Baixe Solution do Livro de Calculo do Anton e outras Exercícios em PDF para Cálculo, somente na Docsity! Physics ACT http://PHYSICSACT.wordpress.com 1 CHAPTER 1 Functions EXERCISE SET 1.1 1. (a) around 1943 (b) 1960; 4200 (c) no; you need the year’s population (d) war; marketing techniques (e) news of health risk; social pressure, antismoking campaigns, increased taxation 2. (a) 1989; $35,600 (b) 1975, 1983; $32,000 (c) the first two years; the curve is steeper (downhill) 3. (a) −2.9,−2.0, 2.35, 2.9 (b) none (c) y = 0 (d) −1.75 ≤ x ≤ 2.15 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2 4. (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 5. (a) x = 2, 4 (b) none (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no maximum value 6. (a) x = 9 (b) none (c) x ≥ 25 (d) ymin = 1; no maximum value 7. (a) Breaks could be caused by war, pestilence, flood, earthquakes, for example. (b) C decreases for eight hours, takes a jump upwards, and then repeats. 8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature change. (b) No; the number is always an integer, so the changes are in movements (jumps) of at least one unit. 9. (a) The side adjacent to the building has length x, so L = x+ 2y. Since A = xy = 1000, L = x+ 2000/x. (b) x > 0 and x must be smaller than the width of the building, which was not given. (c) 120 80 20 80 (d) Lmin ≈ 89.44 ft 10. (a) V = lwh = (6− 2x)(6− 2x)x (b) From the figure it is clear that 0 < x < 3. (c) 20 0 0 3 (d) Vmax ≈ 16 in3 4 Chapter 1 15. (a) V = (8− 2x)(15− 2x)x (b) −∞ < x < +∞,−∞ < V < +∞ (c) 0 < x < 4 (d) minimum value at x = 0 or at x = 4; maximum value somewhere in between (can be approximated by zooming with graphing calculator) 16. (a) x = 3000 tan θ (b) θ = nπ + π/2 for n an integer, −∞ < n <∞ (c) 0 ≤ θ < π/2, 0 ≤ x < +∞ (d) 3000 ft 6000 0 0 6 17. (i) x = 1,−2 causes division by zero (ii) g(x) = x+ 1, all x 18. (i) x = 0 causes division by zero (ii) g(x) = √ x+ 1 for x ≥ 0 19. (a) 25◦F (b) 2◦F (c) −15◦F 20. If v = 48 then −60 = WCI = 1.6T − 55; thus T = (−60 + 55)/1.6 ≈ −3◦F. 21. If v = 8 then −10 = WCI = 91.4 + (91.4− T )(0.0203(8)− 0.304 √ 8− 0.474); thus T = 91.4 + (10 + 91.4)/(0.0203(8)− 0.304 √ 8− 0.474) and T = 5◦F 22. The WCI is given by three formulae, but the first and third don’t work with the data. Hence −15 = WCI = 91.4 + (91.4 − 20)(0.0203v − 0.304√v − 0.474); set x = √v so that v = x2 and obtain 0.0203x2 − 0.304x− 0.474 + (15 + 91.4)/(91.4− 20) = 0. Use the quadratic formula to find the two roots. Square them to get v and discard the spurious solution, leaving v ≈ 25. 23. Let t denote time in minutes after 9:23 AM. Then D(t) = 1000− 20t ft. EXERCISE SET 1.3 1. (e) seems best, though only (a) is bad. -0.5 0.5 y -1 1 x 2. (e) seems best, though only (a) is bad and (b) is not good. -0.5 0.5 y -1 1 x 3. (b) and (c) are good; (a) is very bad. 12 13 14 y -1 1 x Exercise Set 1.3 5 4. (b) and (c) are good; (a) is very bad. -14 -13 -12 -11 y -2 -1 0 1 2 x 5. [−3, 3]× [0, 5] 2 y x 6. [−4, 2]× [0, 3] –2 y x 7. (a) window too narrow, too short (b) window wide enough, but too short (c) good window, good spacing -400 -200 y -5 5 10 20 x (d) window too narrow, too short (e) window too narrow, too short 8. (a) window too narrow (b) window too short (c) good window, good tick spacing -250 -100 50 y -16 -8 -4 4 x (d) window too narrow, too short (e) shows one local minimum only, window too narrow, too short 9. [−5, 14]× [−60, 40] -60 20 40 y -5 5 10 x 10. [6, 12]× [−100, 100] -50 100 y 10 12 x 11. [−0.1, 0.1]× [−3, 3] -2 2 3 y -0.1 x 0.1 12. [−1000, 1000]× [−13, 13] -5 5 10 y -1000 x 1000 13. [−250, 1050]× [−1500000, 600000] -500000 y -1000 1000 x 6 Chapter 1 14. [−3, 20]× [−3500, 3000] -2000 -1000 1000 5 10 15 x y 15. [−2, 2]× [−20, 20] -20 -10 10 20 y x -2 -1 1 2 16. [1.6, 2]× [0, 2] 0.5 1 2 y 1.6 1.7 1.8 2 x 17. depends on graphing utility -2 2 4 6 y -4 2 4 x 18. depends on graphing utility -2 2 4 6 y -4 -2 2 4 x 19. (a) f(x) = √ 16− x2 1 3 -2 2 4 x y (b) f(x) = − √ 16− x2 -3 -1 x y -4 -2 2 4 (c) -2 2 x y -2 2 (d) 1 4 1 4 y x (e) No; the vertical line test fails. 20. (a) y = ±3 √ 1− x2/4 -3 3 -2 2 x y (b) y = ± √ x2 + 1 -4 -2 2 4 -4 -2 2 4 x y Exercise Set 1.4 9 32. (a) y –2 2 1 2 x (b) x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of the curve come together. If a moves past b then a and b switch roles. -3 -1 1 3 1 2 3 x a = 0.5 b = 1.5 y -3 -1 1 3 1 2 3 x a = 1 b = 1.5 y -3 -1 1 3 2 3 x a = 1.5 b = 1.6 y 33. The curve oscillates between the lines y = x and y = −x with increasing rapidity as |x| increases. -30 -10 10 30 y -30 -10 10 30 x 34. The curve oscillates between the lines y = +1 and y = −1, infinitely many times in any neighborhood of x = 0. y –1 4–2 x EXERCISE SET 1.4 1. (a) -1 0 1 y -1 1 2 x (b) 2 1 y 1 2 3 x (c) 1 y -1 1 2 x (d) 2 y -4 -2 2 x 10 Chapter 1 2. (a) x y 1-3 (b) x 1 y 1 -1 (c) x 1 y -1 1 (d) x 1 y 1 -1 3. (a) -1 1 y x -2 -1 1 2 (b) -1 1 y 1 x (c) -1 1 y -1 1 2 3 x (d) -1 1 y -1 1 2 3 x 4. 1 y 2 x 5. Translate right 2 units, and up one unit. 10 -2 2 6 x y 6. Translate left 1 unit, reflect over x-axis, and translate up 2 units. y x –2 1 7. Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis, translate down 3 units. -60 -20 -6 -2 2 6 xy Exercise Set 1.4 11 8. Translate right 3 units, compress vertically by a factor of 12 , and translate up 2 units. y x 2 4 9. y = (x+ 3)2 − 9; translate left 3 units and down 9 units. -5 10 15 -8 -4 -2 2 x y 10. y = (x+ 3)2 − 19; translate left 3 units and down 19 units. y x –5 –4 11. y = −(x− 1)2 + 2; translate right 1 unit, reflect over x-axis, translate up 2 units. -6 -2 2 -2 -1 1 3 x y 12. y = 12 [(x− 1)2 + 2]; translate left 1 unit and up 2 units, compress vertically by a factor of 12 . y x 1 2 1 13. Translate left 1 unit, reflect over x-axis, translate up 3 units. 1 2 2 8 x y 14. Translate right 4 units and up 1 unit. y x 1 4 4 10 15. Compress vertically by a factor of 12 , translate up 1 unit. 2 y 1 2 3 x 16. Stretch vertically by a factor of √ 3 and reflect over x-axis. y x -1 2 17. Translate right 3 units. -10 10 y 4 6 x 18. Translate right 1 unit and reflect over x-axis. y x -4 2 -2 2 19. Translate left 1 unit, reflect over x-axis, translate up 2 units. -6 -2 2 6 y -3 -1 1 2 x 14 Chapter 1 50. (a) g(x) = 3 sinx, h(x) = x2 (b) g(x) = 3x2 + 4x, h(x) = sinx 51. (a) f(x) = x3, g(x) = 1 + sinx, h(x) = x2 (b) f(x) = √ x, g(x) = 1− x, h(x) = 3√x 52. (a) f(x) = 1/x, g(x) = 1− x, h(x) = x2 (b) f(x) = |x|, g(x) = 5 + x, h(x) = 2x 53. y x -4 -3 -2 -1 1 2 -3 -2 -1 1 2 3 54. {−2,−1, 0, 1, 2, 3} 55. Note that f(g(−x)) = f(−g(x)) = f(g(x)), so f(g(x)) is even. f (g(x)) x y 1–3 –1 –1 –3 1 56. Note that g(f(−x)) = g(f(x)), so g(f(x)) is even. x y –1 –2 1 3 –1 1 3 –3 g( f (x)) 57. f(g(x)) = 0 when g(x) = ±2, so x = ±1.4; g(f(x)) = 0 when f(x) = 0, so x = ±2. 58. f(g(x)) = 0 at x = −1 and g(f(x)) = 0 at x = −1 59. 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 60. (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 61. 1/(x+ h)− 1/x h = x− (x+ h) xh(x+ h) = −1 x(x+ h) ; 1/w − 1/x w − x = x− w wx(w − x) = − 1 xw 62. 1/(x+ h)2 − 1/x2 h = x2 − (x+ h)2 x2h(x+ h)2 = − 2x+ h x2(x+ h)2 ; 1/w2 − 1/x2 w − x = x2 − w2 x2w2(w − x) = − x+ w x2w2 63. (a) the origin (b) the x-axis (c) the y-axis (d) none Exercise Set 1.4 15 64. (a) x y (b) x y (c) x y 65. (a) x −3 −2 −1 0 1 2 3 f(x) 1 −5 −1 0 −1 −5 1 (b) x −3 −2 −1 0 1 2 3 f(x) 1 5 −1 0 1 −5 −1 66. (a) x y (b) x y 67. (a) even (b) odd (c) odd (d) neither 68. neither; odd; even 69. (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 (e) f(−x) = (−x) 3 − (−x) 1 + (−x)2 = − x3 + x 1 + x2 = −f(x), odd (f) f(−x) = 2 = f(x), even 70. (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 71. (a) y-axis, because (−x)4 = 2y3 + y gives x4 = 2y3 + y (b) origin, because (−y) = (−x) 3 + (−x)2 gives y = x 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 16 Chapter 1 72. 3 –3 -4 4 73. 2 -2 -3 3 74. (a) Whether we replace x with −x, y with −y, or both, we obtain the same equation, so by Theorem 1.4.3 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. (For graphing it may be helpful to use the tricks that precede Exercise 29 in Section 1.3.) 75. 2 5 y 1 2 x 76. y x 2 2 77. (a) 1 y C x O c o (b) 2 y O x C c o 78. (a) x 1 y 1 -1 (b) x 1 y 1 -1 ‚2-‚2 ‚3 (c) x 1 y -1 3 (d) x y 1 c/2C 79. Yes, e.g. f(x) = xk and g(x) = xn where k and n are integers. 