ 2} in set notation or (−∞, 2] in interval notation, and the range is {y|y = 1 or y ≤ 0} in set notation or (−∞, 0] ∪ {1} in interval notation. 69. For x < 1, the graph is a horizontal line at height y = −1; at x = 1, the value of the function is 0; and for 1 < x ≤ 2, the graph is a line with slope −1 and x-intercept 2. The definition for this piecewise function is therefore if x < 1 −1, 0, if x = 1 f (x) = 2 − x, if 1 < x ≤ 2. The domain of f is {x|x ≤ 2} in set notation or (−∞, 2] in interval notation, and the range is {y|y = −1 or 0 ≤ y < 1} in set notation or {−1} ∪ [0, 1) in interval notation. 10 70. For x < −1, the graph is a horizontal line at height y = 2; at x = −1, the value of the function is 0; and for −1 < x ≤ 1, the graph is a line with slope −2 and y-intercept −1. The definition for this piecewise function is therefore if x < −1 2, 0, if x = −1 f (x) = −2x − 1, if − 1 < x ≤ 1. The domain of f is {x|x ≤ 1} in set notation or (−∞, 1] in interval notation, and the range is {y|y = 2 or − 3 ≤ y < 1} in set notation of [−3, 1) ∪ {2} in interval notation. 71. The definition for this piecewise function is −x3 , 0, f (x) = 2 x , if − 2 < x < 1 if x = 1 if 1 < x ≤ 3. The domain of f is the set {x| − 2 < x ≤ 3} or the interval (−2, 3] , and the range is the set {y| − 1 < y ≤ 9} or the interval (−1, 9] . 72. The definition for this piecewise function is  f (x) = x2 − 1, x3 , if − 3 ≤ x ≤ 0 if 0 < x ≤ 2. The domain of f is the set {x| − 3 ≤ x ≤ 2} or the closed interval [−3, 2] , and the range is the set {y| − 1 ≤ y ≤ 8} or the closed interval [−1, 8] . 73. (a) The cost of manufacturing 100 road bikes is C(100) = 0.004(100)3 − 0.6(100)2 + 250(100) + 100, 500 = 123, 500; the cost of manufacturing 101 road bikes is C(101) = 0.004(101)3 − 0.6(101)2 + 250(101) + 100, 500 ≈ 123, 750.60. The average rate of change of C from 100 to 101 road bikes is therefore 123, 750.60 − 123, 500 C(101) − C(100) ≈ = $250.60 per bike . 101 − 100 1 (b) The cost of manufacturing 500 road bikes is C(500) = 0.004(500)3 − 0.6(500)2 + 250(500) + 100, 500 = 575, 500; the cost of manufacturing 501 road bikes is C(501) = 0.004(501)3 − 0.6(501)2 + 250(501) + 100, 500 ≈ 578, 155.40. The average rate of change of C from 500 to 501 road bikes is therefore 578, 155.40 − 575, 500 C(501) − C(500) ≈ = $2655.40 per bike . 501 − 500 1 (c) Answers will vary. One possible interpretation is that the unit cost per road bike increases as the number of road bikes manufactured increases. 11 74. (a) The cost of producing 500 tons of steel is C(500) = 1 (500)2 + 5(500) + 1500 = 29000; 10 the cost of producing 501 tons of steel is C(501) = 1 (501)2 + 5(501) + 1500 = 29105.1; 10 The average rate of change of C from 500 to 501 tons of steel is therefore 29105.1 − 29000 C(501) − C(500) = = $105.10 per ton . 501 − 500 1 (b) The cost of producing 1000 tons of steel is C(1000) = 1 (1000)2 + 5(1000) + 1500 = 106, 500; 10 the cost of producing 1001 tons of steel is C(1001) = 1 (1001)2 + 5(1001) + 1500 = 106, 705.1; 10 The average rate of change of C from 1000 to 1001 tons of steel is therefore C(1001) − C(1000) 106.705.1 − 106, 500 = = $205.10 per ton . 1001 − 1000 1 (c) Answers will vary. One possible interpretation is that the unit cost per ton of steel increases as the number of tons produced increases. 12 P.2: Library of Functions; Mathematical Modeling Concepts and Vocabulary 1. The function f (x) = x2 is (a) increasing on the interval (0, ∞). 2. True . The floor function f (x) = bxc is an example of a step function. 3. True . The cube function is odd and is increasing on the interval (−∞, ∞). 4. False . The cube root function is odd, but is increasing on the interval (−∞, ∞). 5. False . The domain and range of the reciprocal function are all real nonzero numbers. 6. A number r for which f (r) = 0 is called a zero of the function f . 7. If r is a zero of even multiplicity of a function f , the graph o...
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