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!
 A: 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 $f''(r)\lt 0$.
A: 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 $\phi'(t^*) = f(t^*) = 0$.

A: We can choose either $[0,c]$ or $[c,1]$ to be the domain of $f(x)$ for our purposes. Let us write mean value theorems for each of the domains.

*

*$$f:[0,c]\mapsto\Bbb R, \frac{f(c)-f(0)}{c-0}\space=\space\frac{f(c)}{c}\space=\space f'(a)$$

*$$f:[c,1]\mapsto\Bbb R, \frac{f(1)-f(c)}{1-c}\space=\space f'(b)$$ Now we can switch to the original domain and write $$\frac{f'(b)-f'(a)}{b-a}\space=\space -\frac{f(c)(c+1)}{c(1-c)(b-a)}\space=\space f''(x_0)\space\lt\space 0$$ We are done with the first part, we can cover the second part. The integral vanishes, this means $f(x)$ should take on values below and above the $x$ axis. We can easily deduce that $f(x)$ has at least three roots. $$f(0)\space=\space f(1)\space=\space f(d)\space=\space 0, d\in(0,1)$$By Rolle's Theorem we can deduce that for some particular point $f''(x_1)=0$
