(Solution Manual) Calculus Early Transcendentals

Anton,Bivens,Davis, 10

Math & Statistics

Solution Manual

10th Edition

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Solution Manual for Calculus Early Transcendentals, The downloadable file consists of 1390 pages, and the total size of the file is 14.13 MB. Purchase now only for highly discounted price.

Author(s): Anton, Bivens, Davis

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SOLUTION MANUAL Contents Chapter 0. Before Calculus ................................ ................................ ................................ .............................. 1 Chapter 1. Limits and Continuity ................................ ................................ ............................. 39 Chapter 2. The Derivative ................................ ................................ ................................ ........ 71 Chapter 3. Topics in Differentiation ................................ ................................ ........................ 109 Chapter 4. The Derivative in Graphing and Applications ................................ ...................... 153 Chapter 5. Integration ................................ ................................ ................................ .............. 243 Chapter 6. Applications of the Definite Integral in Geometry, Science, and Engineering… 305 Chapter 7. Principals of Integral Evaluation ................................ ................................ ........................... 363 Chapter 8. Mathematical Modeling with Differential Equatio ns ................................ ..................... 413 Chapter 9. Infinite Series ................................ ................................ ................................ ................................ . 437 Chapter 10. Parametric and Polar Curves; Conic Sections ................................ ................................ .. 485 Chapter 11. Three -Dimensional Space; Vectors ................................ ................................ ....... 545 Chapter 12. Vector -Valued Functions ................................ ................................ ....................... 589 Chapter 13. Partial Derivatives ................................ ................................ ................................ ....................... 627 Chapter 14. Multiple Integrals ................................ ................................ ................................ ......................... 675 Chapter 15. Topics in Vector Calculus ................................ ................................ ................................ .......... 713 Appendix A. Graphing Functions Using Calculators and Computer Algebra Systems ............... 