TITLE: A problem on Mean Value Theorem
QUESTION [8 upvotes]: If $f''(x)$ exists on $[a,b]$ and $f'(a)=f'(b)$, then :
$$f(\frac{a+b}{2})=\frac 1 2[f(a)+f(b)]+\frac{(b-a)^2}{8}f''(c)$$
for some $c\in(a,b)$.
I tried but was unable to think of a function and was unable to use the given condition except for Rolle's Theorem which does not yield anything useful(yet). 
Any hints or help will be appreciated.

REPLY [2 votes]: By the MVT for divided differences the evenly spaced "discrete" approximation to the second derivative equals the real second derivative at some $c\in(a,b)$. We have $$f''(c)=\frac{\dfrac{f(b)-f\left({a+b\over 2}\right)}{(b-a)/2}-\dfrac{f\left({a+b\over 2}\right)-f(a)}{(b-a)/2}}{(b-a)/2}=\frac{4}{(b-a)^2}\left[f(a)+f(b)-2f\left({a+b\over 2}\right)\right]$$ Therefore $$f\left({a+b\over 2}\right)=\frac 1 2[f(a)+f(b)]-\frac{(b-a)^2}{8}f''(c)$$ It appears that I got a negative sign in front of the last term, as did Liu Gang. There may be a typo in the question.