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 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.

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

*$$f:[c,1]\mapsto\Bbb R, \frac{f(1)-f(c)}{1-c}\space=\space f'(b)$$ 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 $f''(x_1)=0$
