calculus - 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!

Answer

For the second problem, we note that if $f$ is constant, then any point satisfies the requirement, so we can suppose $f$ is not constant.

If we can show that we can split $[0,1]$ into two intervals of non-zero length
$[0, t^*], [t^* ,1]$ such that $f(t^*) = 0$, and there exists $t_1 \in (0, t^*)$ and $t_2 \in (t^*,1)$ such that $f'(t_1)=0, f'(t_2) = 0$, we can apply
the mean value theorem to find some $t_3 \in (t_1,t_2)$ such that $f''(t_3 ) = 0$.

Hint:

Let $\phi(t) = \int_0^t f(x) dx$ and note that $\phi(0) = \phi(1) = 0$. Since $f$ is not constant, and $\phi'(t) = f(t)$, we see that $\phi$ is not constant. Then $\phi$ must have a maximum or minimum at $t^* \in (0,1)$, and we see that $\phi'(t^*) = f(t^*) = 0$.