80. If x ≥ 0 then |x| = x and f(x) = g(x). If x < 0 then f(x) = |x|p/q if p is even and f(x) = −|x|p/q if p is odd; in both cases f(x) agrees with g(x). Exercise Set 1.5 19 27. (a) The velocity is the slope, which is 5− (−4) 10− 0 = 9/10 ft/s. (b) x = −4 (c) The line has slope 9/10 and passes through (0,−4), so has equation x = 9t/10− 4; at t = 2, x = −2.2. (d) t = 80/9 28. (a) v = 5− 1 4− 2 = 2 m/s (b) x− 1 = 2(t− 2) or x = 2t− 3 (c) x = −3 29. (a) The acceleration is the slope of the velocity, so a = 3− (−1) 1− 4 = − 4 3 ft/s2. (b) v − 3 = − 43 (t− 1), or v = − 43 t+ 133 (c) v = 133 ft/s 30. (a) The acceleration is the slope of the velocity, so a = 0− 5 10− 0 = − 5 10 = −1 2 ft/s2. (b) v = 5 ft/s (c) v = 4 ft/s (d) t = 4 s 31. (a) It moves (to the left) 6 units with velocity v = −3 cm/s, then remains motionless for 5 s, then moves 3 units to the left with velocity v = −1 cm/s. (b) vave = 0− 9 10− 0 = − 9 10 cm/s (c) Since the motion is in one direction only, the speed is the negative of the velocity, so save = 910 cm/s. 32. It moves right with constant velocity v = 5 km/h; then accelerates; then moves with constant, though increased, velocity again; then slows down. 33. (a) If x1 denotes the final position and x0 the initial position, then v = (x1 − x0)/(t1 − t0) = 0 mi/h, since x1 = x0. (b) If the distance traveled in one direction is d, then the outward journey took t = d/40 h. Thus save = total dist total time = 2d t+ (2/3)t = 80t t+ (2/3)t = 48 mi/h. (c) t+ (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip 34. (a) down, since v < 0 (b) v = 0 at t = 2 (c) It’s constant at 32 ft/s2. 35. (a) v t 20 60 20 80 (b) v = 10t if 0≤ t ≤ 10 100 if 10≤ t ≤ 100 600− 5t if 100≤ t ≤ 120 36. x t 20 Chapter 1 37. (a) y = 20− 15 = 5 when x = 45, so 5 = 45k, k = 1/9, y = x/9 (b) y x 0.2 0.6 2 6 (c) l = 15 + y = 15 + 100(1/9) = 26.11 in. (d) If ymax = 15 then solve 15 = kx = x/9 for x = 135 lb. 38. (a) Since y = 0.2 = (1)k, k = 1/5 and y = x/5 (b) y x 1 2 6 (c) y = 3k = 3/5 so 0.6 ft. (d) ymax = (1/2)3 = 1.5 so solve 1.5 = x/5 for x = 7.5 tons 39. Each increment of 1 in the value of x yields the increment of 1.2 for y, so the relationship is linear. If y = mx+ b then m = 1.2; from x = 0, y = 2, follows b = 2, so y = 1.2x+ 2 40. Each increment of 1 in the value of x yields the increment of −2.1 for y, so the relationship is linear. If y = mx+ b then m = −2.1; from x = 0, y = 10.5 follows b = 10.5, so y = −2.1x+ 10.5 41. (a) With TF as independent variable, we have TC − 100 TF − 212 = 0− 100 32− 212 , so TC = 5 9 (TF − 32). (b) 5/9 (c) Set TF = TC = 59 (TF − 32) and solve for TF : TF = TC = −40◦ (F or C). (d) 37◦ C 42. (a) One degree Celsius is one degree Kelvin, so the slope is the ratio 1/1 = 1. Thus TC = TK − 273.15. (b) TC = 0− 273.15 = −273.15◦ C 43. (a) p− 1 h− 0 = 5.9− 1 50− 0 , or p = 0.098h+ 1 (b) when p = 2, or h = 1/0.098 ≈ 10.20 m 44. (a) R− 123.4 T − 20 = 133.9− 123.4 45− 20 , so R = 0.42T + 115. (b) T = 32.38 ◦C 45. (a) r − 0.80 t− 0 = 0.75− 0.80 4− 0 , so r = −0.0125t+ 0.8 (b) 64 days 46. (a) Let the position at rest be y0. Then y0 + y = y0 + kx; with x = 11 we get y0 + kx = y0 + 11k = 40, and with x = 24 we get y0 + kx = y0 + 24k = 60. Solve to get k = 20/13 and y0 = 300/13. (b) 300/13 + (20/13)W = 30, so W = (390− 300)/20 = 9/2 g. Exercise Set 1.6 21 47. (a) For x trips we have C1 = 2x and C2 = 25 + x/4 C x 20 60 10 20 (b) 2x = 25 + x/4, or x = 100/7, so the commuter pass becomes worthwhile at x = 15. 48. If the student drives x miles, then the total costs would be CA = 4000 + (1.25/20)x and CB = 5500 + (1.25/30)x. Set 4000 + 5x/80 = 5500 + 5x/120 and solve for x = 72, 000 mi. EXERCISE SET 1.6 1. (a) y = 3x+ b (b) y = 3x+ 6 (c) -10 10 y -2 2 x y = 3x + 6 y = 3x + 2 y = 3x - 4 2. Since the slopes are negative reciprocals, y = − 13x+ b. 3. (a) y = mx+ 2 (b) m = tanφ = tan 135◦ = −1, so y = −x+ 2 (c) 1 3 4 5 -2 1 2 x y m = –1 m = 1 m = 1.5 4. (a) y = mx (b) y = m(x− 1) (c) y = −2 +m(x− 1) (d) 2x+ 4y = C 5. (a) The slope is −1. -3 -2 2 5 -1 1 2 x y (b) The y-intercept is y = −1. -4 2 4 -1 1 x y 24 Chapter 1 (b) x y 2 4 -2 -4 x y 2 -2 -2 -4 (c) x y -1 -2 4 -3 -2 2 4 -2 xy 15. (a) -10 -5 5 10 y -2 2 x (b) -2 4 6 y -2 1 2 x (c) -10 -5 5 10 y -2 2 x 16. (a) x y 2 3 (b) x y 2 2 (c) x y 2 2 Exercise Set 1.6 25 17. (a) 2 6 y -3 -1 1 x (b) -80 -20 20 80 y -1 1 2 3 4 5 x (c) -40 -10 y -3 1 x (d) -40 -20 20 40 y 21 3 4 5 x 18. (a) x y 1 1 (b) -2-4-6 2 -2 -1 1 2 x y (c) x y 2 4 -2 -4 (d) x y -2 4 -6 19. (a) -1 0.5 1.5 y -3 -2 1 x (b) -1 -0.5 0.5 1 y 1 3 x (c) -20 10 30 y -1 1 3 x (d) -20 10 30 y -2 1 2 x 26 Chapter 1 20. (a) -10 10 y 1 3 4 x (b) -4 -4 -2 2 4 2 x y (c) -106 106 y -1 1 x (d) 5 10 y -2 2 x 21. y = x2 + 2x = (x+ 1)2 − 1 5 10 y -2 2 x 22. (a) 1 y -2 2 x (b) 1 y -2 2 x 23. (a) N·m (b) k = 20 N·m (c) 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 (d) 10 20 30 P(N/m2) 10 20 V(m3) 24. If the side of the square base is x and the height of the container is y then V = x2y = 100; minimize A = 2x2 + 4xy = 2x2 + 400/x. A graphing utility with a zoom feature suggests that the solution is a cube of side 100 1 3 cm. Exercise Set 1.7 29 (c) A = 5 √ 13/2, θ = tan−1 1 2 √ 3 ; x = 5 √ 13 2 sin ( 2πt+ tan−1 1 2 √ 3 ) 10 -10 ^ 6 46. three; x = 0, x = ±1.8955 3 –3 -3 3 EXERCISE SET 1.7 1. The sum of the squares for the residuals for line I is approximately 12 + 12 + 12 + 02 + 22 + 12 + 12 + 12 = 10, and the same for line II is approximately 02 + (0.4)2 + (1.2)2 + 02 + (2.2)2 + (0.6)2 + (0.2)2 + 02 = 6.84; line II is the regression line. 2. (a) The data appear to be periodic, so a trigonometric model may be appropriate. (b) The data appear to lie near a parabola, so a quadratic model may be appropriate. (c) The data appear to lie near a line, so a linear model may be appropriate. (d) The data appear to be randomly distributed, so no model seems to be appropriate. 3. Least squares line S = 1.5388t− 2842.9, correlation coefficient 0.83409 1980 1983 1986 1989 1992 1995 1998 190 200 210 220 230 240 t S 4. (a) p = 0.0154T + 4.19 (b) 3.42 atm 5. (a) Least squares line p = 0.0146T + 3.98, correlation coefficient 0.9999 (b) p = 3.25 atm (c) T = −272◦C 6. (a) p = 0.0203T + 5.54, r = 0.9999 (b) T = −273 (c) 1.05 p = 0.0203(T )(1.1) + 5.54 and p = 0.0203T + 5.54, subtract to get .05p = 0.1(0.0203T ), p = 2(0.0203T ). But p = 0.0203T + 5.54, equate the right hand sides, get T = 5.54/0.0203 ≈ 273◦C 7. (a) R = 0.00723T + 1.55 (b) T = −214◦C 30 Chapter 1 8. (a) y = 0.0236x+ 4.79 (b) T = −203◦C 9. (a) S = 0.50179w − 0.00643 (b) S = 8, w = 16 lb 10. (a) S = 0.756w − 0.0133 (b) 2S = 0.756(w + 5)− 0.0133, subtract to get S = 5(0.756) = 3.78 11. (a) Let h denote the height in inches and y the number of rebounds per minute. Then y = 0.00630h− 0.266, r = 0.313 (b) 66 67 68 69 70 71 72 73 74 75 0.05 0.1 0.15 0.2 0.25 0.3 h y (c) No, the data points are too widely scattered. 12. (a) Let h denote the height in inches and y the weight in pounds. Then y = 7.73h− 391 (b) 72 76 80 84 88 160 180 200 220 240 260 280 h y (c) 259 lb 13. (a) H ≈ 20000/110 ≈ 181 km/s/Mly (b) One light year is 9.408× 1012 km and t = d v = 1 H = 1 20km/s/Mly = 9.408× 1018km 20km/s = 4.704× 1017 s = 1.492× 1010 years. (c) The Universe would be even older. 14. (a) f = 0.906m+ 5.93 (b) f = 85.7 15. (a) P = 0.322t2 + 0.0671t+ 0.00837 (b) P = 1.43 cm 16. The population is modeled by P = 0.0045833t2− 16.378t+ 14635; in the year 2000 the population would be P = 212, 200, 000. This is far short; in 1990 the population of the US was approximately 250,000,000. 