Just answer d) tell me the detailed procedure for calculating f '(c) = (f (e ^5) -f (1)) /e ^5-1, and why the final answer is 5/e ^5 (e ^5-1)

4) The function f(x) = "has derivatives f' ( x ) = = 1 - In(x) f" ( x ) = 2In(x) - 3 f'll ( x) = 11 - 6 In(x) x2 x3 x 4 a) (5 points) f has a critical point at x = e. Is that critical point a local maximum, a local minimum, or neither? (Don't forget to justify your answer.) file ) = 2 luce ) - 3 By Second Deriv Test , X = @ is a local maximum. b) (5 points) The function f is continuous on [1, es]. By the Extreme Value Theorem, f has an absolute minimum on this interval. Find the x value where it occurs. Explain why the point you found is an absolute minimum. f (e ) = lle) . el Xz 1- In(x ) = 0 and = 0 - least Valve x = e fles ) - bale ) = 5 es only citpt : x = e Must occur at either critical point or Endpoint. At x =_ 1 c) (3 points) Find the average rate of change of the function f on the interval [1, es]. fles ) - f(1 505 - 0 e5 - 1 5 VALUE : es(es- 1) d) (4 points) The function f is continuous on [1, es] and differentiable on (1, es). Suppose that c in the interval (1, es) is the point guaranteed by the Mean Value Theorem. What is the value of f'(c)? Explain your answer. By MVT there is a point c in ( 1, e5 ) with f (c) = f(85) - fli which Equals es( es- 1 ) 5 2 5 - 1 5 f' ( c ) =_ es( es- 1 )