745 Appendix B. Trigonometry Review ................................ ................................ ................................ ................ 753 Appendix C. Solving Polynomial Equations ................................ ................................ ................................ 759 / − — ≈ − √ Before Calculus Exercise Set 0.1 1. (a) −2.9, −2.0, 2.35, 2.9 (b) None (c) y = 0 (d) −1.75 ≤ x ≤ 2.15, x = −3, x = 3 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2 2. (a) x = −1, 4 (b) None (c) y = −1 (d) x = 0, 3, 5 (e) ymax = 9 at x = 6; ymin = −2 at x = 0 3. (a) Yes (b) Yes (c) No (vertical line test fails) (d) No (vertical line test fails) 4. (a) The natural domain of f is x = 1, and for g it is the set of all x. f (x) = g(x) on the intersection of their domains. (b) The domain of f is the set of all x ≥ 0; the domain of g is the same, and f (x) = g(x). 5. (a) 1999, $47,700 (b) 1993, $41,600 (c) The slope between 2000 and 2001 is steeper than the slope between 2001 and 2002, so the median income was declining more rapidly during the first year of the 2-year period. 6. (a) In thousands, approximately 47.7 − 41.6 = 6.1 per yr, or $1017/yr. 6 6 (b) From 1993 to 1996 the median income increased from $41.6K to $44K (K for ‘kilodollars’; all figures approx - imate); the average rate of increase during this time was (44 41.6)/3 K/yr = 2.4/3 K/yr = $800/year. From 1996 to 1999 the average rate of increase was (47 .7 44)/3 K/yr = 3.7/3 K/yr $1233/year. The increase was larger during the last 3 years of the period. (c) 1994 and 2005. 7. (a√) f (0) = 3(0)2 − 2 = −2; f (2) = 3(2)2 − 2 = 10; f (−2) = 3(−2)2 − 2 = 10; f (3) = 3(3)2 − 2 = 25; f (√2) = 3( 2)2 − 2 = 4; f (3t) = 3(3t)2 − 2 = 27t2 − 2. (b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f (−2) = 2(−2) = −4; f (3) = 2(3) = 6; f (√2) = 2√2; f (3t) = 1/(3t) for t > 1 and f (3t) = 6t for t ≤ 1. 3 + 1 8. (a) g(3) = = 2; g(−1) = −1 + 1 = 0; g(π) = π + 1 ; g(−1.1) = −1.1 + 1 = −0.1 = 1 ; g(t2 − 1) = t2 − 1 + 1 3 − t21 −1 − 1 π − 1 −1.1 − 1 −2.1 21 t2 − 1 − 1 = t2 − 2 . (b) g(3) = √3 + 1 = 2; g(−1) = 3; g(π) = √π + 1; g(−1.1) = 3; g(t2 − 1) = 3 if t2 < 2 and g(t2 − 1) = t2 − 1 + 1 = |t| if t2 ≥ 2. 1 2 Chapter 0 −√ 25 − x2, 0 < x ≤ 5 9. (a) Natural domain: x /= 3. Range: y =/ 0. (b) Natural domain: x /= 0. Range: {1, −1}. (c) Natural domain: x ≤ −√3 or x ≥ √3. Range: y ≥ 0. (d) x2 − 2x + 5 = (x − 1)2 + 4 ≥ 4. So G(x) is defined for all x, and is ≥ √4 = 2. Natural domain: all x. Range: y ≥ 2. (e) Natural domain: sin x =/ 1, so x /= (2n+ 1 )π, n = 0, ±1, ±2, .....For such x, −1 ≤ sin x < 1, so 0 < 1−sin x ≤ 2, 2 and 1 ≥ 1 . Range: y ≥ 1 . 1−sin x 2 2 (f) Division by 0 occurs for x = 2. For all other x, x2−4 = x + 2, which is nonnegative for x ≥ −2. Natural √ x−2 √ domain: [−2, 2) ∪ (2, +∞). The range of Range: [0, 2) ∪ (2, +∞). x + 2 is [0, +∞). But we must exclude x = 2, for which x + 2 = 2. 