Let f be a differentiable function with f(0)=0 and f(1)=1, f'(0)=f'(1)=0. Prove that |f''(x)| > 4 for some x in (0,1). (Without invoking integrals) Let $f$ be a twice differentiable function such that $f(0)=0$ and $f(1)=1$.
Also, $f'(0)=f'(1)=0$.
Prove that $f''(x)>4$ for some $x \in (0,1)$.
Any help would​ be appreciated. 
My initial attempts were using Lagrange's Mean Value Theorem first between $f(0)$ and $f(1)$ to show that $f'(t)=1$ for some $t$ in $(0,1)$. 
Now applying LMV between $f'(t)$ and $f'(0)$, and $f'(t)$ and $f'(1)$ I could prove it if t lies in either $(0,1/4]$ or in $[3/4,1)$.
Don't really know where to go next.

Curious about how much work it is to get close to the boundary value. Easier to switch domain and range to $[-1,1]$ and try to keep $|f''|$ not much larger than 2. 
Here is a version of the question with built in scaling: Prove that $|f''(\xi)|\geqslant\frac{4|f(a)-f(b)|}{(b-a)^2}$
Here is an example of a quintic, not bad. Red is the quintic, second derivative is green. As I said, we are taking $-1 \leq x \leq 1.$ Let's see, $a=-1, \; f(a)=-1, \; b = 1, \; f(b) = 1.$ So
$$ \frac{4|f(b)-f(a)|}{(b-a)^2} = 2. $$
 
