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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) =4-12,a =1 32. f(x) =9-12,a = 2 33. f(x) =1-13,a = 2 34. f(x) =2-x3,a=1 35. f(x) = 1/r, 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) = 1/23,a = 2 40. f(x) = 1/r,a = 1 41. f(x) = 1/Vr,a = 4 42. f(x) = -1/VI,a =9 43. f(x) = - 6 5 71 0 =2 44. f(x) = - 2+ 3' a =2 r+2 -, a =2 45. f() = 14 7 46. f(x) = 7+ 1 47. f(x) = rta=2 48. f(x) = r+2 1+5 a=1 49. f(x) = \\5+12,a=1 50. f(x) = V3+12,a = 2 2 51. f(x) = 141 4 7210=2 52. f(x) = = 2+210=1 1 2 53. f(x) = =2 54. f(x) = 55. f(x) = 14 /210=2 56. f(x) = 2+7210=1 r 57. f(x) = 1210=2 58. f(x) = = 2+2:0=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 1 63. f(x) = -,a =1 64. f(x) = a = 2 1 1 65. f(x) = = 2 66. J(@)= = 1 67. f(x) = \\/1+ vr, a = 1 68. f(x) = \\2+ Vr,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: msec (h) = f(a + h) - f(a) h