17. As in Example 4, a possible model is of the form T = D + A sin [ B ( t− C B )] . Since the longest day has 993 minutes and the shortest has 706, take 2A = 993 − 706 = 287 or A = 143.5. The midpoint between the longest and shortest days is 849.5 minutes, so there is a vertical shift of D = 849.5. The period is about 365.25 days, so 2π/B = 365.25 or B = π/183. Note that the sine function takes the value −1 when t − C B = −91.8125, and T is a minimum at about t = 0. Thus the phase shift C B ≈ 91.5. Hence T = 849.5 + 143.5 sin [ π 183 t− π 2 ] is a model for the temperature. Exercise Set 1.8 31 18. As in Example 4, a possible model for the fraction f of illumination is of the form f = D + A sin [ B ( t− C B )] . Since the greatest fraction of illumination is 1 and the least is 0, 2A = 1, A = 1/2. The midpoint of the fraction of illumination is 1/2, so there is a vertical shift of D = 1/2. The period is approximately 30 days, so 2π/B = 30 or B = π/15. The phase shift is approximately 49/2, so C/B = 49/2, and f = 1/2 + 1/2 sin [ π 15 ( t− 49 2 )] 19. t = 0.445 √ d 2.3 0 0 25 20. (a) t = 0.373 r1.5 (b) 238,000 km (c) 1.89 days EXERCISE SET 1.8 1. (a) x+ 1 = t = y − 1, y = x+ 2 2 4 6 y 2 4 t = 0 t = 1 t = 2 t = 3 t = 4 t = 5 x (c) t 0 1 2 3 4 5 x −1 0 1 2 3 4 y 1 2 3 4 5 6 2. (a) x2 + y2 = 1 1 y -1 1 xt = 1 t = 0.5 t = 0.75 t = 0.25 t = 0 (c) t 0 0.2500 0.50 0.7500 1 x 1 0.7071 0.00 −0.7071 −1 y 0 0.7071 1.00 0.7071 0 3. t = (x+ 4)/3; y = 2x+ 10 -8 6 12 x y 4. t = x+ 3; y = 3x+ 2,−3 ≤ x ≤ 0 x y (0, 2) (-3, -7) 34 Chapter 1 23. (a) 3 -5 0 20 (b) o O -1 1 24. (a) 1.7 -1.7 -2.3 2.3 (b) -10 10 ^ 6 25. (a) Eliminate t to get x− x0 x1 − x0 = y − y0 y1 − y0 (b) Set t = 0 to get (x0, y0); t = 1 for (x1, y1). (c) x = 1 + t, y = −2 + 6t (d) x = 2− t, y = 4− 6t 26. (a) x = −3− 2t, y = −4 + 5t, 0 ≤ t ≤ 1 (b) x = at, y = b(1− t), 0 ≤ t ≤ 1 27. (a) |R−P |2 = (x−x0)2+(y−y0)2 = t2[(x1−x0)2+(y1−y0)2] and |Q−P |2 = (x1−x0)2+(y1−y0)2, so r = |R− P | = |Q− P |t = qt. (b) t = 1/2 (c) t = 3/4 28. x = 2 + t, y = −1 + 2t (a) (5/2, 0) (b) (9/4,−1/2) (c) (11/4, 1/2) 29. The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1 coincide. 30. (a) Eliminate t− t0 t1 − t0 to obtain y − y0 x− x0 = y1 − y0 x1 − x0 . (b) from (x0, y0) to (x1, y1) (c) x = 3− 2(t− 1), y = −1 + 5(t− 1) 5 -2 0 5 Exercise Set 1.8 35 31. (a) x− b a = y − d c (b) 1 2 3 y 1 2 3 x 32. (a) If a = 0 the line segment is vertical; if c = 0 it is horizontal. (b) The curve degenerates to the point (b, d). 33. 1 2 y 0.5 1 x 34. x= 1/2− 4t, y = 1/2 for 0≤ t≤ 1/4 x=−1/2, y = 1/2− 4(t− 1/4) for 1/4≤ t≤ 1/2 x=−1/2 + 4(t− 1/2), y =−1/2 for 1/2≤ t≤ 3/4 x= 1/2, y =−1/2 + 4(t− 3/4) for 3/4≤ t≤ 1 35. (a) x = 4 cos t, y = 3 sin t (b) x = −1 + 4 cos t, y = 2 + 3 sin t (c) 3 -3 -4 4 5 -1 -5 3 36. (a) t = x/(v0 cosα), so y = x tanα− gx2/(2v20 cos2 α). (b) 2000 6000 10000 40000 80000 x y 37. (a) From Exercise 36, x = 400 √ 2t, y = 400 √ 2t− 4.9t2. (b) 16,326.53 m (c) 65,306.12 m 36 Chapter 1 38. (a) 15 –15 -25 25 (b) 15 –15 -25 25 (c) 15 –15 -25 25 a = 3, b = 2 15 –15 -25 25 a = 2, b = 3 15 –15 -25 25 a = 2, b = 7 39. Assume that a = 0 and b = 0; eliminate the parameter to get (x − h)2/a2 + (y − k)2/b2 = 1. If |a| = |b| the curve is a circle with center (h, k) and radius |a|; if |a| = |b| the curve is an ellipse with center (h, k) and major axis parallel to the x-axis when |a| > |b|, or major axis parallel to the y-axis when |a| < |b|. (a) ellipses with a fixed center and varying axes of symmetry (b) (assume a = 0 and b = 0) ellipses with varying center and fixed axes of symmetry (c) circles of radius 1 with centers on the line y = x− 1 40. Refer to the diagram to get bθ = aφ, θ = aφ/b but θ − α = φ+ π/2 so α = θ − φ− π/2 = (a/b− 1)φ− π/2 x= (a− b) cosφ− b sinα = (a− b) cosφ+ b cos ( a− b b ) φ, y= (a− b) sinφ− b cosα = (a− b) sinφ− b sin ( a− b b ) φ. x y a ba − b Chapter 1 Supplementary Exercises 39 18. (a) y = cosx− 2 sinx cosx = (1− 2 sinx) cosx, so x = ±π 2 ,±3π 2 , π 6 , 5π 6 ,−7π 6 ,−11π 6 (b) ( ±π 2 , 0 ) , ( ±3π 2 , 0 ) , (π 6 , √ 3/2 ) , ( 5π 6 ,− √ 3/2 ) , ( −7π 6 ,− √ 3/2 ) , ( −11π 6 , √ 3/2 ) 19. (a) If x denotes the distance from A to the base of the tower, and y the distance from B to the base, then x2+d2 = y2. Moreover h = x tanα = y tanβ, so d2 = y2−x2 = h2(cot2 β−cot2 α), h2 = d2 cot2 β − cot2 α = d2 1/ tan2 β − 1/ tan2 α = d2 tan2 α tan2 β tan2 α− tan2 β , which yields the result. (b) 295.72 ft. 20. (a) -20 20 60 y 100 300 t (b) when 2π 365 (t − 101) = 3π 2 , or t = 374.75, which is the same date as t = 9.75, so during the night of January 10th-11th (c) from t = 0 to t = 70.58 and from t = 313.92 to t = 365 (the same date as t = 0) , for a total of about 122 days 21. When x = 0 the value of the green curve is higher than that of the blue curve, therefore the blue curve is given by y = 1 + 2 sinx. The points A,B,C,D are the points of intersection of the two curves, i.e. where 1+2 sinx = 2 sin(x/2)+2 cos(x/2). Let sin(x/2) = p, cos(x/2) = q. Then 2 sinx = 4 sin(x/2) cos(x/2), so the equation which yields the points of intersection becomes 1 + 4pq = 2p+ 2q, 4pq− 2p− 2q+ 1 = 0, (2p− 1)(2q− 1) = 0; thus whenever either sin(x/2) = 1/2 or cos(x/2) = 1/2, i.e. when x/2 = π/6, 5π/6,±π/3. Thus A has coordinates (−2π/3, 1 − √ 3), B has coordinates (π/3, 1 + √ 3), C has coordinates (2π/3, 1 + √ 3), and D has coordinates (5π/3, 1− √ 3). 22. Let y = A+B sin(at+ b). Since the maximum and minimum values of y are 35 and 5, A+B = 35 and A−B = 5, A = 20, B = 15. The period is 12 hours, so 12a = 2π and a = π/6. The maximum occurs at t = 2, so 1 = sin(2a + b) = sin(π/3 + b), π/3 + b = π/2, b = π/2 − π/3 = π/6 and y = 20 + 15 sin(πt/6 + π/6). 23. (a) The circle of radius 1 centered at (a, a2); therefore, the family of all circles of radius 1 with centers on the parabola y = x2. (b) All parabolas which open up, have latus rectum equal to 1 and vertex on the line y = x/2. 24. (a) x = f(1− t), y = g(1− t) 25. -2 1 y -2 1 2 x 40 Chapter 1 26. Let y = ax2 + bx+ c. Then 4a+ 2b+ c = 0, 64a+ 8b+ c = 18, 64a− 8b+ c = 18, from which b = 0 and 60a = 18, or finally y = 310x 2 − 65 . 27. -2 -1 1 2 y -1 1 2 x 28. (a) R = R0 is the R-intercept, R0k is the slope, and T = −1/k is the T -intercept (b) −1/k = −273, or k = 1/273 (c) 1.1 = R0(1 + 20/273), or R0 = 1.025 (d) T = 126.55◦C 29. d = √ (x− 1)2 + (√x− 2)2; d = 9.1 at x = 1.358094 1 2 y 1 2 x 30. d = √ (x− 1)2 + 1/x2; d = 0.82 at x = 1.380278 0.8 1 1.2 1.6 2 0.5 1 2 3 y x 31. w = 63.9V , w = 63.9πh2(5/2− h/3); h = 0.48 ft when w = 108 lb 32. (a) 1000 3000 W 1 3 5 h (b) w = 63.9πh2(5/2− h/3); at h = 5/2, w = 2091.12 lb 33. (a) 100 200 N 10 30 50 t (b) N = 80 when t = 9.35 yrs (c) 220 sheep 34. (a) T v 10 20 (b) T = 17◦F, 27◦F, 32◦F Chapter 1 Supplementary Exercises 41 35. (a) -20 20 WCI 20 40 v (b) T = 3◦F, −11◦F, −18◦F, −22◦F (c) v = 35, 19, 12, 7 mi/h 36. The domain is the set of all x, the range is −0.1746 ≤ y ≤ 0.1227. 37. The domain is the set −0.7245 ≤ x ≤ 1.2207, the range is −1.0551 ≤ y ≤ 1.4902. 38. (a) The potato is done in the interval 27.65 < t < 32.71. (b) 91.54 min. 39. (a) 5 20 v 1 2 3 4 5 t (b) As t→∞, (0.273)t → 0, and thus v → 24.61 ft/s. (c) For large t the velocity approaches c. (d) No; but it comes very close (arbitrarily close). (e) 3.013 s 40. (a) y = −0.01716428571x+ 1.433827619 41. (a) 1.90 1.92 1.94 1.96 1.98 2.00 2.02 2.04 2.06 2.08 2.10 3.4161 3.4639 3.5100 3.5543 3.5967 3.6372 3.6756 3.7119 3.7459 3.7775 3.8068 (b) y = 1.9589x− 0.2910 (c) y − 3.6372 = 1.9589(x− 2), or y = 1.9589x− 0.2806 (d) As one zooms in on the point (2, f(2)) the two curves seem to converge to one line. 3.8 3.4 1.9 2.2 42. (a) −0.10 −0.08 −0.06 −0.04 −0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.9950 0.9968 0.9982 0.9992 0.9998 1.0000 0.9998 0.9992 0.9982 0.9968 0.9950 (b) y = − 12x2 + 1 44 CHAPTER 2 Limits and Continuity EXERCISE SET 2.1 1. (a) −1 (b) 3 (c) does not exist (d) 1 (e) −1 (f) 3 2. (a) 2 (b) 0 (c) does not exist (d) 2 (e) 0 (f) 2 3. (a) 1 (b) 1 (c) 1 (d) 1 (e) −∞ (f) +∞ 4. (a) 3 (b) 3 (c) 3 (d) 3 (e) +∞ (f) +∞ 5. (a) 0 (b) 0 (c) 0 (d) 3 (e) +∞ (f) +∞ 6. (a) 2 (b) 2 (c) 2 (d) 3 (e) −∞ (f) +∞ 7. (a) −∞ (b) +∞ (c) does not exist (d) undef (e) 2 (f) 0 8. (a) +∞ (b) +∞ (c) +∞ (d) undef (e) 0 (f) −1 9. (a) −∞ (b) −∞ (c) −∞ (d) 1 (e) 1 (f) 2 10. (a) 1 (b) −∞ (c) does not exist (d) −2 (e) +∞ (f) +∞ 11. (a) 0 (b) 0 (c) 0 (d) 0 (e) does not exist (f) does not exist 12. (a) 3 (b) 3 (c) 3 (d) 3 (e) does not exist (f) 0 13. for all x0 = −4 14. for all x0 = −6, 3 19. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.999 0.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337 1 0 0 2 The limit is 1/3. Exercise Set 2.1 45 (b) 2 1.5 1.1 1.01 1.001 1.0001 0.4286 1.0526 6.344 66.33 666.3 6666.3 50 0 1 2 The limit is +∞. (c) 0 0.5 0.9 0.99 0.999 0.9999 −1 −1.7143 −7.0111 −67.001 −667.0 −6667.0 0 -50 0 1 The limit is −∞. 20. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25 0.5359 0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.4721 0.6 0 -0.25 0.25 The limit is 1/2. (b) 0.25 0.1 0.001 0.0001 8.4721 20.488 2000.5 20001 100 0 0 0.25 The limit is +∞. 46 Chapter 2 (c) −0.25 −0.1 −0.001 −0.0001 −7.4641 −19.487 −1999.5 −20000 0 -100 -0.25 0 The limit is −∞. 21. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.25 2.7266 2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.7266 3 2 -0.25 0.25 The limit is 3. (b) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001 1 1.7552 6.2161 54.87 541.1 −0.1415 −4.536 −53.19 −539.5 60 -60 -1.5 0 The limit does not exist. 