Only number 76
= Open App 74 Chapter 2: Derivatives and simplify as much as possible. 101-104, you will have to guess the value of the limit us- (b) Find the slope of the tangent line m = lim mec(r) ing a computer/calculator and the graphical method (See by evaluating the limit. Find this limit exactly using the Example 2.1.6). algebraic method, c) Write an equation for the tangent line to f at (c) Write an equation for the tangent line to f at (a, f(a)). Optional: Check your work by graphing the (a, f(a)). Optional: Check your work by graphing the function f and the tangent line at the given point. function f and the tangent line at the given point. NOTE: In 105-106, the function f is piecewise-defined, so in evaluating the limit, you will have to use one-sided 31. f(x) =4-12,a=1 32. f(x) = 9-x2, a =2 limits. (See Example 2.1.7.) Also in this situation the 33. f(x) =1-23,a = 2 34. f(x) =2-x3,a=1 tangent line may fail to exist, but tangent lines on each 35. f(x) = 1/x, a =2 36. f(x) = 2/x, a = 1 side of where the graph is pieced together may exist. If so, for part (c) write equations for these two lines and 37. f(x) = 1/z,a = 3 38. f(x) = 1/r , a = 2 check your work by graphing them and along with f. 39. f(x) = 1/x3,a = 2 40. f(x) = 1/x3, a =1 41. f(x) = 1/Vr, a = 4 42. f(x) = -1/Vr, a =9 69. f(x) = x2 - 3x, a = 2 70. f(x) = x' - 4r, a =1 43. f(x) = - 6 710 =2 44. f(x) = 5 71. f(x) = 13 -4x2 + 12, a = 1 r + 2 r + 3 70 =2 72. f(x) = 13 -31+2, a=0 45. f(x) = I -10 =2 46. f(I) = -.0=1 73. f(x) = Vr+1, a =1 74. f(x) = Vx -2, a =5 1+r 75. f(x) = 1/r, a = 2 76. f(x) = 3/r, a = 1 47. f(x) = = 48. f(x) = I+2 77. f(x) = 2/(r+3), a = 2 78. f(x) = 3/(r+2), a = 3 49. f(x) = \\5+12,a = 1 50. f(x) = V3+12,a = 2 79. f(x) = 1/x2, a = 2 80. f(x) = 3/x2, a = 1 81. f(x) = 1/x3, a = 2 82. f(x) = 1/r',a= 1 51. f(x) = = 2 1+ 2 0 =2 52. f(x) = 2+ 21 83. f(x) = 1/VI, a = 4 84. f(x) = -1/Vr, a =9 53. f(x) = 1 2 1+ 73,4=2 54. f(x) = 85. f(x) = =,a=2 6 86. f(x) = ,a=2 5 2+13 2 87. f(x) = -,1 =2 88. f(x) = - .0=3 55. f(x) =- 1+ ,2 0=2 56. f(x) = 2+210 =1 1+x 89. f(x) = 90. f(x) = .1 =3 57. f(x) = 210=2 58. f(I) = ,7,4=1 2+ 7 2 59. f(x) = vx2 + 3x, a = 160. f(x) = Vx2 +5x,a = 1 91. f(x) = 1+ 7?' 92. f(x) = 1 2 61. f(x) = Vx3 + 1,a=1 62. f(x) = Vx3 +2,a=1 93. f(x) = 1+ 7310 =2 94. f(I) = 63. f(x) = 1 , a =1 64. f(x) = 95. f(x) = 1+ x?' ,a =2 96. f(x) = 2+210=1 65. f(x) = V1+ 2 0=2 66. f(x) = V4+ 12' 10 20 97. f(x) = 1 + vr 98. f(@)= 2+ VI Fa=1 67. f(x) = \\1+ Vx, a = 1 68. f(x) = \\2+ Vi,a =1 99. f(x) = \\1+ Vz,a=1 100. f(z) = \\2+ Vz,a = 1 Algebraically Calculating Slopes of Tangent Lines 101. f(r) = (10Inx)/x, a = 2 102. f(x) = In(x + 2), a =1 (Newton Quotients h Variable Version) 103. f(x) = re ', a=1 104. f(r) =re , a=1 In Exercises 69-106 , for the given function f and the 105. f(z) = 4-x2 if x -1 1 0=-1 (a) Write the expression for the secant-slope function in the h-variable form: 106. f(z) = -3 if x 50 msee (h) = J(ath) - f(a)
