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Unformatted text preview: 8) 18. f (x) = (x−1 + x−2 ) 19. 12x(3x2 + 1) 21. d d (5x − 3) (1) − (1) (5x − 3) 5 dy dx dx = =− ; y (1) = −5/4 dx (5x − 3)2 (5x − 3)2 22. 23. 24. 25. 26. d d (3x3 + 27) + (3x3 + 27) (x−1 + x−2 ) dx dx = (x−1 + x−2 )(9x2 ) + (3x3 + 27)(−x−2 − 2x−3 ) = 3 + 6x − 27x−2 − 54x−3 20. f (x) = x10 + 4x6 + 4x2 , f (x) = 10x9 + 24x5 + 8x √ d√ d ( x + 2) (3) − 3 ( x + 2) √√ dy dx√ dx = = −3/(2 x( x + 2)2 ; y (1) = −3/18 = −1/6 2 dx ( x + 2) d d (2t + 1) (3t) − (3t) (2t + 1) dx (2t + 1)(3) − (3t)(2) 3 dt dt = = = dt (2t + 1)2 (2t + 1)2 (2t + 1)2 d d (3t) (t2 + 1) − (t2 + 1) (3t) dx (3t)(2t) − (t2 + 1)(3) t2 − 1 dt dt = = = 2 2 dt (3t) 9t 3t2 d d (x + 3) (2x − 1) − (2x − 1) (x + 3) dy (x + 3)(2) − (2x − 1)(1) 7 dy dx dx = = = ; 2 2 2 dx dx (x + 3) (x + 3) (x + 3) = x=1 7 16 d d (x2 − 5) (4x + 1) − (4x + 1) (x2 − 5) (x2 − 5)(4) − (4x + 1)(2x) 4x2 + 2x + 20 dy dx dx = = =− ; dx (x2 − 5)2 (x2 − 5)2 (x2 − 5)2 13 dy = dx x=1 8 Exercise Set 3.3 74 dy = dx 3x + 2 x = 27. 3x + 2 x dy dx 28. −5x−6 + x−5 + 1 3x + 2 x x(3) − (3x + 2)(1) = x2 3x + 2 x x=1 1 2 = (2 − 1) + 0(14 − 2) = 4 2 (a) 30. 2π 31. dV = 4πr2 dr 3πr2 (b) dV dr 32. = 4π (5)2 = 100π g (x) = √ 1 1 xf (x) + √ f (x), g (4) = (2)(−5) + (3) = −37/4 4 2x g (x) = xf (x) − f (x) (4)(−5) − 3 = −23/16 , g (4) = x2 16 (a) g (x) = 6x − 5f (x), g (3) = 6(3) − 5(4) = −2 2f (x) − (2x + 1)f (x) 2(−2) − 7(4) , g (3) = = −8 2 (x) f (−2)2 (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) 38. −2α−2 + 1 r=5 (a) (b) g (x) = 37. 2 x2 = 5(−5) + 2(−2) = −29 (b) 36. − x=1 29. 32t 35. −5x−6 + x−5 + 1 x−1 d d x−1 dy = (2x7 − x2 ) + (2x7 − x2 ) dx dx x + 1 x + 1 dx (x + 1)(1) − (x − 1)(1) x−1 (14x6 − 2x) + = (2x7 − x2 ) 2 (x + 1) x+1 2 x−1 (14x6 − 2x); + = (2x7 − x2 ) · 2 (x + 1) x+1 dy dx 33. d d x−5 + 1 + x−5 + 1 dx dx F (x) = [g (x)f (x) − f (x)g (x)]/g 2 (x), F (2) = [(1)(4) − (−1)(−5)]/(1)2 = −1 2 dy 2 (1 + x)(−1) − (1 − x)(1) 1 dy =− , = − and y = − for x = 2 so an equation of = dx (1 + x)2 (1 + x)2 dx x=2 9 3 2 2 1 1 = − (x − 2), or y = − x + . the tangent line is y − − 3 9 9 9 39. y − 2 = 5(x + 3), y = 5x + 17 40. d 1 1 λ0 + 6λ5 d λλ0 + λ6 (λλ0 + λ6 ) = = (λ0 + 6λ5 ) = dλ 2 − λ0 2 − λ0 dλ 2 − λ0 2 − λ0 41. (a) dy/dx = 21x2 − 10x + 1, d2 y/dx2 = 42x − 10 (b) dy/dx = 24x − 2, d2 y/dx2 = 24 (c) dy/dx = −1/x2 , d2 y/dx2 = 2/x3 (d) y = 35x5 − 16x3 − 3x, dy/dx = 175x4 − 48x2 − 3, d2 y/dx2 = 700x3 − 96x ; 75 Chapter 3 y= y = 2x4 + 3x3 − 10x − 15, y = 8x3 + 9x2 − 10, y = 24x2 + 18x (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 (a) dy/dx = 10x − 4, d2 y/dx2 = 10, d3 y/dx3 = 0 (b) dy/dx = −6x−3 − 4x−2 + 1, d2 y/dx2 = 18x−4 + 8x−3 , d3 y/dx3 = −72x−5 − 24x−4 (c) dy/dx = 4ax3 + 2bx, d2 y/dx2 = 12ax2 + 2b, d3 y/dx3 = 24ax (a) f (x) = 6x, f (x) = 6, f (x) = 0, f (2) = 0 (b) 45. y = 3, y = 0 (d) 44. y = 28x6 − 15x2 + 2, y = 168x5 − 30x (c) 43. (a) (b) 42. dy d2 y d2 y = 30x4 − 8x, = 120x3 − 8, dx dx2 dx2 (c) 2 4 ,y =− 3 5x2 5x = 112 x=1 d2 d3 d4 d x−3 = −3x−4 , x−3 = 12x−5 , x−3 = −60x−6 , x−3 = 360x−7 , dx dx2 dx3 dx4 d4 = 360 x−3 dx4 x=1 (a) y = 16x3 + 6x2 , y = 48x2 + 12x, y = 96x + 12, y (0) = 12 (b) 46. y = 6x−4 , d4 y dx4 47. dy d2 y d3 y d4 y = −24x−5 , = 120x−6 , = −720x−7 , = 5040x−8 , dx dx2 dx3 dx4 = 5040 x=1 y = 3x2 + 3, y = 6x, and y = 6 so y + xy − 2y = 6 + x(6x) − 2(3x2 + 3) = 6 + 6x2 − 6x2 − 6 = 0 48. y = x−1 , y = −x−2 , y = 2x−3 so x3 y + x2 y − xy = x3 (2x−3 ) + x2 (−x−2 ) − x(x−1 ) = 2 − 1 − 1 = 0 49. 51. F (x) = xf (x) + f (x), F (x) = xf (x) + f (x) + f (x) = xf (x) + 2f (x) dy = 0, but The graph has a horizontal tangent at points where dx dy = x2 − 3x + 2 = (x − 1)(x − 2) = 0 if x = 1, 2. The dx corresponding values of y are 5/6 and 2/3 so the tangent line is horizontal at (1, 5/6) and (2, 2/3). 1.5 0 3 0 Exercise Set 3.3 52. 76 dy dy 9 − x2 ; =2 = 0 when x2 = 9 so x = ±3. The points are dx (x + 9)2 dx (3, 1/6) and (−3, −1/6). 0.2 -5.5 5.5 -0.2 53. f (1) ≈ 54. 0.999699 − (−1) f (1.01) − f (1) = = 0.0301, and by differentiation, f (1) = 3(1)2 − 3 = 0 0.01 0.01 f (1) ≈ 1.01504 − 1 f (1.01) − f (1) = = 1.504, and by differentiation, 0.01 0.01 √ x x+ √ = 1.5 f (1) = 2 x x=1 55. f (1) = 0 57. 56. f (1) = 1 The y -intercept is −2 so the point (0, −2) is on the graph; −2 = a(0)2 + b(0) + c, c = −2. The x-intercept is 1 so the point (1,0) is on the graph; 0 = a + b − 2. The slope is dy/dx = 2ax + b; at x = 0 the slope is b so b = −1, thus a...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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