22. (a) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001 1.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.0000 1.5 1 -1.5 0 The limit is 1. Exercise Set 2.2 49 27. It appears that lim t→+∞ n(t) = +∞, and lim t→+∞ e(t) = c. 28. (a) It is the initial temperature of the oven. (b) It is the ambient temperature, i.e. the temperature of the room. 29. (a) lim t→0+ sin t t (b) lim t→0+ t− 1 t + 1 (c) lim t→0− (1 + 2t)1/t 30. (a) lim t→0+ cosπt πt (b) lim t→0+ 1 t + 1 (c) lim t→0− ( 1 + 2 t )t 31. lim x→−∞ f(x) = L and lim x→+∞ = L 32. (a) no (b) yes; tanx and secx at x = nπ + π/2, and cotx and cscx at x = nπ, n = 0,±1,±2, . . . 33. (a) The limit appears to be 3. 3.5 2.5 –1 1 (b) The limit appears to be 3. 3.5 2.5 –0.001 0.001 (c) The limit does not exist. 3.5 2.5 –0.000001 0.000001 35. (a) The plot over the interval [−a, a] becomes subject to catastrophic subtraction if a is small enough (the size depending on the machine). (c) It does not. EXERCISE SET 2.2 1. (a) 7 (b) π (c) −6 (d) 36 2. (a) 1 (b) −1 (c) 1 (d) −1 3. (a) −6 (b) 13 (c) −8 (d) 16 (e) 2 (f) −1/2 (g) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. 50 Chapter 2 4. (a) 0 (b) The limit doesn’t exist because lim f doesn’t exist and lim g does. (c) 0 (d) 3 (e) 0 (f) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. (g) The limit doesn’t exist because √ f(x) is not defined for 0 ≤ x < 2. (h) 1 5. 0 6. 3/4 7. 8 8. −3 9. 4 10. 12 11. −4/5 12. 0 13. 3/2 14. 4/3 15. +∞ 16. −∞ 17. does not exist 18. +∞ 19. −∞ 20. does not exist 21. +∞ 22. −∞ 23. does not exist 24. −∞ 25. +∞ 26. does not exist 27. +∞ 28. +∞ 29. 6 30. 4 32. −19 33. (a) 2 (b) 2 (c) 2 34. (a) −2 (b) 0 (c) does not exist 35. (a) 3 (b) y x 4 1 36. (a) −6 (b) F (x) = x− 3 37. (a) Theorem 2.2.2(a) doesn’t apply; moreover one cannot add/subtract infinities. (b) lim x→0+ ( 1 x − 1 x2 ) = lim x→0+ ( x− 1 x2 ) = −∞ 38. lim x→0− ( 1 x + 1 x2 ) = lim x→0− x + 1 x2 = +∞ 39. lim x→0 x x (√ x + 4 + 2 ) = 1 4 40. lim x→0 x2 x (√ x + 4 + 2 ) = 0 41. The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any preassigned real number. For example, let q(x) = x− x0 and let p(x) = a(x− x0)n where n takes on the values 0, 1, 2. Exercise Set 2.3 51 EXERCISE SET 2.3 1. (a) −3 (b) −∞ 2. (a) 1 (b) −1 3. (a) −12 (b) 21 (c) −15 (d) 25 (e) 2 (f) −3/5 (g) 0 (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. 4. (a) 20 (b) 0 (c) +∞ (d) −∞ (e) −421/3 (f) −6/7 (g) 7 (h) −7/12 5. +∞ 6. 5 7. −∞ 8. +∞ 9. +∞ 10. +∞ 11. 3/2 12. 5/2 13. 0 14. 0 15. 0 16. 5/3 17. −51/3/2 18. 3 √ 3/2 19. − √ 5 20. √ 5 21. 1/ √ 6 22. −1/ √ 6 23. √ 3 24. √ 3 25. −∞ 26. +∞ 27. −1/7 28. 4/7 29. (a) +∞ (b) −5 30. (a) 0 (b) −6 31. lim x→+∞ ( √ x2 + 3− x) √ x2 + 3 + x√ x2 + 3 + x = lim x→+∞ 3√ x2 + 3 + x = 0 32. lim x→+∞ ( √ x2 − 3x− x) √ x2 − 3x + x√ x2 − 3x + x = limx→+∞ −3x√ x2 − 3x + x = −3/2 33. lim x→+∞ (√ x2 + ax− x ) √x2 + ax + x√ x2 + ax + x = lim x→+∞ ax√ x2 + ax + x = a/2 34. lim x→+∞ (√ x2 + ax− √ x2 + bx ) √x2 + ax +√x2 + bx√ x2 + ax + √ x2 + bx = lim x→+∞ (a− b)x√ x2 + ax + √ x2 + bx = a− b 2 35. lim x→+∞ p(x) = (−1)n∞ and lim x→−∞ p(x) = +∞ 36. If m > n the limits are both zero. If m = n the limits are both 1. If n > m the limits are (−1)n+m∞ and +∞, respectively. 37. If m > n the limits are both zero. If m = n the limits are both equal to am, the leading coefficient of p. If n > m the limits are ±∞ where the sign depends on the sign of am and whether n is even or odd. 38. (a) p(x) = q(x) = x (b) p(x) = x, q(x) = x2 (c) p(x) = x2, q(x) = x (d) p(x) = x + 3, q(x) = x 54 Chapter 2 6. √ 5x + 1 = 3.5 at x = 2.25, √ 5x + 1 = 4.5 at x = 3.85, so δ = min(3− 2.25, 3.85− 3) = 0.75 5 0 2 4 7. With the TRACE feature of a calculator we discover that (to five decimal places) (0.87000, 1.80274) and (1.13000, 2.19301) belong to the graph. Set x0 = 0.87 and x1 = 1.13. Since the graph of f(x) rises from left to right, we see that if x0 < x < x1 then 1.80274 < f(x) < 2.19301, and therefore 1.8 < f(x) < 2.2. So we can take δ = 0.13. 8. From a calculator plot we conjecture that lim x→1 f(x) = 2. Using the TRACE feature we see that the points (±0.2, 1.94709) belong to the graph. Thus if −0.2 < x < 0.2 then 1.95 < f(x) ≤ 2 and hence |f(x)− L| < 0.05 < 0.1 = . 9. |2x− 8| = 2|x− 4| < 0.1 if |x− 4| < 0.05, δ = 0.05 10. |x/2 + 1| = (1/2)|x− (−2)| < 0.1 if |x + 2| < 0.2, δ = 0.2 11. |7x + 5− (−2)| = 7|x− (−1)| < 0.01 if |x + 1| < 1700 , δ = 1700 12. |5x− 2− 13| = 5|x− 3| < 0.01 if |x− 3| < 1500 , δ = 1500 13. ∣∣∣∣x2 − 4x− 2 − 4 ∣∣∣∣ = ∣∣∣∣x2 − 4− 4x + 8x− 2 ∣∣∣∣ = |x− 2| < 0.05 if |x− 2| < 0.05, δ = 0.05 14. ∣∣∣∣x2 − 1x + 1 − (−2) ∣∣∣∣ = ∣∣∣∣x2 − 1 + 2x + 2x + 1 ∣∣∣∣ = |x + 1| < 0.05 if |x + 1| < 0.05, δ = 0.05 15. if δ < 1 then ∣∣x2 − 16∣∣ = |x− 4||x + 4| < 9|x− 4| < 0.001 if |x− 4| < 19000 , δ = 19000 16. if δ < 1 then |√x − 3| ∣∣∣∣ √ x + 3√ x + 3 ∣∣∣∣ = |x− 9||√x + 3| < |x− 9|√8 + 3 < 14 |x − 9| < 0.001 if |x − 9| < 0.004, δ = 0.004 17. if δ ≤ 1 then ∣∣∣∣ 1x − 15 ∣∣∣∣ = |x− 5|5|x| ≤ |x− 5|20 < 0.05 if |x− 5| < 1, δ = 1 18. |x− 0| = |x| < 0.05 if |x| < 0.05, δ = 0.05 19. |3x− 15| = 3|x− 5| < if |x− 5| < 13, δ = 13 20. |(4x− 5)− 7| = |4x− 12| = 4|x− 3| < if |x− 3| < 14, δ = 14 21. |2x− 7− (−3)| = 2|x− 2| < if |x− 2| < 12, δ = 12 22. |2− 3x− 5| = 3|x + 1| < if |x + 1| < 13, δ = 13 23. ∣∣∣∣x2 + xx − 1 ∣∣∣∣ = |x| < if |x| < , δ = Exercise Set 2.4 55 24. ∣∣∣∣x2 − 9x + 3 − (−6) ∣∣∣∣ = |x + 3| < if |x + 3| < , δ = 25. if δ < 1 then |2x2 − 2| = 2|x− 1||x + 1| < 6|x− 1| < if |x− 1| < 16, δ = min(1, 16) 26. if δ < 1 then |x2 − 5− 4| = |x− 3||x + 3| < 7|x− 3| < if |x− 3| < 17, δ = min(1, 17) 27. if δ < 1 6 then ∣∣∣∣ 1x − 3 ∣∣∣∣ = 3|x− 13 ||x| < 18 ∣∣∣∣x− 13 ∣∣∣∣ < if ∣∣∣∣x− 13 ∣∣∣∣ < 118, δ = min ( 1 6 , 1 18 ) 28. If δ < 12 and |x− (−2)| < δ then − 52 < x < − 32 , x + 1 < − 12 , |x + 1| > 12 ; then∣∣∣∣ 1x + 1 − (−1) ∣∣∣∣ = |x + 2||x + 1| < 2|x + 2| < if |x + 2| < 12, δ = min ( 1 2 , 1 2 ) 29. |√x− 2| = ∣∣∣∣(√x− 2) √ x + 2√ x + 2 ∣∣∣∣ = ∣∣∣∣ x− 4√x + 2 ∣∣∣∣ < 12 |x− 4| < if |x− 4| < 2, δ = 2 30. |√x + 3− 3| ∣∣∣∣ √ x + 3 + 3√ x + 3 + 3 ∣∣∣∣ = |x− 6|√x + 3 + 3 ≤ 13 |x− 6| < if |x− 6| < 3, δ = 3 31. |f(x)− 3| = |x + 2− 3| = |x− 1| < if 0 < |x− 1| < , δ = 32. If δ < 1 then |(x2 + 3x− 1)− 9| = |(x− 2)(x + 5)| < 8|x− 2| < if |x− 2| < 18, δ = min (1, 18) 33. (a) |f(x)− L| = 1 x2 < 0.1 if x > √ 10, N = √ 10 (b) |f(x)− L| = ∣∣∣∣ xx + 1 − 1 ∣∣∣∣ = ∣∣∣∣ 1x + 1 ∣∣∣∣ < 0.01 if x + 1 > 100, N = 99 (c) |f(x)− L| = ∣∣∣∣ 1x3 ∣∣∣∣ < 11000 if |x| > 10, x < −10, N = −10 (d) |f(x) − L| = ∣∣∣∣ xx + 1 − 1 ∣∣∣∣ = ∣∣∣∣ 1x + 1 ∣∣∣∣ < 0.01 if |x + 1| > 100, −x − 1 > 100, x < −101, N = −101 34. (a) ∣∣∣∣ 1x3 ∣∣∣∣ < 0.1, x > 101/3, N = 101/3 (b) ∣∣∣∣ 1x3 ∣∣∣∣ < 0.01, x > 1001/3, N = 1001/3 (c) ∣∣∣∣ 1x3 ∣∣∣∣ < 0.001, x > 10, N = 10 35. (a) x21 1 + x21 = 1− , x1 = − √ 1− ; x22 1 + x22 = 1− , x2 = √ 1− (b) N = √ 1− (c) N = − √ 1− 36. (a) x1 = −1/3; x2 = 1/3 (b) N = 1/3 (c) N = −1/3 37. 1 x2 < 0.01 if |x| > 10, N = 10 38. 1 x + 2 < 0.005 if |x + 2| > 200, x > 198, N = 198 39. ∣∣∣∣ xx + 1 − 1 ∣∣∣∣ = ∣∣∣∣ 1x + 1 ∣∣∣∣ < 0.001 if |x + 1| > 1000, x > 999, N = 999 56 Chapter 2 40. ∣∣∣∣4x− 12x + 5 − 2 ∣∣∣∣ = ∣∣∣∣ 112x + 5 ∣∣∣∣ < 0.1 if |2x + 5| > 110, 2x > 105, N = 52.5 41. ∣∣∣∣ 1x + 2 − 0 ∣∣∣∣ < 0.005 if |x + 2| > 200, −x− 2 > 200, x < −202, N = −202 42. ∣∣∣∣ 1x2 ∣∣∣∣ < 0.01 if |x| > 10, −x > 10, x < −10, N = −10 43. ∣∣∣∣4x− 12x + 5 − 2 ∣∣∣∣ = ∣∣∣∣ 112x + 5 ∣∣∣∣ < 0.1 if |2x+5| > 110, −2x−5 > 110, 2x < −115, x < −57.5, N = −57.5 44. ∣∣∣∣ xx + 1 − 1 ∣∣∣∣ = ∣∣∣∣ 1x + 1 ∣∣∣∣ < 0.001 if |x + 1| > 1000, −x− 1 > 1000, x < −1001, N = −1001 45. ∣∣∣∣ 1x2 ∣∣∣∣ < if |x| > 1√ , N = 1√ 46. ∣∣∣∣ 1x ∣∣∣∣ < if |x| > 1 , −x > 1 , x < −1 , N = −1 47. ∣∣∣∣ 1x + 2 ∣∣∣∣ < if |x + 2| > 1 , −x− 2 < 1 , x > −2− 1 , N = −2− 1 48. ∣∣∣∣ 1x + 2 ∣∣∣∣ < if |x + 2| > 1 , x + 2 > 1 , x > 1 − 2, N = 1 − 2 49. ∣∣∣∣ xx + 1 − 1 ∣∣∣∣ = ∣∣∣∣ 1x + 1 ∣∣∣∣ < if |x + 1| > 1 , x > 1 − 1, N = 1 − 1 50. ∣∣∣∣ xx + 1 − 1 ∣∣∣∣ = ∣∣∣∣ 1x + 1 ∣∣∣∣ < if |x + 1| > 1 , −x− 1 > 1 , x < −1− 1 , N = −1− 1 51. ∣∣∣∣4x− 12x + 5 − 2 ∣∣∣∣ = ∣∣∣∣ 112x + 5 ∣∣∣∣ < if |2x + 5| > 11 , −2x− 5 > 11 , 2x < −11 − 5, x < −112 − 52 , N = −5 2 − 11 2 52. ∣∣∣∣4x− 12x + 5 − 2 ∣∣∣∣ = ∣∣∣∣ 112x + 5 ∣∣∣∣ < if |2x + 5| > 11 , 2x > 11 − 5, x > 112 − 52 , N = 112 − 52 53. (a) 1 x2 > 100 if |x| < 1 10 (b) 1 |x− 1| > 1000 if |x− 1| < 1 1000 (c) −1 (x− 3)2 < −1000 if |x− 3| < 1 10 √ 10 (d) − 1 x4 < −10000 if x4 < 1 10000 , |x| < 1 10 54. (a) 1 (x− 1)2 > 10 if and only if |x− 1| < 1√ 10 (b) 1 (x− 1)2 > 1000 if and only if |x− 1| < 1 10 √ 10 (c) 1 (x− 1)2 > 100000 if and only if |x− 1| < 1 100 √ 10 55. if M > 0 then 1 (x− 3)2 > M , 0 < (x− 3) 2 < 1 M , 0 < |x− 3| < 1√ M , δ = 1√ M Exercise Set 2.5 59 12. (a) no; disasters (war, flood, famine, pestilence, for example) can cause discontinuities (b) continuous (c) not usually continuous; see Exercise 11 (d) continuous 13. none 14. none 15. none 16. f is not defined at x = ±1 17. f is not defined at x = ±4 18. f is not defined at x = −7± √ 57 2 19. f is not defined at x = ±3 20. f is not defined at x = 0,−4 21. none 22. f is not defined at x = 0,−3 23. none; f(x) = 2x + 3 is continuous on x < 4 and f(x) = 7 + 16 x is continuous on 4 < x; lim x→4− f(x) = lim x→4+ f(x) = f(4) = 11 so f is continuous at x = 4 24. lim x→1 f(x) does not exist so f is discontinuous at x = 1 25. (a) f is continuous for x < 1, and for x > 1; lim x→1− f(x) = 5, lim x→1+ f(x) = k, so if k = 5 then f is continuous for all x (b) f is continuous for x < 2, and for x > 2; lim x→2− f(x) = 4k, lim x→2+ f(x) = 4 +k, so if 4k = 4 +k, k = 4/3 then f is continuous for all x 26. (a) no, f is not defined at x = 2 (b) no, f is not defined for x ≤ 2 (c) yes (d) no, f is not defined for x ≤ 2 27. (a) y x c (b) y x c 28. (a) f(c) = lim x→c f(x) (b) lim x→1 f(x) = 2 lim x→1 g(x) = 1 y x 1 1-1 y x 1 1 (c) Define f(1) = 2 and redefine g(1) = 1. 60 Chapter 2 29. (a) x = 0, lim x→0− f(x) = −1 = +1 = lim x→0+ f(x) so the discontinuity is not removable (b) x = −3; define f(−3) = −3 = lim x→−3 f(x), then the discontinuity is removable (c) f is undefined at x = ±2; at x = 2, lim x→2 f(x) = 1, so define f(2) = 1 and f becomes continuous there; at x = −2, lim x→−2 does not exist, so the discontinuity is not removable 30. (a) f is not defined at x = 2; lim x→2 f(x) = lim x→2 x + 2 x2 + 2x + 4 = 1 3 , so define f(2) = 1 3 and f becomes continuous there (b) lim x→2− f(x) = 1 = 4 = lim x→2+ f(x), so f has a nonremovable discontinuity at x = 2 (c) lim x→1 f(x) = 8 = f(1), so f has a removable discontinuity at x = 1 31. (a) discontinuity at x = 1/2, not removable; at x = −3, removable y x -5 5 5 (b) 2x2 + 5x− 3 = (2x− 1)(x + 3) 32. (a) there appears to be one discontinuity near x = −1.52 4 –4 -3 3 (b) one discontinuity at x = −1.52 33. For x > 0, f(x) = x3/5 = (x3)1/5 is the composition (Theorem 2.4.6) of the two continuous functions g(x) = x3 and h(x) = x1/5 and is thus continuous. For x < 0, f(x) = f(−x) which is the composition of the continuous functions f(x) (for positive x) and the continuous function y = −x. Hence f(−x) is continuous for all x > 0. At x = 0, f(0) = lim x→0 f(x) = 0. 34. x4 + 7x2 + 1 ≥ 1 > 0, thus f(x) is the composition of the polynomial x4 + 7x2 + 1, the square root√ x, and the function 1/x and is therefore continuous by Theorem 2.5.6. 35. (a) Let f(x) = k for x = c and f(c) = 0; g(x) = l for x = c and g(c) = 0. If k = −l then f + g is continuous; otherwise it’s not. (b) f(x) = k for x = c, f(c) = 1; g(x) = l = 0 for x = c, g(c) = 1. If kl = 1, then fg is continuous; otherwise it’s not. 36. A rational function is the quotient f(x)/g(x) of two polynomials f(x) and g(x). By Theorem 2.5.2 f and g are continuous everywhere; by Theorem 2.5.3 f/g is continuous except when g(x) = 0. Exercise Set 2.5 61 37. Since f and g are continuous at x = c we know that lim x→c f(x) = f(c) and lim x→c g(x) = g(c). In the following we use Theorem 2.2.2. (a) f(c) + g(c) = lim x→c f(x) + lim x→c g(x) = lim x→c (f(x) + g(x)) so f + g is continuous at x = c. (b) same as (a) except the + sign becomes a − sign (c) f(c) g(c) = lim x→c f(x) lim x→c g(x) = lim x→c f(x) g(x) so f g is continuous at x = c 38. h(x) = f(x)− g(x) satisfies h(a) > 0, h(b) < 0. Use the Intermediate Value Theorem or Theorem 2.5.9. 39. Of course such a function must be discontinuous. Let f(x) = 1 on 0 ≤ x < 1, and f(x) = −1 on 1 ≤ x ≤ 2. 40. A square whose diagonal has length r has area f(r) = r2/2. Note that f(r) = r2/2 < πr2/2 < 2r2 = f(2r). By the Intermediate Value Theorem there must be a value c between r and 2r such that f(c) = πr2/2, i.e. a square of diagonal c whose area is πr2/2. 41. The cone has volume πr2h/3. The function V (r) = πr2h (for variable r and fixed h) gives the volume of a right circular cylinder of height h and radius r, and satisfies V (0) < πr2h/3 < V (r). By the Intermediate Value Theorem there is a value c between 0 and r such that V (c) = πr2h/3, so the cylinder of radius c (and height h) has volume equal to that of the cone. 42. If f(x) = x3 − 4x + 1 then f(0) = 1, f(1) = −2. Use Theorem 2.5.9. 43. If f(x) = x3 + x2 − 2x then f(−1) = 2, f(1) = 0. Use the Intermediate Value Theorem. 44. Since lim x→−∞ p(x) = −∞ and lim x→+∞ p(x) = +∞ (or vice versa, if the leading coefficient of p is negative), it follows that for M = −1 there corresponds N1 < 0, and for M = 1 there is N2 > 0, such that p(x) < −1 for x < N1 and p(x) > 1 for x > N2. Choose x1 < N1 and x2 > N2 and use Theorem 2.5.9 on the interval [x1, x2] to find a solution of p(x) = 0. 45. For the negative root, use intervals on the x-axis as follows: [−2,−1]; since f(−1.3) < 0 and f(−1.2) > 0, the midpoint x = −1.25 of [−1.3,−1.2] is the required approximation of the root. For the positive root use the interval [0, 1]; since f(0.7) < 0 and f(0.8) > 0, the midpoint x = 0.75 of [0.7, 0.8] is the required approximation. 46. x = −1.25 and x = 0.75. 10 -5 -2 -1 1 -1 0.7 0.8 47. For the negative root, use intervals on the x-axis as follows: [−2,−1]; since f(−1.7) < 0 and f(−1.6) > 0, use the interval [−1.7,−1.6]. Since f(−1.61) < 0 and f(−1.60) > 0 the midpoint x = −1.605 of [−1.61,−1.60] is the required approximation of the root. For the positive root use the interval [1, 2]; since f(1.3) > 0 and f(1.4) < 0, use the interval [1.3, 1.4]. Since f(1.37) > 0 and f(1.38) < 0, the midpoint x = 1.375 of [1.37, 1.38] is the required approximation. 64 Chapter 2 39. −0.9 −0.99 −0.999 −0.9999 −0.99999 −1.1 −1.01 −1.001 −1.0001 −1.00001 0.405086 0.340050 0.334001 0.333400 0.333340 0.271536 0.326717 0.332667 0.333267 0.333327 The limit is 1/3. 40. k = f(0) = lim x→0 sin 3x x = 3 lim x→0 sin 3x 3x = 3, so k = 3 41. lim x→0− f(x) = k lim x→0 sin kx kx cos kx = k, lim x→0+ f(x) = 2k2, so k = 2k2, k = 1 2 42. No; sinx/|x| has unequal one-sided limits. 43. (a) lim t→0+ sin t t = 1 (b) lim t→0− 1− cos t t = 0 (Theorem 2.6.3) (c) sin(π − t) = sin t, so lim x→π π − x sinx = lim t→0 t sin t = 1 44. cos (π 2 − t ) = sin t, so lim x→2 cos(π/x) x− 2 = limt→0 (π − 2t) sin t 4t = lim t→0 π − 2t 4 lim t→0 sin t t = π 4 45. t = x− 1; sin(πx) = sin(πt + π) = − sinπt; and lim x→1 sin(πx) x− 1 = − limt→0 sinπt t = −π 46. t = x− π/4; tanx− 1 = 2 sin t cos t− sin t ; limx→π/4 tanx− 1 x− π/4 = limt→0 2 sin t t(cos t− sin t) = 2 47. −|x| ≤ x cos ( 50π x ) ≤ |x| 48. −x2 ≤ x2 sin ( 50π 3 √ x ) ≤ x2 49. lim x→0 f(x) = 1 by the Squeezing Theorem -1 0 1 -1 1 x y y = cos x y = 1 – x2 y = f (x) 50. lim x→+∞ f(x) = 0 by the Squeezing Theorem y x -1 4 51. Let g(x) = − 1 x and h(x) = 1 x ; thus lim x→+∞ sinx x = 0 by the Squeezing Theorem. 52. y x y x Exercise Set 2.6 65 53. (a) sinx = sin t where x is measured in degrees, t is measured in radians and t = πx 180 . Thus lim x→0 sinx x = lim t→0 sin t (180t/π) = π 180 . 54. cosx = cos t where x is measured in degrees, t in radians, and t = πx 180 . Thus lim x→0 1− cosx x = lim t→0 1− cos t (180t/π) = 0. 55. (a) sin 10◦ = 0.17365 (b) sin 10◦ = sin π 18 ≈ π 18 = 0.17453 56. (a) cos θ = cos 2α = 1− 2 sin2(θ/2) ≈ 1− 2(θ/2)2 = 1− 12θ2 (b) cos 10◦ = 0.98481 (c) cos 10◦ = 1− 1 2 ( π 18 )2 ≈ 0.98477 57. (a) 0.08749 (b) tan 5◦ ≈ π 36 = 0.08727 58. (a) h = 52.55 ft (b) Since α is small, tanα◦ ≈ πα 180 is a good approximation. (c) h ≈ 52.36 ft 59. (a) Let f(x) = x − cosx; f(0) = −1, f (π/2) = π/2. By the IVT there must be a solution of f(x) = 0. (b) y x 0 0.5 1 1.5 y = cos x c/2 y = x (c) 0.739 60. (a) f(x) = x + sinx − 1; f(0) = −1, f (π/6) = π/6 − 1/2 > 0. By the IVT there must be a solution of f(x) = 0. (b) y x 0 0.5 c/6 y = x y = 1 – sin x (c) x = 0.511 61. (a) There is symmetry about the equatorial plane. (b) Let g(φ) be the given function. Then g(38) < 9.8 and g(39) > 9.8, so by the Intermediate Value Theorem there is a value c between 38 and 39 for which g(c) = 9.8 exactly. 