10. (a) Natural domain: x ≤ 3. Range: y ≥ 0. (b) Natural domain: −2 ≤ x ≤ 2. Range: 0 ≤ y ≤ 2. (c) Natural domain: x ≥ 0. Range: y ≥ 3. (d) Natural domain: all x. Range: all y. (e) Natural domain: all x. Range: −3 ≤ y ≤ 3. √(f ) For √x to exist, we must have x ≥ 0. For H(x) to exist, we must also have sin √x =/ 0, which is eq√uivalent to x πn√for n = 0, 1, 2, . . .. Natural domain: x > 0, x /= (πn)2 for n = 1, 2, . . .. For such x, 0 < | sin x| ≤ 1, so 0 < (sin x)2 ≤ 1 and H(x) ≥ 1. Range: y ≥ 1. 11. (a) The curve is broken whenever someone is born or someone dies. (b) C decreases for eight hours, increases rapidly (but continuously), and then repeats. 12. (a) Yes. The temperature may change quickly under some conditions, but not instantaneously. (b) No; the number is always an integer, so the changes are in movements (jumps) of at least one unit. 13. 14. 15. Yes. y = √ 25 − x2. 16. Yes. y = −√ 25 − x2. 17. Yes. y = √ 25 − x2, −5 ≤ x ≤ 0 h t t Exercise Set 0.2 3 ≤ — − ≤ 1 − 2x, x < 0 7 − 2x, x < −1 18. No; the vertical line x = 0 meets the graph twice. 19. False. E.g. the graph of x2 − 1 crosses the x-axis at x = 1 and x = −1. 20. True. This is Definition 0.1.5. 21. False. The range also includes 0. 22. False. The domain of g only includes those x for which f (x) > 0. 23. (a) x = 2, 4 (b) None (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no maximum value. 24. (a) x = 9 (b) None (c) x ≥ 25 (d) ymin = 1; no maximum value. 25. The cosine of θ is (L − h)/L (side adjacent over hypotenuse), so h = L(1 − cos θ). 26. The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse), so L = 20 sin(θ/2). 27. (a) If x < 0, then |x| = −x so f (x) = −x + 3x + 1 = 2x + 1. If x ≥ 0, then |x| = x so f (x) = x + 3x + 1 = 4x + 1; (b) If x < 0, then |x| = −x and |x − 1| = 1 − x so g(x) = −x + (1 − x) = 1 − 2x. If 0 ≤ x < 1, then |x| = x and |x − 1| = 1 − x so g(x) = x + (1 − x) = 1. If x ≥ 1, then |x| = x and |x − 1| = x − 1 so g(x) = x + (x − 1) = 2x − 1; g(x) = 1, 0 x < 1 2x − 1, x ≥ 1 28. (a) If x < 5/2, then |2x − 5| = 5 − 2x so f (x) = 3 + (5 − 2x) = 8 − 2x. If x ≥ 5/2, then |2x − 5| = 2x − 5 so f (x) = 3 + (2x − 5) = 2x − 2; f (x) = 8 − 2x, x < 5/2 2x − 2, x ≥ 5/2 (b) If x < −1, then |x − 2| = 2 − x and |x + 1| = −x − 1 so g(x) = 3(2 − x) − (−x − 1) = 7 − 2x. If −1 ≤ x < 2, then |x − 2| = 2 − x and |x + 1| = x + 1 so g(x) = 3(2 − x) − (x + 1) = 5 − 4x. If x ≥ 2, then |x − 2| = x − 2 and |x + 1| = x + 1 so g(x) = 3(x − 2) − (x + 1) = 2x − 7; g(x) = 5 4x, 1 x < 2 2x − 7, x ≥ 2 29. (a) V = (8 − 2x)(15 − 2x)x (b) 0 < x < 4 100 0 4 (c) 0 0 < V ≤ 91, approximately (d) As x increases, V increases and then decreases; the maximum value occurs when x is about 1.7. 