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.
 A: 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.
A: 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
$$ \frac{4|f(b)-f(a)|}{(b-a)^2} = 2. $$
 
A: I will use the fact the if $f(0) \leq g(0)$, $f'(0) \leq g'(0)$ and $f''(t) \leq g''(t)$ when $0 < t < x$ then $f(x) \leq g(x)$ with equality iff $f = g$.
Suppose that $f$ is twice differentiable, $f(0) = 0$, $f'(0) = 0$ and that $f''(t) \leq 4$ for all $t \in (0, 1).$ Then, taking $g(t) = 2t^2$ in the above fact, we have $f(x) \leq 2x^2$ and especially $f(\frac12) \leq \frac12$ with equality iff $f(t) = 2t^2$ for $t \in [0, \frac12]$. Applying the fact on $\hat f(t) = 1 - f(1-t)$ and $\hat g(t) = 1 - g(1-t)$ gives $\hat f(\frac12) \leq \hat g(\frac12)$, i.e. $f(\frac12) \geq \frac12$ with equality iff $f(t) = 1-2t^2$ for $t \in [\frac12, 1]$. 
Thus, for $f$ to be continuous at $\frac12$ we must have $f(t) = 2t^2$ for $t \in [0, \frac12]$ and $f(t) = 1-2t^2$ for $t \in [\frac12, 1]$. However, $f''(\frac12)$ is not defined for this function, which contradicts our assumption.
Therefore, some assumption must be invalid. The assumptions that $f$ is twice differentiable, and that $f(0)=0$ and $f'(0)=0$, are given. Therefore the assumption $f''(t) \leq 4$ for all $t \in (0, 1)$ must be taken as invalid. Thus there must be some point $x \in (0, 1)$ where $f''(x) > 4$.
