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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-x3,a =2 34. f(x) =2-r,a=1 35. f(x) = 1/x, a = 2 36. f(x) = 2/x,a =1 37. f(x) = 1/x ,a = 3 38. f(x) = 1/r , a = 2 39. f(x) = 1/23,a = 2 40. f(x) = 1/x3,a = 1 41. f(x) = 1/Vr,a = 4 42. f(x) = -1/VI, a = 9 6 5 43. f(x) = - , a =2 44. f(x) =-,a=2 r 45. f(x) = 1+1 -, 0 = 2 46. f(x) =,0=1 2+r 47. f(x) = = r+1 -,a =2 48. f(x) = r+2 I + 3' I + 5' -,a=1 49. f(x) = \\5+x2,a=1 50. f(x) = V3+12,a = 2 1 2 51. f(x) = 7210=2 52. f(:) = , 210=1 2 53. f(r)= 1 1+ 23 3,0= 2 54. f(@) = 2+23:a=1 55. f(x) = 1+ 2 0=2 56. f(x) = 2+72:0=1 57. f(I) 1+ 2 0 =2 58. f(3) = 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) = =1 64. f(x) = 1 VA- T V9 - 12 1 65. f(x) = 1 V1+ = 2 66. f(x) = V 4+ 12 ,a=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