Differentiability and finding point c 
Assume that a function $f : ]0, 3[ \rightarrow \mathbb{R}$ is differentiable on $]0, 3[$. Assume also that $f'(1) = 4$ and $f'(2) = 6$. Show that there exists an $c \in ]1, 2[$, for which $f'(c) = 5$ holds.
Hint: Investigate the function $g(x) = f(x) -5x$.
Notice: You cannot use Bolzano's theorem here since $f'(x)$ may not be continuous.

Can we show this using one of the other familiar theorems or is there something I'm missing?

Consider the function $g(x)=f(x)-5x$. You are given that $g$ is differentiable and $g'(x)=f'(x)-5$, so
$$
g'(1)=-1,\ \ \ \ g'(2)=1. 
$$
Because $g$ is continuous on $[1,2]$, it achieves a minimum at some point $c\in[1,2]$. From $g'(1)<0$ we get that $c\ne 1$ (because there have to exist points $r>1$ with $f(r)-f(1)<0$ for $g'(1)$ to be negative) and similarly $c\ne2$. So $c\in(1,2)$ and then, being a local minimum of a differentiable function, $g'(c)=0$. So $f'(c)=5$.
