Suppose $f(0) = f(1) = 0$ and $f(x_0) = 1$. Show that there is $\rho$ with $\lvert f'(\rho) \rvert > 2$. Suppose that $f : [0; 1] \rightarrow \mathbb{R}$ is continous and differentiable on $(0,1)$, that $f(0) = f(1) = 0$, and that $\exists_{x_0 \in (0; 1)} f(x_0) = 1$.
Prove that $\exists_{\rho \in (0;1)}|f'(\rho)| > 2$
Specifically, I cannot prove that $\exists_{\rho \in (0;1)}|f'(\rho)| > 2$ and not just $\exists_{\rho \in (0;1)}|f'(\rho)| \ge 2$ - it gets problematic when $\rho = \frac{1}{2}$.
There's been two similar questions asked about this problem, but none of the answers seem to be complete:
Suppose $f(0) = f(1) = 0$ and $f(x_0) = 1$. Show that there is $\rho$ with $\lvert f'(\rho) \rvert \geq 2$.
https://math.stackexchange.com/questions/1747287/f0-f1-0-f-frac12-1-find-such-c-that-fc-2
Unfortunately, I cannot post comments to these answers as my rank is too low.
 A: Suppose that $|f'(x)|\le 2$ holds for all $x\in (0,1)$. Select arbitrary $x^*\in(0,\frac{1}{2})$, then
$$
\left|\frac{f(x^*)-f(0)}{x^*-0}\right| = \left|\frac{f(x^*)}{x^*}\right| \le 2
$$
and so
$$
|f(x^*)|\le 2x^*
$$
In similar way,
$$
\left|\frac{f(\frac{1}{2})-f(x^*)}{\frac{1}{2}-x^*}\right| \le 2
$$
and so
$$
1-|f(x^*)|\le |1-f(x^*)| \le 1-2x^*
$$
Since $2x^* \le |f(x^*)| \le 2x^*$, $|f(x^*)|=2x^*$ and so $f(x)=2x$ for all $x\in(0,\frac{1}{2})$. We can also show that $f(x)=2-2x$ for all $x\in(\frac{1}{2},1)$. However,
$$
f(x)=\begin{cases}
2x,&x\in[0,\frac{1}{2})\\
2-2x,&x\in[\frac{1}{2},1]
\end{cases}
$$
is not differentiable at $x=\frac{1}{2}$. By contradiction, we get the conclusion.
A: If $0<x_0<1/2$ or $1/2<x_0<1$, it can be proved easily by mean value theorem. So let's assume $x_0=1/2$.
Assume that $|f'(\rho)|\le2$ $\forall \rho\in(0,1)$. Then again by mean value theorem, for $x\in[0,1/2)$, $f(x) \le 2x$. For $x\in (1/2,1]$, $f(x) \le -2x+2$. Then it can be shown that $f$ can't be differentiable at $x=1/2$.
In fact
$$\partial_-f(1/2)=\lim_{h\rightarrow 0^-}\frac{f(1/2+h)-f(1/2)}{h} \ge \lim_{h\rightarrow 0^-}\frac{2(1/2+h)-1}{h}=2$$
while
$$\partial_+f(1/2)=\lim_{h\rightarrow 0^+}\frac{f(1/2+h)-f(1/2)}{h} \le \lim_{h\rightarrow 0^+}\frac{-2(1/2+h)+2-1}{h}=-2$$
