Let $f\in C([0,1])$ be differentiable on $(0,1)$. Suppose that $f(0)=f(1)=0$ and that there is an $x_0\in(0,1)$ where $f(x_0)=1$. I have to prove that there is a $c\in(0,1)$ satisfying $|f^\prime(c)|>2$.
This is what I have done:
$f(x)=\int^x_0f^\prime(t)dt=-\int^1_xf^\prime(t)dt$. Now suppose in the seek of a contradiction that $|f^\prime(x)|\leq2$ for every $x\in(0,1)$. Then we have $|f(x)|\leq\int^x_0|f^\prime(t)|dt\leq 2x$ and $|f(x)|\leq\int^1_x|f^\prime(t)|dt\leq 2(1-x)$. And so $|f(x)|\leq\mathrm{min}\{2x,2(1-x)\}$. Evaluating this inequality in $x_0$ we obtain $1\leq2\mathrm{min}\{x_0,1-x_0\}$ and this is a contradiction if $x_0\neq1/2$. Now, if $x_0=1/2$ we can easily reduce to the case $|f(x)|<1$ if $x\neq1/2$. But I can't continue the proof in this case. Could you help me ,please?