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74 Chapter 2: Derivatives and simplify as much as possible. 10 (b) Find the slope of the tangent line m = lim msec(I) ing Ex by evaluating the limit. Find this limit exactly using the algebraic method, (c) (c) Write an equation for the tangent line to f at (a, f(a)). Optional: Check your work by graphing the ful function f and the tangent line at the given point. N So 31. f(x) =4-12,a =1 32. f(x) =9-12,a =2 lin 33. f(x) =1-23,a = 2 34. f(x) =2-13,a=1 ta sid 35. f(x) = 1/x, a = 2 36. f(x) = 2/r,a =1 SO 37. f(x) = 1/x ,a = 3 38. f(x) = 1/r, a = 2 ch 39. f(x) = 1/23,a = 2 40. f(x) = 1/x3, a =1 41. f(x) = 1/ Vz,a = 4 42. f(x) = -1/VI,a =9 69 71 43. f(x) = - 6 5 74910=2 44. f(x) = - 72 45. f(x) = -, a =2 1+ x 46. f(x) = - -0=1 73 2+ 1' 75 47. f(r) a+1 5, a =2 48. f(x) = x+2 I + 3 2 +5 , a=1 77 49. f(x) = \\5+12,a =1 50. f(x) = V3+12,a =2 79 81 51. f(x) = 1 1+ 2 0 =2 52. f(x) = 2 83 1 2 85 53. f(x) = 14 73 0= 2 54. f(1) = 2+r 87 55. f(x) =- 1472 0= 2 56. f(x) =71 2 4=1 r 89 57. f(x) = 14 72 0=2 58. f(1) = 2+210=1 59. f(x) = vx2 + 3r, a = 160. f(x) = Vx2 + 5r,a=1 9 61. f(x) = Vx3 +1,a=1 62. f(x) = Vx3 +2,a=1 93 63. f(x) = - 1 , a = 1 64. f(x) = = 2 95 1 65. f(x) = VI+ 2?' , a = 2 66. f(x) = V 4+ 1 2 :, a =1 97 67. f(x) = \\/1+ vr, a = 1 68. f(x) = \\2+ vr,a =1 99 Algebraically Calculating Slopes of Tangent Lines 10 (Newton Quotients h Variable Version) 10 In Exercises 69-106 , for the given function f and the given point a, do the following: 10 (a) Write the expression for the secant-slope function in the h-variable form: 10 more (h) = (ath) - f(a) h and simplify as much as possible