66 Chapter 2 62. (a) does not exist (b) the limit is zero (c) For part (a) consider the fact that given any δ > 0 there are infinitely many rational numbers x satisfying |x| < δ and there are infinitely many irrational numbers satisfying the same condition. Thus if the limit were to exist, it could not be zero because of the rational numbers, and it could not be 1 because of the irrational numbers, and it could not be anything else because of all the numbers. Hence the limit cannot exist. For part (b) use the Squeezing Theorem with +x and −x as the ‘squeezers’. CHAPTER 2 SUPPLEMENTARY EXERCISES 1. (a) 1 (b) no limit (c) no limit (d) 1 (e) 3 (f) 0 (g) 0 (h) 2 (i) 1/2 2. (a) f(x) = 2x/(x− 1) (b) y x 10 10 4. f(x) = −1 for a ≤ x < a + b 2 and f(x) = 1 for a + b 2 ≤ x ≤ b 5. (a) 0.222 . . . , 0.24390, 0.24938, 0.24994, 0.24999, 0.25000; for x = 2, f(x) = 1 x + 2 , so the limit is 1/4. (b) 1.15782, 4.22793, 4.00213, 4.00002, 4.00000, 4.00000; to prove, use tan 4x x = sin 4x x cos 4x = 4 cos 4x sin 4x 4x , the limit is 4. 6. (a) y = 0 (b) none (c) y = 2 7. (a) x 1 0.1 0.01 0.001 0.0001 0.00001 0.000001 f(x) 1.000 0.443 0.409 0.406 0.406 0.405 0.405 (b) y x 0.5 -1 1 8. (a) 0.4 amperes (b) [0.3947, 0.4054] (c) [ 3 7.5 + δ , 3 7.5− δ ] (d) 0.0187 (e) It becomes infinite. Chapter 2 Supplementary Exercises 69 39. (a) x3 − x− 1 = 0, x3 = x + 1, x = 3√x + 1. (b) y x -1 2 -1 1 (c) y x x1 x2 x3 (d) 1, 1.26, 1.31, 1.322, 1.324, 1.3246, 1.3247 40. (a) y x 10 20 -1 1 32 x1 x2 (b) 0,−1,−2,−9,−730 41. x = 5 √ x + 2; 1.267168 42. x = cosx; 0.739085 (after 33 iterations!). 70 CHAPTER 3 The Derivative EXERCISE SET 3.1 1. (a) mtan = (50− 10)/(15− 5) = 40/10 = 4 m/s (b) t (s) 4 10 20 v (m/s) 2. (a) (10− 10)/(3− 0) = 0 cm/s (b) t = 0, t = 2, and t = 4.2 (horizontal tangent line) (c) maximum: t = 1 (slope > 0) minimum: t = 3 (slope < 0) (d) (3− 18)/(4− 2) = −7.5 cm/s (slope of estimated tangent line to curve at t = 3) 3. From the figure: t s t0 t1 t2 (a) The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 is greater than that at t2. (b) The initial velocity is 0 because the slope of a horizontal line is 0. (c) The particle is speeding up because the slope increases as t increases from t0 to t1. (d) The particle is slowing down because the slope decreases as t increases from t1 to t2. 4. t s t0 t1 5. It is a straight line with slope equal to the velocity. 6. (a) decreasing (slope of tangent line decreases with increasing time) (b) increasing (slope of tangent line increases with increasing time) (c) increasing (slope of tangent line increases with increasing time) (d) decreasing (slope of tangent line decreases with increasing time) 7. (a) msec = f(4)− f(3) 4− 3 = (4)2/2− (3)2/2 1 = 7 2 (b) mtan = lim x1→3 f(x1)− f(3) x1 − 3 = lim x1→3 x21/2− 9/2 x1 − 3 = lim x1→3 x21 − 9 2(x1 − 3) = lim x1→3 (x1 + 3)(x1 − 3) 2(x1 − 3) = lim x1→3 x1 + 3 2 = 3 Exercise Set 3.1 71 (c) mtan = lim x1→x0 f(x1)− f(x0) x1 − x0 = lim x1→x0 x21/2− x20/2 x1 − x0 = lim x1→x0 x21 − x20 2(x1 − x0) = lim x1→x0 x1 + x0 2 = x0 (d) x y Tangent Secant 5 10 8. (a) msec = f(2)− f(1) 2− 1 = 23 − 13 1 = 7 (b) mtan = lim x1→1 f(x1)− f(1) x1 − 1 = lim x1→1 x31 − 1 x1 − 1 = lim x1→1 (x1 − 1)(x21 + x1 + 1) x1 − 1 = lim x1→1 (x21 + x1 + 1) = 3 (c) mtan = lim x1→x0 f(x1)− f(x0) x1 − x0 = lim x1→x0 x31 − x30 x1 − x0 = lim x1→x0 (x21 + x1x0 + x 2 0) = 3x20 (d) x y Secant Tangent 5 9 9. (a) msec = f(3)− f(2) 3− 2 = 1/3− 1/2 1 = − 1 6 (b) mtan = lim x1→2 f(x1)− f(2) x1 − 2 = lim x1→2 1/x1 − 1/2 x1 − 2 = lim x1→2 2− x1 2x1(x1 − 2) = lim x1→2 −1 2x1 = − 1 4 (c) mtan = lim x1→x0 f(x1)− f(x0) x1 − x0 = lim x1→x0 1/x1 − 1/x0 x1 − x0 = lim x1→x0 x0 − x1 x0x1(x1 − x0) = lim x1→x0 −1 x0x1 = − 1 x20 (d) x y Secant Tangent1 4 10. (a) msec = f(2)− f(1) 2− 1 = 1/4− 1 1 = − 3 4 (b) mtan = lim x1→1 f(x1)− f(1) x1 − 1 = lim x1→1 1/x21 − 1 x1 − 1 = lim x1→1 1− x21 x21(x1 − 1) = lim x1→1 −(x1 + 1) x21 = −2 74 Chapter 3 5. y x 1 6. y x 7. y − (−1) = 5(x− 3), y = 5x− 16 8. y − 3 = −4(x+ 2), y = −4x− 5 9. f ′(x) = lim w→x f(w)− f(x) w − x = limw→x 3w2 − 3x2 w − x = limw→x 3(w + x) = 6x; f(3) = 3(3) 2 = 27, f ′(3) = 18 so y − 27 = 18(x− 3), y = 18x− 27 10. f ′(x) = lim w→x f(w)− f(x) w − x = limw→x w4 − x4 w − x = limw→x(w 3 + w2x+ wx2 + x3) = 4x3; f(−2) = (−2)4 = 16, f ′(−2) = −32 so y − 16 = −32(x+ 2), y = −32x− 48 11. f ′(x) = lim w→x f(w)− f(x) w − x = limw→x w3 − x3 w − x = limw→x(w 2 + wx+ x2) = 3x2; f(0) = 03 = 0, f ′(0) = 0 so y − 0 = (0)(x− 0), y = 0 12. f ′(x) = lim w→x f(w)− f(x) w − x = limw→x 2w3 + 1− (2x3 + 1) w − x = limw→x 2(w 2 + wx + x2) = 6x2; f(−1) = 2(−1)3 + 1 = −1, f ′(−1) = 6 so y + 1 = 6(x+ 1), y = 6x+ 5 13. f ′(x) = lim w→x f(w)− f(x) w − x = limw→x √ w + 1−√x+ 1 w − x = lim w→x √ w + 1−√x+ 1 w − x √ w + 1 + √ x+ 1√ w + 1 + √ x+ 1 = lim w→x 1(√ w + 1 + √ x+ 1 ) = 1 2 √ x+ 1 ; f(8) = √ 8 + 1 = 3, f ′(8) = 1 6 so y − 3 = 1 6 (x− 8), y = 1 6 x+ 5 3 14. f ′(x)= lim w→x f(w)− f(x) w − x = limw→x √ 2w + 1−√2x+ 1 w − x = lim w→x 2√ 2w + 1 + √ 2x+ 1 = lim w→x 2√ 9 + 2h+ 3 = 1√ 2x+ 1 f(4) = √ 2(4) + 1 = √ 9 = 3, f ′(4) = 1/3 so y − 3 = 1 3 (x− 4), y = 1 3 x+ 5 3 15. f ′(x)= lim ∆x→0 1 x+∆x − 1 x ∆x = lim ∆x→0 x− (x+∆x) x(x+∆x) ∆x = lim ∆x→0 −∆x x∆x(x+∆x) = lim ∆x→0 − 1 x(x+∆x) = − 1 x2 Exercise Set 3.2 75 16. f ′(x)= lim ∆x→0 1 (x+∆x) + 1 − 1 x+ 1 ∆x = lim ∆x→0 (x+ 1)− (x+∆x+ 1) (x+ 1)(x+∆x+ 1) ∆x = lim ∆x→0 x+ 1− x−∆x− 1 ∆x(x+ 1)(x+∆x+ 1) = lim ∆x→0 −∆x ∆x(x+ 1)(x+∆x+ 1) = lim ∆x→0 −1 (x+ 1)(x+∆x+ 1) = − 1 (x+ 1)2 17. f ′(x)= lim ∆x→0 [a(x+∆x)2 + b]− [ax2 + b] ∆x = lim ∆x→0 ax2 + 2ax∆x+ a(∆x)2 + b− ax2 − b ∆x = lim ∆x→0 2ax∆x+ a(∆x)2 ∆x = lim ∆x→0 (2ax+ a∆x) = 2ax 18. f ′(x)= lim ∆x→0 (x+∆x)2 − (x+∆x)− (x2 − x) ∆x = lim ∆x→0 2x∆x+∆x2 −∆x ∆x = lim ∆x→0 (2x− 1 + ∆x) = 2x− 1 19. f ′(x)= lim ∆x→0 1√ x+∆x − 1√ x ∆x = lim ∆x→0 √ x−√x+∆x ∆x √ x √ x+∆x = lim ∆x→0 x− (x+∆x) ∆x √ x √ x+∆x( √ x+ √ x+∆x) = lim ∆x→0 −1√ x √ x+∆x( √ x+ √ x+∆x) = − 1 2x3/2 20. f ′(x)= lim ∆x→0 1 (x+∆x)2 − 1 x2 ∆x = lim ∆x→0 x2 − (x+∆x)2 x2(x+∆x)2 ∆x = lim ∆x→0 x2 − x2 − 2x∆x−∆x2 x2∆x(x+∆x)2 = lim ∆x→0 −2x∆x−∆x2 x2∆x(x+∆x)2 = lim ∆x→0 −2x−∆x x2(x+∆x)2 = − 2 x3 21. f ′(t)= lim h→0 f(t+ h)− f(t) h = lim h→0 [4(t+ h)2 + (t+ h)]− [4t2 + t] h = lim h→0 4t2 + 8th+ 4h2 + t+ h− 4t2 − t h = lim h→0 8th+ 4h2 + h h = lim h→0 (8t+ 4h+ 1) = 8t+ 1 22. dV dr = lim h→0 4 3 π(r + h)3 − 4 3 πr3 h = lim h→0 4 3 π(r3 + 3r2h+ 3rh2 + h3 − r3) h = lim h→0 4 3 π(3r2 + 3rh+ h2) = 4πr2 23. (a) D (b) F (c) B (d) C (e) A (f) E 76 Chapter 3 24. Any function of the form f(x) = x + k has slope 1, and thus the derivative must be equal to 1 everywhere. y x 1 -2 3 -2 25. (a) x y (b) x y -1 (c) x y 1 2 26. (a) x y (b) x y (c) x y 27. (a) f(x) = x2 and a = 3 (b) f(x) = √ x and a = 1 28. (a) f(x) = x7 and a = 1 (b) f(x) = cosx and a = π 29. dy dx = lim h→0 [4(x+ h)2 + 1]− [4x2 + 1] h = lim h→0 4x2 + 8xh+ 4h2 + 1− 4x2 − 1 h = lim h→0 (8x+ 4h) = 8x dy dx ∣∣∣∣ x=1 = 8(1) = 8 Exercise Set 3.3 79 48. Let x = y = 0 to get f(0) = f(0) + f(0) + 0 so f(0) = 0. f ′(x) = lim h→0 f(x+ h)− f(x) h , but f(x+ h) = f(x) + f(h) + 5xh so f(x+ h)− f(x) = f(h) + 5xh and f ′(x) = lim h→0 f(h) + 5xh h = lim h→0 ( f(h) h + 5x ) = 3 + 5x. 49. f ′(x) = lim h→0 f(x+ h)− f(x) h = lim h→0 f(x)f(h)− f(x) h = lim h→0 f(x)[f(h)− 1] h = f(x) lim h→0 f(h)− f(0) h = f(x)f ′(0) = f(x) EXERCISE SET 3.3 1. 28x6 2. −36x11 3. 24x7 + 2 4. 2x3 5. 0 6. √ 2 7. −1 3 (7x6 + 2) 8. 2 5 x 9. 3ax2 + 2bx+ c 10. 1 a ( 2x+ 1 b ) 11. 24x−9 + 1/ √ x 12. −42x−7 − 5 2 √ x 13. −3x−4 − 7x−8 14. 1 2 √ x − 1 x2 15. f ′(x)= (3x2 + 6) d dx ( 2x− 1 4 ) + ( 2x− 1 4 ) d dx (3x2 + 6) = (3x2 + 6)(2) + ( 2x− 1 4 ) (6x) = 18x2 − 3 2 x+ 12 16. f ′(x)= (2− x− 3x3) d dx (7 + x5) + (7 + x5) d dx (2− x− 3x3) = (2− x− 3x3)(5x4) + (7 + x5)(−1− 9x2) = −24x7 − 6x5 + 10x4 − 63x2 − 7 17. f ′(x)= (x3 + 7x2 − 8) d dx (2x−3 + x−4) + (2x−3 + x−4) d dx (x3 + 7x2 − 8) = (x3 + 7x2 − 8)(−6x−4 − 4x−5) + (2x−3 + x−4)(3x2 + 14x) = −15x−2 − 14x−3 + 48x−4 + 32x−5 18. f ′(x)= (x−1 + x−2) d dx (3x3 + 27) + (3x3 + 27) d dx (x−1 + x−2) = (x−1 + x−2)(9x2) + (3x3 + 27)(−x−2 − 2x−3) = 3 + 6x− 27x−2 − 54x−3 19. 