30. (a) V = (6 − 2x)2x (b) 0 < x < 3 f (x) = 2x + 1, x < 0 4x + 1, x ≥ 0 4 Chapter 0 ≈ π 20 0 3 (c) 0 0 < V ≤ 16, approximately (d) As x increases, V increases and then decreases; the maximum value occurs when x is about 1. 31. (a) The side adjacent to the building has length x, so L = x + 2y. (b) A = xy = 1000, so L = x + 2000 /x. 120 20 (c) 0 < x ≤ 100 (d) 80 80 x ≈ 44.72 ft, y ≈ 22.36 ft 6000 0 6 32. (a) x = 3000 tan θ (b) 0 ≤ θ < π/2 (c) 3000 ft 0 33. (a) V = 500 = πr2h, so h = 500 . Then C = (0.02)(2) πr2 + (0.01)2 πrh = 0.04πr2 + 0.02πr 500 = 0.04πr2 + 10 ; πr2 Cmin ≈ 4.39 cents at r ≈ 3.4 cm, h ≈ 13.7 cm. πr2 r (b) C = (0.02)(2)(2 r)2 +(0.01)2 πrh = 0.16r2 + 10 . Since 0.04π < 0.16, the top and bottom now get more weight. r Since they cost more, we diminish their sizes in the solution, and the cans become taller. (c) r ≈ 3.1 cm, h ≈ 16.0 cm, C ≈ 4.76 cents. 34. (a) The length of a track with straightaways of length L and semicircles of radius r is P = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π 1222 .65 ft. Since this is less than 1320 ft (a quarter -mile), a solution is possible. 450 (b) 0 100 0 P = 2L + 2πr = 1320 and 2r = 2x + 160, so L = (1320 − 2πr)/2 = (1320 − 2π(80 + x))/2 = 660 − 80π − πx. (c) The shortest straightaway is L = 360, so we solve the equation 360 = 660 − 80π − πx to obtain x = 300 − 80 ≈ Exercise Set 0.2 5 15.49 ft. 6 Chapter 0 y x –2 1 y 1 x 1 3 y 1 x –2 –1 1 2 –1 (d) The longest straightaway occurs when x = 0, so L = 660 − 80π ≈ 408.67 ft. 35. (i) x = 1, −2 causes division by zero. (ii) g(x) = x + 1, all x. 36. (i) x = 0 causes division by zero. (ii) g(x) = |x| + 1, all x. 37. (a) 25◦F (b) 13◦F (c) 5◦F 38. If v = 48 then −60 = WCT ≈ 1.4157 T − 30.6763; thus T ≈ −21◦F when WCT = −60. 39. If v = 48 then −60 = WCT ≈ 1.4157 T − 30.6763; thus T ≈ 15◦F when WCT = −10. 40. The WCT is given by two formulae, but the first doesn’t work with the data. Hence 5 = WCT = −27.2v0.16 +48.17 and v ≈ 18mi/h. Exercise Set 0.2 y 2 1 1. (a) (b) x 1 2 3 (c) (d) –4 –2 2 2. (a) (b) (c) (d) 3. (a) (b) y 1 x –1 1 2 –1 y 1 x –1 1 2 y 2 x y 1 x 1 y 1 x –1 1 y 1 x 1 –1 Exercise Set 0.2 7 1 y x –1 1 2 3 –1 2 (c) (d) 4. 5. Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis, translate down 3 units. 6. Translate right 3 units, compress vertically by a factor of 1 , and translate up 2 units. 7. y = (x + 3)2 − 9; translate left 3 units and down 9 units. y 1 x –1 1 2 3 –1 1 y x -3 -2 -1 1 2 3 -1 –6 y –2 2 –20 x 6 –60 y 2 x 4 8 Chapter 0 2 2 2 8. y = 1 [(x − 1)2 + 2]; translate right 1 unit and up 2 units, compress vertically by a factor of 1 9. Translate left 1 unit, reflect over x-axis, translate up 3 units. 