Prove that there exist $c \in (0,1)$ such that $|f'(c)|>2$ Let $f$ be continuous on $[0,1]$ and differentiable on $(0,1)$, with $f(0) = f(1) = 0$ and $f(k)=1$ for some $k\in (0,1)$. Prove that there exist $c \in (0,1)$ such that $|f'(c)|>2$
I tried the taylor expansion of $f$ at $x=k$,ie $f(x) = f(k) + f'(\xi)(x-k) = 1+f'(\xi)(x-k)$.
$f(0) = 1+ f'(\xi_0)(-k) = 1+ f'(\xi_1)(1-k) = f(1) = 0$
But i am not sure how to show the desired inequality. Any hints?
 A: Use the mean value theorem:
$$f'(c_0)=\frac{f(k)-f(0)}{k-0}=\frac{1}{k}\\f'(c_1)=\frac{f(1)-f(k)}{1-k}=\frac{-1}{1-k}$$
Now, there are 3 cases: $\begin{cases}k\in(0,\frac12)\\k=\frac12\\k\in(\frac12,1)\end{cases}$

In case 1, $f'(c_0)>\frac1{\frac12}=2$

In case 3, $f'(c_1)<-\frac1{1-\frac12}=-2\implies |f'(c_1)|>2$

In case 2, both $f'(c_0),|f'(c_1)|$ are equal to 2, if every point between $0$ to $k$ and between $k$ to $1$ has slope of $2$ in one side and $-2$ in the other then $f'(k)$ is undefined​, so I know that it can't be, but I also know that the average is $2$ in one side and $-2$ in the other, I know that because the average value of a function is: $$\frac1{b-a}\int_a^b g(x) dx$$ put in this $b=\frac12,a=0$ or $b=1, a=\frac12$, and $g(x)=f'(x)$ and you will get $2$ or $-2$
So there exist at leas one point between $0$ and $k$ that has slope is greater than $2$ and at least one point between $k$ and $1$ that has slope that is less than $-2$
A: It is easy to dispose of the cases $k<1/2$ and $k>1/2$ using the equations $$f(0)+kf'(c_{1})=f(k)=1=f(k)=f(1)+(k-1)f'(c_{2})$$ The case when $k=1/2$ requires a bit more analysis. If $f'(1/2)>0$ then there is a point $k'>1/2$ such that $f(k') >f(1/2)=1$ and then $$1<f(k')=f(1)+(k'-1)f'(c_{3})$$ so that $|f'(c_{3})|>2$. Similarly we can deal with the case $f'(1/2)<0$. 
Let's consider the case when $f'(1/2)=0$ and $f(1/2)=1$ is the maximum value of $f$ in $[0,1]$. If $f'(c) \leq 2$ for all $c\in(0,1/2)$ then we can see that $$f(x)=f(0)+xf'(c)\leq 2x,f(x)=f(1/2)+(x-1/2)f'(d)\geq 1+2(x-1/2)=2x$$ so that $f(x) =2x$ for all $x\in[0,1/2]$ and this contradicts $f'(1/2)=0$. Thus we must have some point $c\in(0,1/2)$ for which $f'(c) >2$.
