A problem on Mean Value Theorem 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.

there exists $x_0$ and $y_0$ in $[a,b]$ such that
\begin{align}
f(a) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(a - \frac{a+b}{2}) + \frac{1}{2}f''(x_0)(a-\frac{a+b}{2})^2 \\
f(b) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(b - \frac{a+b}{2}) + \frac{1}{2}f''(y_0)(b-\frac{a+b}{2})^2 
\end{align}
Multiplying by $\frac{1}{2}$ and summing together these two equations gives
\begin{align}
f(\frac{a+b}{2}) = \frac{1}{2}[f(a) + f(b)] -  \frac{f''(x_0) + f''(y_0)}{2}\frac{(a-b)^2}{8}
\end{align}
Use Darboux's theorem to find a $c \in [a,b]$, such that $f''(c) = \frac{f''(x_0) + f''(y_0)}{2}$
The above part $\textbf{didn't}$ use the assumption $f'(a) = f'(b)$, thus the conclusion is different, instead of + , I have a -. I correct in the following:
there exists $x_1$ and $y_1$ in $[a,b]$ such that
\begin{align}
f(\frac{a+b}{2}) = f(a) + f'(a)(\frac{a+b}{2} - a) + \frac{1}{2}f''(x_1)(a-\frac{a+b}{2})^2 \\
f(\frac{a+b}{2}) = f(b) + f'(b)(\frac{a+b}{2} - b) + \frac{1}{2}f''(y_1)(b-\frac{a+b}{2})^2 
\end{align}
Since $f'(a) = f'(b)$,  multiplying by $\frac{1}{2}$ and summing together these two equations gives
\begin{align}
f(\frac{a+b}{2}) = \frac{1}{2}[f(a) + f(b)] +  \frac{f''(x_1) + f''(y_1)}{2}\frac{(a-b)^2}{8}
\end{align}
The same theorem allows to get the conclusion.
