TITLE: prove that $f(c)=\frac{1}{2}(c-a)(c-b)f''(\xi)$ for $\xi \in (a,b)$
QUESTION [6 upvotes]: A function $f(x)$ is continuous on $[a,b]$ and $f''(x)$ exists for all $x\in (a,b)$. IF $c\in (a,b)$ $f(a)=f(b)=0$ prove that $f(c)=\frac{1}{2}(c-a)(c-b)f''(\xi)$ for $\xi \in (a,b)$.

I have no idea to solve it. I know that I have to use Lagrange MVT. Please help.

REPLY [4 votes]: Consider $g:x\mapsto (b-a)f(x)-\frac 12 (a-b)(b-x)(x-a)C$
You can find $C$ such that $g(c)=0$. 
Note that $g(a)=g(b)=g(c)=0$, so there's some $\xi$ such that $g''(\xi)=0$.