10. Translate right 4 units and up 1 unit. 11. Compress vertically by a factor of 1 , translate up 1 unit. 12. Stretch vertically by a factor of √3 and reflect over x-axis. 13. Translate right 3 units. y 2 1 x 1 0 1 2 3 4 -1 01 2 3 4 5 6 7 8 9 10 y 2 x 1 2 3 y x 2 –1 Exercise Set 0.2 9 10 –10 14. Translate right 1 unit and reflect over x-axis. 15. Translate left 1 unit, reflect over x-axis, translate up 2 units. 16. y = 1 − 1/x; reflect over x-axis, translate up 1 unit. 17. Translate left 2 units and down 2 units. 18. Translate right 3 units, reflect over x-axis, translate up 1 unit. y 19. Stretch vertically by a factor of 2, translate right 1/2 unit and up 1 unit. y x –4 –2 –2 y x 4 6 1 –1 5 y 2 x –2 4 –4 y 6 x –3 –2 –6 1 2 y 5 x 2 –5 x 10 Chapter 0 y x 20. y = |x − 2|; translate right 2 units. 21. Stretch vertically by a factor of 2, reflect over x-axis, translate up 1 unit. 22. Translate right 2 units and down 3 units. 23. Translate left 1 unit and up 2 units. 24. Translate right 2 units, reflect over x-axis. 1 –1 4 y 4 2 x 2 y 2 1 x 2 4 y 3 1 x –2 –1 2 y x 2 –2 y 3 1 x –3 –1 1 Exercise Set 0.2 11 | | y 2 25. (a) x –1 1 (b) y = 0 if x ≤ 0 2x if 0 < x 26. 27. (f + g)(x) = 3√x − 1, x ≥ 1; (f − g)(x) = √x − 1, x ≥ 1; (fg)(x) = 2x − 2, x ≥ 1; (f/g)(x) = 2, x > 1 28. (f + g)(x) = (2x2 + 1)/[x(x2 + 1)], all x /= 0; (f − g)(x) = −1/[x(x2 + 1)], all x 0; (fg)(x) = 1/(x2 + 1), all x 0; (f/g)(x) = x2/(x2 + 1), all x /= 0 29. (a) 3 (b) 9 (c) 2 (d) 2 (e) √2 + h (f) (3 + h)3 + 1 30. (a) √5s + 2 (b) √√x + 2 (c) 3√5x (d) 1/√x (e) (g) 1/√4 x (h) |x − 1| (i) √x + h 31. (f ◦ g)(x) = 1 − x, x ≤ 1; (g ◦ f )(x) = √ 1 − x2, |x| ≤ 1. 32. (f ◦ g)(x) = √√ x2 + 3 − 3, |x| ≥ √6; (g ◦ f )(x) = √x, x ≥ 3. √4 x (f) 0, x ≥ 0 1 1 1 1 33. (f ◦ g)(x) = 1 − 2x , x 2 , 1; (g ◦ f )(x) = − 2x − 2 , x 0, 1. x 1 34. (f ◦ g)(x) = x2 + 1 , x /= 0; (g ◦ f )(x) = x + x, x /= 0. 35. (f ◦ g ◦ h)(x) = x−6 + 1. x 36. (f ◦ g ◦ h)(x) = 1 + x . 37. (a) g(x) = √x, h(x) = x + 2 (b) g(x) = |x|, h(x) = x2 − 3x + 5 38. (a) g(x) = x + 1, h(x) = x2 (b) g(x) = 1/x, h(x) = x − 3 39. (a) g(x) = x2, h(x) = sin x (b) g(x) = 3/x, h(x) = 5 + cos x 40. (a) g(x) = 3 sin x, h(x) = x2 (b) g(x) = 3x2 + 4x, h(x) = sin x 41. (a) g(x) = (1 + x)3, h(x) = sin(x2) (b) g(x) = √ 1 − x, h(x) = √3 x 42. (a) g(x) = 1 , h(x) = x2 (b) g(x) = 5 + x , h(x) = 2x y x 2 –5 12 Chapter 0 1 − x Exercise Set 0.2 13 43. True, by Definition 0.2.1. 44. False. The domain consists of all x in the domain of g such that g(x) is in the domain of f . 45. True, by Theorem 0.2.3(a). 46. False. The graph of y = f (x + 2) + 3 is obtained by translating the graph of y = f (x) left 2 units and up 3 units. 47. 48. {−2, −1, 0, 1, 2, 3} 49. Note that f (g(−x)) = f (−g(x)) = f (g(x)), so f (g(x)) is even. 50. Note that g(f (−x)) = g(f (x)), so g(f (x)) is even. 51. f (g(x)) = 0 when g(x) = ±2, so x ≈ ±1.5; g(f (x)) = 0 when f (x) = 0, so x = ±2. 52. f (g(x)) = 0 at x = −1 and g(f (x)) = 0 at x = −1. 