Let $f$ be a $C^2([a,b];\mathbb{R})$ function such that $f(a)=f(b)=0$, with $f(c)>0$ for a $c\in(a,b)$. Prove that there's at least one $d\in(a,b)$ such that $f''(d)<0$.
What I've tried is to define a function in $[a,b]$ $$F(x)=\begin{cases} \displaystyle \frac{f(x)-f(c)}{x-c} & x\neq c\\ f'(c) & x=c \end{cases}$$
and apply MVT but I have no information about $f'(c)$. Then I see $f(a)=f(b)=0$ and think of Rolle's Theorem, but without any results. Any hint please?
On
The function must have a maximum, which is attained at some point $p$ and, since $f(c)>0$ for some $c$, $f(p)>0$. But then $f'(p)=0$. Then $\frac{f(b)-f(p)}{b-p}<0$. So, there is some $q\in(p,b)$ such that $f'(q)=\frac{f(b)-f(p)}{b-p}<0$. And there is some $d\in(p,q)$ such that $f''(d)=\frac{f'(q)-f'(p)}{q-p}=\frac{f'(q)}{q-p}<0$.
I did it! Let's split the interval $[a,b]$ in $[a,c]\cup[c,b]$. $f$ continuous in $[a,c]$, derivable in $(a,c)$ then the MVT states that theres a $\xi_1\in(a,c)$ such that $$f'(\xi_1)=\frac{f(c)-f(a)}{c-a}=\frac{f(c)}{c-a}>0$$ The same applies for the interval $[c,b]$, theres a $\xi_2\in(c,b)$ such that $$f'(\xi_2)=\frac{f(b)-f(c)}{b-c}=-\frac{f(c)}{b-c}<0$$ Let $g=f'$. $g$ is continuous in $[\xi_1,\xi_2]$, as $f$ is twice derivable, then $g$ is derivable in $(\xi_1,\xi_2)$ then, the MVT states there's a $d\in(\xi_1,\xi_2)\subset(a,b)$ such that $$g'(d)=f''(d)=\frac{g(\xi_2)-g(\xi_1)}{\xi_2-\xi_1}=\frac{f'(\xi_2)-f'(\xi_1)}{\xi_2-\xi_1}<0$$