12x(3x2 + 1) 20. f(x) = x10 + 4x6 + 4x2, f ′(x) = 10x9 + 24x5 + 8x 21. dy dx = (5x− 3) d dx (1)− (1) d dx (5x− 3) (5x− 3)2 = − 5 (5x− 3)2 ; y ′(1) = −5/4 80 Chapter 3 22. dy dx = ( √ x+ 2) d dx (3)− 3 d dx ( √ x+ 2) ( √ x+ 2)2 = −3/(2√x(√x+ 2)2); y′(1) = −3/18 = −1/6 23. dx dt = (2t+ 1) d dt (3t)− (3t) d dt (2t+ 1) (2t+ 1)2 = (2t+ 1)(3)− (3t)(2) (2t+ 1)2 = 3 (2t+ 1)2 24. dx dt = (3t) d dt (t2 + 1)− (t2 + 1) d dt (3t) (3t)2 = (3t)(2t)− (t2 + 1)(3) 9t2 = t2 − 1 3t2 25. dy dx = (x+ 3) d dx (2x− 1)− (2x− 1) d dx (x+ 3) (x+ 3)2 = (x+ 3)(2)− (2x− 1)(1) (x+ 3)2 = 7 (x+ 3)2 ; dy dx ∣∣∣∣ x=1 = 7 16 26. dy dx = (x2 − 5) d dx (4x+ 1)− (4x+ 1) d dx (x2 − 5) (x2 − 5)2 = (x2 − 5)(4)− (4x+ 1)(2x) (x2 − 5)2 = − 4x2 + 2x+ 20 (x2 − 5)2 ; dy dx ∣∣∣∣ x=1 = 13 8 27. dy dx = ( 3x+ 2 x ) d dx ( x−5 + 1 ) + ( x−5 + 1 ) d dx ( 3x+ 2 x ) = ( 3x+ 2 x )(−5x−6)+ (x−5 + 1) [x(3)− (3x+ 2)(1) x2 ] = ( 3x+ 2 x )(−5x−6)+ (x−5 + 1)(− 2 x2 ) ; dy dx ∣∣∣∣ x=1 = 5(−5) + 2(−2) = −29 28. dy dx = (2x7 − x2) d dx ( x− 1 x+ 1 ) + ( x− 1 x+ 1 ) d dx (2x7 − x2) = (2x7 − x2) [ (x+ 1)(1)− (x− 1)(1) (x+ 1)2 ] + ( x− 1 x+ 1 ) (14x6 − 2x) = (2x7 − x2) · 2 (x+ 1)2 + ( x− 1 x+ 1 ) (14x6 − 2x); dy dx ∣∣∣∣ x=1 = (2− 1)2 4 + 0(14− 2) = 1 2 29. f ′(1) ≈ f(1.01)− f(1) 0.01 = 0.999699− (−1) 0.01 = 0.0301, and by differentiation, f ′(1) = 3(1)2 − 3 = 0 30. f ′(1) ≈ f(1.01)− f(1) 0.01 = 1.01504− 1 0.01 = 1.504, and by differentiation, f ′(1) = (√ x+ x 2 √ x ) ∣∣∣ x=1 = 1.5 Exercise Set 3.3 81 31. f ′(1) = 0 32. f ′(1) = 1 33. 32t 34. 2π 35. 3πr2 36. −2α−2 + 1 37. (a) dV dr = 4πr2 (b) dV dr ∣∣∣∣ r=5 = 4π(5)2 = 100π 38. d dλ [ λλ0 + λ6 2− λ0 ] = 1 2− λ0 d dλ (λλ0 + λ6) = 1 2− λ0 (λ0 + 6λ5) = λ0 + 6λ5 2− λ0 39. (a) g′(x) = √ xf ′(x) + 1 2 √ x f(x), g′(4) = (2)(−5) + 1 4 (3) = −37/4 (b) g′(x) = xf ′(x)− f(x) x2 , g′(4) = (4)(−5)− 3 16 = −23/16 40. (a) g′(x) = 6x− 5f ′(x), g′(3) = 6(3)− 5(4) = −2 (b) g′(x) = 2f(x)− (2x+ 1)f ′(x) f2(x) , g′(3) = 2(−2)− 7(4) (−2)2 = −8 41. (a) F ′(x) = 5f ′(x) + 2g′(x), F ′(2) = 5(4) + 2(−5) = 10 (b) F ′(x) = f ′(x)− 3g′(x), F ′(2) = 4− 3(−5) = 19 (c) F ′(x) = f(x)g′(x) + g(x)f ′(x), F ′(2) = (−1)(−5) + (1)(4) = 9 (d) F ′(x) = [g(x)f ′(x)− f(x)g′(x)]/g2(x), F ′(2) = [(1)(4)− (−1)(−5)]/(1)2 = −1 42. (a) F ′(x) = 6f ′(x)− 5g′(x), F ′(π) = 6(−1)− 5(2) = −16 (b) F ′(x) = f(x) + g(x) + x(f ′(x) + g′(x)), F ′(π) = 10− 3 + π(−1 + 2) = 7 + π (c) F ′(x) = 2f(x)g′(x) + 2f ′(x)g(x) = 2(20) + 2(3) = 46 (d) F ′(x) = (4 + g(x))f ′(x)− f(x)g′(x) (4 + g(x))2 = (4− 3)(−1)− 10(2) (4− 3)2 = −21 43. y − 2 = 5(x+ 3), y = 5x+ 17 44. dy dx = (1 + x)(−1)− (1− x)(1) (1 + x)2 = − 2 (1 + x)2 , dy dx ∣∣∣∣ x=2 = −2 9 and y = −1 3 for x = 2 so an equation of the tangent line is y − ( −1 3 ) = −2 9 (x− 2), or y = −2 9 x+ 1 9 . 45. (a) dy/dx = 21x2 − 10x+ 1, d2y/dx2 = 42x− 10 (b) dy/dx = 24x− 2, d2y/dx2 = 24 (c) dy/dx = −1/x2, d2y/dx2 = 2/x3 (d) y = 35x5 − 16x3 − 3x, dy/dx = 175x4 − 48x2 − 3, d2y/dx2 = 700x3 − 96x 46. (a) y′ = 28x6 − 15x2 + 2, y′′ = 168x5 − 30x (b) y′ = 3, y′′ = 0 (c) y′ = 2 5x2 , y′′ = − 4 5x3 (d) y = 2x4 + 3x3 − 10x− 15, y′ = 8x3 + 9x2 − 10, y′′ = 24x2 + 18x 47. (a) y′ = −5x−6 + 5x4, y′′ = 30x−7 + 20x3, y′′′ = −210x−8 + 60x2 (b) y = x−1, y′ = −x−2, y′′ = 2x−3, y′′′ = −6x−4 (c) y′ = 3ax2 + b, y′′ = 6ax, y′′′ = 6a 84 Chapter 3 69. f ′(x) = 1 + 1/x2 > 0 for all x = 0 6 -6 6 -6 70. f ′(x) = −5 x 2 − 4 (x2 + 4)2 ; f ′(x) > 0 when x2 < 4, i.e. on −2 < x < 2 1.5 -1.5 -5 5 71. (f · g · h)′ = [(f · g) · h]′ = (f · g)h′ + h(f · g)′ = (f · g)h′ + h[fg′ + f ′g] = fgh′ + fg′h+ f ′gh 72. (f1f2 · · · fn)′ = (f ′1f2 · · · fn) + (f1f ′2 · · · fn) + · · ·+ (f1f2 · · · f ′n) 73. (a) 2(1 + x−1)(x−3 + 7) + (2x+ 1)(−x−2)(x−3 + 7) + (2x+ 1)(1 + x−1)(−3x−4) (b) (x7 + 2x− 3)3 = (x7 + 2x− 3)(x7 + 2x− 3)(x7 + 2x− 3) so d dx (x7 + 2x− 3)3 = (7x6 + 2)(x7 + 2x− 3)(x7 + 2x− 3) +(x7 + 2x− 3)(7x6 + 2)(x7 + 2x− 3) +(x7 + 2x− 3)(x7 + 2x− 3)(7x6 + 2) = 3(7x6 + 2)(x7 + 2x− 3)2 74. (a) −5x−6(x2 + 2x)(4− 3x)(2x9 + 1) + x−5(2x+ 2)(4− 3x)(2x9 + 1) +x−5(x2 + 2x)(−3)(2x9 + 1) + x−5(x2 + 2x)(4− 3x)(18x8) (b) (x2 + 1)50 = (x2 + 1)(x2 + 1) · · · (x2 + 1), where (x2 + 1) occurs 50 times so d dx (x2 + 1)50 = [(2x)(x2 + 1) · · · (x2 + 1)] + [(x2 + 1)(2x) · · · (x2 + 1)] + · · ·+ [(x2 + 1)(x2 + 1) · · · (2x)] = 2x(x2 + 1)49 + 2x(x2 + 1)49 + · · ·+ 2x(x2 + 1)49 = 100x(x2 + 1)49 because 2x(x2 + 1)49 occurs 50 times. 75. f is continuous at 1 because lim x→1− f(x) = lim x→1+ f(x) = f(1); also lim x→1− f ′(x) = lim x→1− (2x + 1) = 3 and lim x→1+ f ′(x) = lim x→1+ 3 = 3 so f is differentiable at 1. 76. f is not continuous at x = 9 because lim x→9− f(x) = −63 and lim x→9+ f(x) = 36. f cannot be differentiable at x = 9, for if it were, then f would also be continuous, which it is not. 77. f is continuous at 1 because lim x→1− f(x) = lim x→1+ f(x) = f(1), also lim x→1− f ′(x) = lim x→1− 2x = 2 and lim x→1+ f ′(x) = lim x→1+ 1 2 √ x = 1 2 so f is not differentiable at 1. 78. f is continuous at 1/2 because lim x→1/2− f(x) = lim x→1/2+ f(x) = f(1/2), also lim x→1/2− f ′(x) = lim x→1/2− 3x2 = 3/4 and lim x→1/2+ f ′(x) = lim x→1/2+ 3x/2 = 3/4 so f ′(1/2) = 3/4, and f is differentiable at x = 1/2. Exercise Set 3.4 85 79. (a) f(x) = 3x − 2 if x ≥ 2/3, f(x) = −3x + 2 if x < 2/3 so f is differentiable everywhere except perhaps at 2/3. f is continuous at 2/3, also lim x→2/3− f ′(x) = lim x→2/3− (−3) = −3 and lim x→2/3+ f ′(x) = lim x→2/3+ (3) = 3 so f is not differentiable at x = 2/3. (b) f(x) = x2 − 4 if |x| ≥ 2, f(x) = −x2 + 4 if |x| < 2 so f is differentiable everywhere except perhaps at ±2. f is continuous at −2 and 2, also lim x→2− f ′(x) = lim x→2− (−2x) = −4 and lim x→2+ f ′(x) = lim x→2+ (2x) = 4 so f is not differentiable at x = 2. Similarly, f is not differentiable at x = −2. 80. (a) f ′(x) = −(1)x−2, f ′′(x) = (2 · 1)x−3, f ′′′(x) = −(3 · 2 · 1)x−4 f (n)(x) = (−1)nn(n− 1)(n− 2) · · · 1 xn+1 (b) f ′(x) = −2x−3, f ′′(x) = (3 · 2)x−4, f ′′′(x) = −(4 · 3 · 2)x−5 f (n)(x) = (−1)n (n+ 1)(n)(n− 1) · · · 2 xn+2 81. (a) d2 dx2 [cf(x)] = d dx [ d dx [cf(x)] ] = d dx [ c d dx [f(x)] ] = c d dx [ d dx [f(x)] ] = c d2 dx2 [f(x)] d2 dx2 [f(x) + g(x)] = d dx [ d dx [f(x) + g(x)] ] = d dx [ d dx [f(x)] + d dx [g(x)] ] = d2 dx2 [f(x)] + d2 dx2 [g(x)] (b) yes, by repeated application of the procedure illustrated in Part (a) 82. (f · g)′ = fg′ + gf ′, (f · g)′′ = fg′′ + g′f ′ + gf ′′ + f ′g′ = f ′′g + 2f ′g′ + fg′′ 83. (a) f ′(x) = nxn−1, f ′′(x) = n(n− 1)xn−2, f ′′′(x) = n(n− 1)(n− 2)xn−3, . . ., f (n)(x) = n(n− 1)(n− 2) · · · 1 (b) from Part (a), f (k)(x) = k(k − 1)(k − 2) · · · 1 so f (k+1)(x) = 0 thus f (n)(x) = 0 if n > k (c) from Parts (a) and (b), f (n)(x) = ann(n− 1)(n− 2) · · · 1 84. lim h→0 f ′(2 + h)− f ′(2) h = f ′′(2); f ′(x) = 8x7 − 2, f ′′(x) = 56x6, so f ′′(2) = 56(26) = 3584. 85. (a) If a function is differentiable at a point then it is continuous at that point, thus f ′ is continuous on (a, b) and consequently so is f . (b) f and all its derivatives up to f (n−1)(x) are continuous on (a, b) EXERCISE SET 3.4 1. f ′(x) = −2 sinx− 3 cosx 2. f ′(x) = sinx(− sinx) + cosx(cosx) = cos2 x− sin2 x = cos 2x 3. f ′(x) = x(cosx)− (sinx)(1) x2 = x cosx− sinx x2 86 Chapter 3 4. f ′(x) = x2(− sinx) + (cosx)(2x) = −x2 sinx+ 2x cosx 5. f ′(x) = x3(cosx) + (sinx)(3x2)− 5(− sinx) = x3 cosx+ (3x2 + 5) sinx 6. f(x) = cotx x (because cosx sinx = cotx), f ′(x) = x(− csc2 x)− (cotx)(1) x2 = −x csc 2 x+ cotx x2 7. f ′(x) = secx tanx− √ 2 sec2 x 8. f ′(x) = (x2 + 1) secx tanx+ (secx)(2x) = (x2 + 1) secx tanx+ 2x secx 9. f ′(x) = secx(sec2 x) + (tanx)(secx tanx) = sec3 x+ secx tan2 x 10. f ′(x)= (1 + tanx)(secx tanx)− (secx)(sec2 x) (1 + tanx)2 = secx tanx+ secx tan2 x− sec3 x (1 + tanx)2 = secx(tanx+ tan2 x− sec2 x) (1 + tanx)2 = secx(tanx− 1) (1 + tanx)2 11. f ′(x) = (cscx)(− csc2 x) + (cotx)(− cscx cotx) = − csc3 x− cscx cot2 x 12. f ′(x) = 1 + 4 cscx cotx− 2 csc2 x 13. f ′(x) = (1 + cscx)(− csc2 x)− cotx(0− cscx cotx) (1 + cscx)2 = cscx(− cscx− csc2 x+ cot2 x) (1 + cscx)2 but 1 + cot2 x = csc2 x (identity) thus cot2 x− csc2 x = −1 so f ′(x) = cscx(− cscx− 1) (1 + cscx)2 = − cscx 1 + cscx 14. f ′(x) = tanx(− cscx cotx)− cscx(sec2 x) tan2 x = −cscx(1 + sec 2 x) tan2 x 15. f(x) = sin2 x+ cos2 x = 1 (identity) so f ′(x) = 0 16. f(x) = 1 cotx = tanx, so f ′(x) = sec2 x 17. f(x)= tanx 1 + x tanx (because sinx secx = (sinx)(1/ cosx) = tanx), f ′(x)= (1 + x tanx)(sec2 x)− tanx[x(sec2 x) + (tanx)(1)] (1 + x tanx)2 = sec2 x− tan2 x (1 + x tanx)2 = 1 (1 + x tanx)2 (because sec2 x− tan2 x = 1) 18. f(x)= (x2 + 1) cotx 3− cotx (because cosx cscx = (cosx)(1/ sinx) = cotx), f ′(x)= (3− cotx)[2x cotx− (x2 + 1) csc2 x]− (x2 + 1) cotx csc2 x (3− cotx)2 = 6x cotx− 2x cot2 x− 3(x2 + 1) csc2 x (3− cotx)2 Exercise Set 3.5 89 39. f ′(x) = − sinx, f ′′(x) = − cosx, f ′′′(x) = sinx, and f (4)(x) = cosx with higher order derivatives repeating this pattern, so f (n)(x) = sinx for n = 3, 7, 11, . . . 