TITLE: If $f(0)=0 ,f(1)=1 ,f'(0)=f'(1)=0$, then $|f''(x)|>4$
QUESTION [5 upvotes]: Let $ f $ be a twice differentiable function with $ f (0) = 0, f (1) = 1$ and $ f '(0) = f' (1) = 0 $, then $ 4 \leq | f'' (x ) | $,  for some $ x \in [0,1] $.
I tried using the mean value theorem for derivatives with
$ \dfrac{f '(1)-f' (0)}{f (1)-f (0) } = f '(c) = 0$
so there is a value at which $ f ''(c) = 0 $, but that did not help me.
Any ideas?
Any help is appreciated.

REPLY [5 votes]: Let $g(x)=f'(x)$. Then we know $\int_0^1 g(x)\,dx=1$ and $g(0)=g(1)=0$.
If $|g'(x)|<4$ everywhere, then due to the MVT $g(x)$ must be smaller than $4x$ and also smaller than $4(1-x)$. So $g$ is entirely within the triangle with corners $(0,0)$, $(\frac 12,2)$, $(1,0)$. Can its integral then be large enough?