Proving $f''(c)<0$ in the given condition. 
A twice differentiable function $f(x)$ defined on $[a,b]$ such that $f(a)=f(b)=0$,. $f(c)>0$ for all $a<c<b$, prove that there exist at least one value $c$ in $(a,b)$ such that $f''(c)<0$.

I tried by  starting with Rolle's theorem and got there exist a point '$s$'  between $a$ and $b$ as $f'(s)=0$. I don't know how to proceed further.
 A: Let $c$ be any fixed point in $(a, b)$. Then by Mean Value Theorem we can see that that there is a $d\in(a, c)$ such that $f'(d) =(f(c) - f(a)) /(c-a) >0$ and there is an $e\in(c, b) $ such that $f'(e) =(f(b) - f(c)) /(b-c) <0$. And thus by Mean Value Theorem we have $\xi\in (d, e) $ such that $f''(\xi) =(f'(e) - f'(d))/(e-d) <0$.

Note that we need to have just one single point $c\in(a, b) $ such that $f(c)>0$ for the above argument to work. The question has unnecessarily give that $f$ is positive on whole interval $(a, b) $. 
A: Since $f(a)=0$ and $f(c)>0$, we can conclude that the right derivative $f'(a) \ge 0$.
Since $f(b)=0$ and $f(c)>0$, we can conclude that the left derivative $f'(b) \le 0$.
Suppose $f''(x) \ge 0$ for all $x \in (a,b)$, then $f'(a) \le f'(x) \le f'(b)$, which implies $f'(x)=0$ for all $x\in [a,b]$, which implies $f(x)=0$ for all $x\in [a,b]$, a contradiction.
Another proof:
Choose a $c \in (a,b)$. We know that $f(c)>0$.
Then there exists $p\in (a,c)$ so that $f'(p)>0$, and $q \in (c,b)$ so that $f'(q)<0$.
Then there exists $r \in (p,q)$ so that $f''(r) < 0$
A: Sketch:


*

*By the extreme value theorem, let $c$ be such that $f(c)$ is maximal (actually any $c$ between $a$ and $b$ will do by the given conditions).  Observe that $f(c)>0$.

*By the mean value theorem, there is some $x$ such that $a<x<c$ and $f'(x)>0$.  Similarly, there is some $y$ such that $c<y<b$ and $f'(y)<0$.

*By the mean value theorem, there is some $z$ between $x$ and $y$ such that $f''(z)=\frac{f(y)-f(x)}{y-x}$.  Observe that the numerator is negative and the denominator is positive.  Therefore, $f''(z)<0$.
