Determine if there is a value, c, -8<c<12 such that h’(c)=-1

Question

Let h be a twice differentiable function such that h(-8)=12 and h(12=-8. Let f be a function defined such that f(x)=h(h(x)).

Answer 1

Are you sure you copied the whole problem correctly? The problem defines a function f(x) = h(h(x)), but we never use this function f. And we're given that h is "twice differentiable", but we never use the "twice". All of this is suspicious.

Maybe the problem is

"Determine if there is a value c, -8 < c < 12, such that f'(c) = 1."

Then

f(-8) = h(h(-8)) = h(12) = -8

f(12) = h(h(12) = h(-8) = 12,

so the average slope of f from x=-8 to x=12 is 1.

The Chain Rule says that f'(x) = h'(h(x))h'(x), so f is differentiable on (-8,12),

so the Mean Value Theorem says that there is indeed a c, -8 < c < 12 with f'(c) = 1.

I'm still not satisfied with this solution because we never used the "twice differentiability" of h.

Answer 2

Since h is differentiable on the interval, and therefore also continuous, the Mean Value Theorem applies to h.

And, on [-8 , 12] , we have the avg. rate of change of h = (-8 - 12) / (12 - (-8)) = - 1, the MVT guarantees an x-value, c, with -8 < c < 12 , such that h^'(c) = -1.