53. 3(x + h)2 − 5 − (3x2 − 5) h 6xh + 3h2 = h = 6x + 3h; 3w2 − 5 − (3x2 − 5) w − x = 3(w − x)(w + x) w − x = 3w + 3x. 54. (x + h)2 + 6(x + h) − (x2 + 6x) h 2xh + h2 + 6h = h = 2x + h + 6; w2 + 6w − (x2 + 6x) w − x = w + x + 6. 55. 1/(x + h) − 1/x = x − (x + h) = −1 ; 1/w − 1/x = x − w = − 1 . h xh(x + h) x(x + h) w − x wx(w − x) xw 1/(x + h)2 − 1/x2 x2 − (x + h)2 2x + h 1/w2 − 1/x2 x2 − w2 x + w 56. h = x2h(x + h)2 = − x2(x + h)2 ; w − x = x2w2(w − x) = − x2w2 . 2 y x –2 2 –2 –4 y f (g(x)) 1 x –3 –1 –1 1 –3 y 3 g( f (x)) 1 x –3 –1 –1 –2 1 3 14 Chapter 0 57. Neither; odd; even. Exercise Set 0.2 15 y x 2 2 2 2 58. (a) (b) 59. (a) (b) (c) 60. (a) (b) 61. (a) Even. (b) Odd. 62. (a) Odd. (b) Neither. 63. (a) f (−x) = (−x)2 = x2 = f (x), even. (b) f (−x) = (−x)3 = −x3 = −f (x), odd. (c) f (−x) = | − x| = |x| = f (x), even. (d) f (−x) = −x + 1, neither. (−x)5 − (−x) x5 − x (e) f (−x) = 1 + (−x)2 = − 1 + x2 = −f (x), odd. (f) f (−x) = 2 = f (x), even. 64. (a) g(−x) = f (−x) + f (x) = f (x) + f (−x) = g(x), so g is even. (b) h(−x) = f (−x) − f (x) = − f (x) − f (−x) = −h(x), so h is odd. 65. In Exercise 64 it was shown that g is an even function, and h is odd. Moreover by inspection f (x) = g(x) + h(x) for all x, so f is the sum of an even function and an odd function. 66. (a) x-axis, because x = 5(−y)2 + 9 gives x = 5y2 + 9. (b) x-axis, y-axis, and origin, because x2 − 2(−y)2 = 3, (−x)2 − 2y2 = 3, and (−x)2 − 2(−y)2 = 3 all give x2 − 2y2 = 3. (c) Origin, because (−x)(−y) = 5 gives xy = 5. 67. (a) y-axis, because (−x)4 = 2y3 + y gives x4 = 2y3 + y. (b) Origin, because (−y) = (−x) gives y = x . 3 + (−x)2 3 + x2 (c) x-axis, y-axis, and origin because (−y)2 = |x| − 5, y2 = | − x| − 5, and (−y)2 = | − x| − 5 all give y2 = |x| − 5. y x y x y x y x x −3 −2 −1 0 1 2 3 f (x) 1 −5 −1 0 −1 −5 1 x −3 −2 −1 0 1 2 3 f (x) 1 5 −1 0 1 −5 −1 16 Chapter 0 y 5 2 x — − 3 –4 4 68. –3 2 –3 3 69. –2 70. (a) Whether we replace x with x, y with y, or both, we obtain the same equation, so by Theorem 0.2.3 the graph is symmetric about the x-axis, the y-axis and the origin. (b) y = (1 − x2/3)3/2. (c) For quadrant II, the same; for III and IV use y = −(1 − x2/3)3/2. 71. 1 2 72. 73. (a) (b) y 1 74. (a) (b) – ‚2 –1 x 1 ‚2‚3 y 2 x 2 y 1 x y 2 x y 1 x –1 1 Exercise Set 0.2 17 y 3 x –1 1 3 3 9 − x0 we get y = ±√ 9 − x2 . (c) (d) 75. Yes, e.g. f (x) = xk and g(x) = xn where k and n are integers. Exercise Set 0.3 1. (a) y = 3x + b (b) y = 3x + 6 (c) 2. Since the slopes are negative reciprocals, y = − 1 x + b. 3. (a) y = mx + 2 (b) m = tan φ = tan 135◦ = −1, so y = −x + 2 (c) 4. (a) y = mx (b) y = m(x − 1) (c) y = −2 + m(x − 1) (d) 2x + 4y = C 5. Let the line be tangent to the circle at the point (x0, y0) where x2 + y2 = 9. The slope of the tangent line is the 0 0 negative reciprocal of y0/x0 (why?), so m = −x0/y0 and y = −(x0/y0)x + b. Substituting the point (x0, y0) as √ 2 9 − x0x 6. Solve the simultaneous equations to get the point (−2, 1/3) of intersection. Then y = 1 + m(x + 2). 