How to show [f(a+h)-f(a-h)]/h=f'(a+ch)+f'(a-ch)?

Question

Given
[f(a+h)-f(a-h)]/h=f'(a+ch)+f'(a-ch)
where f is differentiable on the reals, a is in the reals and h>0
How could you show that there's a c in (0,1) such that the above is true?

Answer 1

Define a function g(x) = f(a+x) - f(a-x)

By the mean value theorem, there exists a c in the range [0, h] such that g'(c) = (g(h) - g(0))/h

http://en.wikipedia.org/wiki/Mean_value_theorem

The rest of the details you can fill in for yourself.