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74 Chapter 2: Derivatives and simplify as much as possible. (b) Find the slope of the tangent line m = lim msec(I) by evaluating the limit. Find this limit exactly using the algebraic method, (c) Write an equation for the tangent line to f at (a, f(a)). Optional: Check your work by graphing the function f and the tangent line at the given point. 31. f(x) =4-x2,a=1 32. f(x) =9-x2,a=2 33. f(x) =1-13,a =2 34. f(x) =2-13,a=1 35. f(x) = 1/x, a = 2 36. f(x) = 2/r,a = 1 37. f(x) = 1/1 ,a = 3 38. f(x) = 1/r , a = 2 39. f(x) = 1/x3,a = 2 40. f(x) = 1/r', a =1 41. f(x) = 1/Vz,a = 4 42. f(x) = -1/Vr,a =9 6 5 43. f(x) = - -,a =2 44. f(x) = = 7,0 =2 r +2 2 +3 45. f(x) = 14 1 46. f(3)=1,=1 47. f(x) = , a =2 48. f(x)= I+2 r + 3 I +5' a=1 49. f(x) = \\5+12,a =1 50. f(x) = V3+12,a = 2 51. f(x) = 1+ 2 0=2 52. f(x) = = 2 1 53. f(x) = =,a=2 54. f(x) =- 2 2+23:0=1 55. f(x)= + 2,a= 2 56. f(x) = 74 214=1 57. f(1) = 472 0=2 58. f(I) =, 2,=1 59. f(x) = vz? + 3r, a =160. f(x) = Vx2 +5r,a = 1 61. f(x) = Vx3 + 1,a=1 62. f(x) = Vx3 +2,a=1 63. f(x) = - 1 , a = 1 64. f(x) =- 1 V9 - 17 :, a = 2 1 1 65. f(x) = V I+ 1 a= 2 66. f(1) = VA+ 7?' a =1 67. f(x) = \\/1+ vr, a = 1 68. f(x) = \\2+ Vx,a =1 Algebraically Calculating Slopes of Tangent Lines (Newton Quotients h Variable Version) In Exercises 69-106 , for the given function f and the given point a, do the following: (a) Write the expression for the secant-slope function in the h-variable form: mace (h) = fath) - f(a) h