7. The x-intercept is x = 10 so that with depreciation at 10% per year the final value is always zero, and hence y = m(x − 10). The y-intercept is the original value. y 1 x y 10 y = 3x + 6 y = 3x + 2 y = 3x – 4 x –2 2 –10 y 5 4 3 y =1.5x +2 y =-x +2 y =x +2 1 x -2 1 2 well as y0 = ± 0 y x 2 6 10 18 Chapter 0 y 2 x -2 2 -2 8. A line through (6, −1) has the form y + 1 = m(x − 6). The intercepts are x = 6 + 1/m and y = −6m − 1. Set −(6 + 1/m)(6m + 1) = 3, or 36m2 + 15m + 1 = (12m + 1)(3 m + 1) = 0 with roots m = −1/12, −1/3; thus y + 1 = −(1/3)(x − 6) and y + 1 = −(1/12)(x − 6). 9. (a) The slope is −1. (b) The y-intercept is y = −1. (c) They pass through the point (−4, 2). (d) The x-intercept is x = 1. 10. (a) Horizontal lines. (b) The y-intercept is y = −1/2. (c) The x-intercept is x = −1/2. 4 2 y x –1 1 –4 y 6 2 x –6 –4 –2 y 3 1 x 1 2 –1 –3 y x y 2 x 2 – 2 y 1 x 1 Exercise Set 0.2 19 y 4 2 x –4 2 4 –2 y 2 1 x –1 2 3 –2 y 1 x (d) They pass through (−1, 1). 11. (a) VI (b) IV (c) III (d) V (e) I (f) II 12. In all cases k must be positive, or negative values would appear in the chart. Only kx−3 decreases, so that must be f (x). Next, kx2 grows faster than kx3/2, so that would be g(x), which grows faster than h(x) (to see this, consider ratios of successive values of the functions). Finally, experimentation (a spreadsheet is handy) for values of k yields (approximately) f (x) = 10x−3, g(x) = x2/2, h(x) = 2x1.5. 13. (a) (b) –4 2 4 (c) 1 2 3 y (–1, 1) 1 x –2 1 y 30 10 x –2 1 2 –10 –30 y x –2 –10 1 2 –40 y 10 6 x 20 Chapter 0 14. (a) y x –2 (b) (c) 15. (a) (b) (c) 16. (a) (b) (c) y 40 x 2 –40 y 80 40 x –2 2 y 4 x –2 2 –4 y 4 x –3 –1 –2 y x –2 2 4 –2 y 10 5 x –2 2 –5 –10 y 6 4 x –2 1 2 –2 y 10 5 x –2 2 –5 –10 y 3 x 2 y 2 x 4 -2 y 2 x 2 –2 –4 2 Exercise Set 0.2 21 √ √ the reflection of the graph of y = x across the y-axis. y 17. (a) –3 –1 1 (b) (c) (d) 18. (a) (b) (c) (d) 19. y = x2 + 2x = (x + 1)2 − 1. 20. (a) The part of the graph of y =√√ |x| with x ≥ 0 is the same as the graph of y = √x. The part with x ≤ 0 is -3 -2 -1 0 1 2 3 (b) The part of the graph of y = 3 |x| with x ≥ 0 is the same as the part of the graph of y = √3 x with x ≥ 0. The part with x ≤ 0 is the reflection of the graph of y = 3 x with x ≥ 0 across the y-axis. -3 -2 -1 0 1 2 3 21. (a) N·m (b) k = 20 N·m (c) y x – y 6 2 x y 80 20 x –20 1 2 3 4 5 –80 y x –3 –10 –40 y 1 x 1 y 4 x –2 2 –4 12 8 -7 -6 -5 -4 -3 -2 -1 0 1 40 20 x –20 –40 1 2 3 4 5 2 1 6 –2 –1 –2 2 V (L) 0.25 0.5 1.0 1.5 2.0 P (N/m2) 80 × 103 40 × 103 20 × 103 13.3 × 103 10 × 103 y 3 2 x y 3 2 x