Proving using mean value theorem Let f be a function continuous on [0, 1] and twice differentiable on (0, 1).
a) Suppose that f(0) = f(1)  =0 and f(c) > 0 for some c ∈ (0,1).
Prove that there exists $x_0$ ∈ (0,1) such that f′′($x_0$) < 0.)
b) Suppose that $$\int_{0}^{1}f(x)\,\mathrm dx=f(0) = f(1) = 0.$$
Prove that there exists a number x$_0$ ∈ (0,1) such that f′′(x$_0$) = 0.
How do I solve the above two questions? I have tried using Mean Value Theorem but it gives me zero when I differentiate for the first time. Not sure how I can get the second derivative to be less than zero.
Any help is much appreciated! Thanks!

We solve the first problem only. The function $f$ reaches a (positive) maximum at some point $p$ strictly between $0$ and $1$. At any such $p$, we have $f'(p)=0$.
Since $f(1)=0$, by the Mean Value Theorem there is a $q$ strictly between $p$ and $1$ such that $f'(q)\lt 0$.
Thus by the Mean Value Theorem there is a point $r$ strictly between $p$ and $q$ such that $f''(r)\lt 0$.
