Show your work/reasoning. Be neat and orderly. Solve on scratch paper first if n

Question

Show your work/reasoning. Be neat and orderly. Solve on scratch paper first if necessary 1. (20 pts) Use Calculus, not a graphing calculator, to locate any critical numbers and the absolute extrema (give x- and y-coord) of the function f (x)J (x-2) 2/3 1 on 0,50 You need to explicitly state x -1 what the critical numbers and absolute extrema are (if any). 2. (10 pts) Let f(x) arctan x) on [0,1]. If the Mean Value Theorem can be applied, find all values c in the f(l)- f(0) interval (0,1) such that f'(c) 1-0 3. Given the graph of y f(x), which is differentiable everywhere and where grid lines are 1/2 unit apart estimate the following: (a) (3 pts)f'(1/2) (b) (5 pts) All values of x where f (x) 30 (c) (8 pts) D, f f(x) least two reasons for your labels.

Explanation / Answer

3) From the graph we can see that the function has zeros at x=-3/2, -1/4(multiciplity 2), 1/2, 2(multiciplity 2)

so we can assum the function to be

f(x)=a(x+1.5)(x+1/4)^2(x-1/2)(x-2)^2

Now, we see that at x=-5/4, f(x)=9/4

Using this in the given equation we see that

9/4=a(-5/4+1.5)(-5/4+1/4)^2(-5/4-1/2)(-5/4-2)^2

a=-0.486898

Thus, the function f(x) becomes,

f(x)=-0.486898(x+1.5)(x+1/4)^2(x-1/2)(x-2)^2

a) Now, f’(x)=[-0.486898(x+1/4)^2(x-1/2)(x-2)^2]+[-0.486898(x+1.5)2(x+1/4)(x-1/2)(x-2)^2]+[-0.486898(x+1.5)(x+1/4)^2(x-2)^2]+[-0.486898(x+1.5)(x+1/4)^2(x-1/2)2(x-2)]

Then, we have, at x=1/2

f(1/2)=0 and, f’(1/2)=-1.23246

b) f’(x)=0 at the points where the slope of the tangent line is horizontal. So, from the graph we can see that f’(x)=0 at x=-5/4, -1/4, +1/4, 5/4, 2

c) We have, f’(0)=0.517329, f’(1/2)=0,

Dx{f[f(x)]}=f’[f(x)]*f’(x)

Then at x=1/2

Dx{f[f(x)]}|x=1/2=f’[f(1/2)]*f’(1/2)=f’(0)*(-1.23246)=0.517329*(-1.23246)=-0.63758729934

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