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.

Note that
$$1 = |f(1) - f(0)| \leqslant |f(1/2) - f(0)| + |f(1/2) - f(1)|,$$
and by Taylor's theorem there exist $c_1 \in (0,1/2)$ and $c_2 \in (1/2,1)$ such that
$$f(1/2) = f(0) + \frac{1}{2} f''(c_1)\left(\frac{1}{2}\right)^2, \\ f(1/2) = f(1) + \frac{1}{2} f''(c_2)\left(\frac{1}{2}\right)^2$$
Hence,
$$1 \leqslant \frac{1}{8} (\, |f''(c_1)| + |f''(c_2)|\,) \leqslant \frac{1}{4} \max (|f''(c_1)|, |f''(c_2)|)$$
Your result follows.
