# Calculus

posted by
**Lin**
.

Suppose f(x) is a differentiable function with f(-1)=2 and f(2)=-1. The differentiable function g(x) is defined by the formula g(x)=f(f(x))'

A. Compute g(-1) and g(2). Explain why g(x)=0 must have at least one solution A between -1 and 2.

B. Compute g'(-1) and g'(2) in terms of values of f and f'. Verify that g'(-1) = g'(2). Explain why g"(x)=0 must have at least one solution B between -1 and 2.

C. Suppose now that f(x)=Cx^2 +D. Find values of C and D so that f(-1)=2 and f(2)=-1. Compute g(x)=f(f(x)) directly for those values of C and D and use algebra on the resulting formulas for g(x) and g"(x) to find numbers A and B between -1 and 2 so that g(A)=0 and g"(B)=0. The "abstract" assertions of a) and b) should be verified.