40. (a) lim h→0 tanh h = lim h→0 ( sinh cosh ) h = lim h→0 ( sinh h ) cosh = 1 1 = 1 (b) d dx [tanx] = lim h→0 tan(x+ h)− tanx h = lim h→0 tanx+ tanh 1− tanx tanh − tanx h = lim h→0 tanx+ tanh− tanx+ tan2 x tanh h(1− tanx tanh) = limh→0 tanh(1 + tan2 x) h(1− tanx tanh) = lim h→0 tanh sec2 x h(1− tanx tanh) = sec 2 x lim h→0 tanh h 1− tanx tanh = sec2 x lim h→0 tanh h lim h→0 (1− tanx tanh) = sec 2 x 41. lim x→0 tan(x+ y)− tan y x = lim h→0 tan(y + h)− tan y h = d dy (tan y) = sec2 y 43. Let t be the radian measure, then h = 180 π t and cosh = cos t, sinh = sin t. (a) lim h→0 cosh− 1 h = lim t→0 cos t− 1 180t/π = π 180 lim t→0 cos t− 1 t = 0 (b) lim h→0 sinh h = lim t→0 sin t 180t/π = π 180 lim t→0 sin t t = π 180 (c) d dx [sinx] = sinx lim h→0 cosh− 1 h + cosx lim h→0 sinh h = (sinx)(0) + (cosx)(π/180) = π 180 cosx EXERCISE SET 3.5 1. (f ◦ g)′(x) = f ′(g(x))g′(x) so (f ◦ g)′(0) = f ′(g(0))g′(0) = f ′(0)(3) = (2)(3) = 6 2. (f ◦ g)′(2) = f ′(g(2))g′(2) = 5(−3) = −15 3. (a) (f ◦ g)(x) = f(g(x)) = (2x− 3)5 and (f ◦ g)′(x) = f ′(g(x)g′(x) = 5(2x− 3)4(2) = 10(2x− 3)4 (b) (g ◦ f)(x) = g(f(x)) = 2x5 − 3 and (g ◦ f)′(x) = g′(f(x))f ′(x) = 2(5x4) = 10x4 4. (a) (f ◦ g)(x) = 5√4 + cosx and (f ◦ g)′(x) = f ′(g(x))g′(x) = 5 2 √ 4 + cosx (− sinx) (b) (g ◦ f)(x) = 4 + cos(5√x) and (g ◦ f)′(x) = g′(f(x))f ′(x) = − sin(5√x) 5 2 √ x 5. (a) F ′(x) = f ′(g(x))g′(x) = f ′(g(3))g′(3) = −1(7) = −7 (b) G′(x) = g′(f(x))f ′(x) = g′(f(3))f ′(3) = 4(−2) = −8 6. (a) F ′(x) = f ′(g(x))g′(x), F ′(−1) = f ′(g(−1))g′(−1) = f ′(2)(−3) = (4)(−3) = −12 (b) G′(x) = g′(f(x))f ′(x), G′(−1) = g′(f(−1))f ′(−1) = −5(3) = −15 90 Chapter 3 7. f ′(x) = 37(x3 + 2x)36 d dx (x3 + 2x) = 37(x3 + 2x)36(3x2 + 2) 8. f ′(x) = 6(3x2 +2x− 1)5 d dx (3x2 +2x− 1) = 6(3x2 +2x− 1)5(6x+2) = 12(3x2 +2x− 1)5(3x+1) 9. f ′(x) = −2 ( x3 − 7 x )−3 d dx ( x3 − 7 x ) = −2 ( x3 − 7 x )−3( 3x2 + 7 x2 ) 10. f(x) = (x5 − x+ 1)−9, f ′(x) = −9(x5 − x+ 1)−10 d dx (x5 − x+ 1) = −9(x5 − x+ 1)−10(5x4 − 1) = − 9(5x 4 − 1) (x5 − x+ 1)10 11. f(x) = 4(3x2 − 2x+ 1)−3, f ′(x) = −12(3x2 − 2x+ 1)−4 d dx (3x2 − 2x+ 1) = −12(3x2 − 2x+ 1)−4(6x− 2) = 24(1− 3x) (3x2 − 2x+ 1)4 12. f ′(x) = 1 2 √ x3 − 2x+ 5 d dx (x3 − 2x+ 5) = 3x 2 − 2 2 √ x3 − 2x+ 5 13. f ′(x) = 1 2 √ 4 + 3 √ x d dx (4 + 3 √ x) = 3 4 √ x √ 4 + 3 √ x 14. f ′(x) = 3 sin2 x d dx (sinx) = 3 sin2 x cosx 15. f ′(x) = cos(x3) d dx (x3) = 3x2 cos(x3) 16. f ′(x) = 2 cos(3 √ x) d dx [cos(3 √ x)] = −2 cos(3√x) sin(3√x) d dx (3 √ x) = −3 cos(3 √ x) sin(3 √ x)√ x 17. f ′(x) = 20 cos4 x d dx (cosx) = 20 cos4 x(− sinx) = −20 cos4 x sinx 18. f ′(x) = − csc(x3) cot(x3) d dx (x3) = −3x2 csc(x3) cot(x3) 19. f ′(x) = cos(1/x2) d dx (1/x2) = − 2 x3 cos(1/x2) 20. f ′(x) = 4 tan3(x3) d dx [tan(x3)] = 4 tan3(x3) sec2(x3) d dx (x3) = 12x2 tan3(x3) sec2(x3) 21. f ′(x) = 4 sec(x7) d dx [sec(x7)] = 4 sec(x7) sec(x7) tan(x7) d dx (x7) = 28x6 sec2(x7) tan(x7) 22. f ′(x)= 3 cos2 ( x x+ 1 ) d dx cos ( x x+ 1 ) = 3 cos2 ( x x+ 1 )[ − sin ( x x+ 1 )] (x+ 1)(1)− x(1) (x+ 1)2 = − 3 (x+ 1)2 cos2 ( x x+ 1 ) sin ( x x+ 1 ) 23. f ′(x) = 1 2 √ cos(5x) d dx [cos(5x)] = − 5 sin(5x) 2 √ cos(5x) 24. f ′(x) = 1 2 √ 3x− sin2(4x) d dx [3x− sin2(4x)] = 3− 8 sin(4x) cos(4x) 2 √ 3x− sin2(4x) Exercise Set 3.5 91 25. f ′(x)= −3 [x+ csc(x3 + 3)]−4 d dx [ x+ csc(x3 + 3) ] = −3 [x+ csc(x3 + 3)]−4 [1− csc(x3 + 3) cot(x3 + 3) d dx (x3 + 3) ] = −3 [x+ csc(x3 + 3)]−4 [1− 3x2 csc(x3 + 3) cot(x3 + 3)] 26. f ′(x)= −4 [x4 − sec(4x2 − 2)]−5 d dx [ x4 − sec(4x2 − 2)] = −4 [x4 − sec(4x2 − 2)]−5 [4x3 − sec(4x2 − 2) tan(4x2 − 2) d dx (4x2 − 2) ] = −16x [x4 − sec(4x2 − 2)]−5 [x2 − 2 sec(4x2 − 2) tan(4x2 − 2)] 27. dy dx = x3(2 sin 5x) d dx (sin 5x) + 3x2 sin2 5x = 10x3 sin 5x cos 5x+ 3x2 sin2 5x 28. dy dx = √ x [ 3 tan2( √ x) sec2( √ x) 1 2 √ x ] + 1 2 √ x tan3( √ x) = 3 2 tan2( √ x) sec2( √ x) + 1 2 √ x tan3( √ x) 29. dy dx = x5 sec ( 1 x ) tan ( 1 x ) d dx ( 1 x ) + sec ( 1 x ) (5x4) = x5 sec ( 1 x ) tan ( 1 x )( − 1 x2 ) + 5x4 sec ( 1 x ) = −x3 sec ( 1 x ) tan ( 1 x ) + 5x4 sec ( 1 x ) 30. dy dx = sec(3x+ 1) cosx− 3 sinx sec(3x+ 1) tan(3x+ 1) sec2(3x+ 1) = cosx cos(3x+ 1)− 3 sinx sin(3x+ 1) 31. dy dx = − sin(cosx) d dx (cosx) = − sin(cosx)(− sinx) = sin(cosx) sinx 32. dy dx = cos(tan 3x) d dx (tan 3x) = 3 sec2 3x cos(tan 3x) 33. dy dx = 3 cos2(sin 2x) d dx [cos(sin 2x)] = 3 cos2(sin 2x)[− sin(sin 2x)] d dx (sin 2x) = −6 cos2(sin 2x) sin(sin 2x) cos 2x 34. dy dx = (1− cotx2)(−2x cscx2 cotx2)− (1 + cscx2)(2x csc2 x2) (1− cotx2)2 = −2x cscx 2 1 + cotx 2 + cscx2 (1− cotx2)2 35. dy dx = (5x+ 8)1312(x3 + 7x)11 d dx (x3 + 7x) + (x3 + 7x)1213(5x+ 8)12 d dx (5x+ 8) = 12(5x+ 8)13(x3 + 7x)11(3x2 + 7) + 65(x3 + 7x)12(5x+ 8)12 36. dy dx = (2x− 5)23(x2 + 4)2(2x) + (x2 + 4)32(2x− 5)(2) = 6x(2x− 5)2(x2 + 4)2 + 4(2x− 5)(x2 + 4)3 = 2(2x− 5)(x2 + 4)2(8x2 − 15x+ 8) 94 Chapter 3 59. (a) 2 -2 2 -2 (c) f ′(x) = x −x√ 4− x2 + √ 4− x2 = 4− 2x 2 √ 4− x2 2 -2 2 -6 (d) f(1) = √ 3 and f ′(1) = 2√ 3 so the tangent line has the equation y − √ 3 = 2√ 3 (x− 1). 3 0 0 2 60. (a) 0.5 0 ^ 6 (c) f ′(x) = 2x cos(x2) cosx− sinx sin(x2) 1.2 -1.2 ^ 6 (d) f(1) = sin 1 cos 1 and f ′(1) = 2 cos2 1− sin2 1, so the tangent line has the equation y − sin 1 cos 1 = (2 cos2 1− sin2 1)(x− 1). 0.8 0 ^ 6 61. (a) dy/dt = −Aω sinωt, d2y/dt2 = −Aω2 cosωt = −ω2y (b) one complete oscillation occurs when ωt increases over an interval of length 2π, or if t increases over an interval of length 2π/ω (c) f = 1/T (d) amplitude = 0.6 cm, T = 2π/15 s/oscillation, f = 15/(2π) oscillations/s 62. dy/dt = 3A cos 3t, d2y/dt2 = −9A sin 3t, so −9A sin 3t+ 2A sin 3t = 4 sin 3t, −7A sin 3t = 4 sin 3t,−7A = 4, A = −4/7 Exercise Set 3.5 95 63. (a) p ≈ 10 lb/in2, dp/dh ≈ −2 lb/in2/mi (b) dp dt = dp dh dh dt ≈ (−2)(0.3) = −0.6 lb/in2/s 64. (a) F = 45 cos θ + 0.3 sin θ , dF dθ = −45(− sin θ + 0.3 cos θ) (cos θ + 0.3 sin θ)2 ; if θ = 30◦, then dF/dθ ≈ 10.5 lb/rad ≈ 0.18 lb/deg (b) dF dt = dF dθ dθ dt ≈ (0.18)(−0.5) = −0.09 lb/s 65. With u = sinx, d dx (| sinx|) = d dx (|u|) = d du (|u|)du dx = d du (|u|) cosx = { cosx, u > 0 − cosx, u < 0 = { cosx, sinx > 0 − cosx, sinx < 0 = { cosx, 0 < x < π − cosx, −π < x < 0 66. d dx (cosx) = d dx [sin(π/2− x)] = − cos(π/2− x) = − sinx 67. (a) For x = 0, |f(x)| ≤ |x|, and lim x→0 |x| = 0, so by the Squeezing Theorem, lim x→0 f(x) = 0. (b) If f ′(0) were to exist, then the limit f(x)− f(0) x− 0 = sin(1/x) would have to exist, but it doesn’t. (c) for x = 0, f ′(x) = x ( cos 1 x )( − 1 x2 ) + sin 1 x = − 1 x cos 1 x + sin 1 x (d) lim x→0 f(x)− f(0) x− 0 = limx→0 sin(1/x), which does not exist, thus f ′(0) does not exist. 68. (a) −x2 ≤ x2 sin(1/x) ≤ x2, so by the Squeezing Theorem lim x→0 f(x) = 0. (b) f ′(0) = lim x→0 f(x)− f(0) x− 0 = limx→0x sin(1/x) = 0 by Exercise 67, Part a. (c) For x = 0, f ′(x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2) = 2x sin(1/x)− cos(1/x) (d) If f ′(x) were continuous at x = 0 then so would cos(1/x) = f ′(x) − 2x sin(1/x) be, since 2x sin(1/x) is continuous there. But cos(1/x) oscillates at x = 0. 69. (a) g′(x) = 3[f(x)]2f ′(x), g′(2) = 3[f(2)]2f ′(2) = 3(1)2(7) = 21 (b) h′(x) = f ′(x3)(3x2), h′(2) = f ′(8)(12) = (−3)(12) = −36 70. F ′(x) = f ′(g(x))g′(x) = √ 3(x2 − 1) + 4(2x) = 2x √ 3x2 + 1 71. F ′(x) = f ′(g(x))g′(x) = f ′( √ 3x− 1) 3 2 √ 3x− 1 = √ 3x− 1 (3x− 1) + 1 3 2 √ 3x− 1 = 1 2x 72. d dx [f(x2)] = f ′(x2)(2x), thus f ′(x2)(2x) = x2 so f ′(x2) = x/2 if x = 0 73. d dx [f(3x)] = f ′(3x) d dx (3x) = 3f ′(3x) = 6x, so f ′(3x) = 2x. Let u = 3x to get f ′(u) = 2 3 u; d dx [f(x)] = f ′(x) = 2 3 x. 96 Chapter 3 74. (a) If f(−x) = f(x), then d dx [f(−x)] = d dx [f(x)], f ′(−x)(−1) = f ′(x), f ′(−x) = −f ′(x) so f ′ is odd. (b) If f(−x) = −f(x), then d dx [f(−x)] = − d dx [f(x)], f ′(−x)(−1) = −f ′(x), f ′(−x) = f ′(x) so f ′ is even. 75. For an even function, the graph is symmetric about the y-axis; the slope of the tangent line at (a, f(a)) is the negative of the slope of the tangent line at (−a, f(−a)). For an odd function, the graph is symmetric about the origin; the slope of the tangent line at (a, f(a)) is the same as the slope of the tangent line at (−a, f(−a)). y x f (x ) f ' (x ) y x f (x ) f ' (x ) 76. dy dx = dy du du dv dv dw dw dx 77. d dx [f(g(h(x)))]= d dx [f(g(u))], u = h(x) = d du [f(g(u))] du dx = f ′(g(u))g′(u) du dx = f ′(g(h(x)))g′(h(x))h′(x) EXERCISE SET 3.6 1. y = (2x− 5)1/3; dy/dx = 2 3 (2x− 5)−2/3 2. dy/dx = 1 3 [ 2 + tan(x2) ]−2/3 sec2(x2)(2x) = 2 3 x sec2(x2) [ 2 + tan(x2) ]−2/3 3. dy/dx = 3 2 [ x− 1 x+ 2 ]1/2 d dx [ x− 1 x+ 2 ] = 9 2(x+ 2)2 [ x− 1 x+ 2 ]1/2 4. dy/dx = 1 2 [ x2 + 1 x2 − 5 ]−1/2 d dx [ x2 + 1 x2 − 5 ] = 1 2 [ x2 + 1 x2 − 5 ]−1/2 −12x (x2 − 5)2 = − 6x (x2 − 5)2 [ x2 + 1 x2 − 5 ]−1/2 5. dy/dx = x3 ( −2 3 ) (5x2 + 1)−5/3(10x) + 3x2(5x2 + 1)−2/3 = 1 3 x2(5x2 + 1)−5/3(25x2 + 9) 6. dy/dx = x2 4 3 (3− 2x)1/3(−2)− (3− 2x)4/3(2x) x4 = 2(3− 2x)1/3(2x− 9) 3x3 7. dy/dx = 5 2 [sin(3/x)]3/2[cos(3/x)](−3/x2) = −15[sin(3/x)] 3/2 cos(3/x) 2x2