Question: Going through Larson's *Problem Solving Through Problems*, I am asked to give a geometric interpretation of the result below. I have been sketching it, and only got so far as to note that there must be a local minimum or maximum at $c$, although I cannot get very much further then that. 

Let $f$ be differentiable with $f'$ continuous on $[a,b]$. 

If $\exists \text{ } c\in(a,b]$ such that $f'(c)=0$, then we can find $d\in(a,b)$ such that $$f'(d)= \frac{f(d)-f(a)}{b-a}$$

Answer: 
The other cases are just not of interest to get the contradiction, they still exist. Assume 
$$
\frac{f(x)-f(c)}{x-c}>ε\text{ for }|x-c|<δ
$$
Then for $x>c$ we get $x-c>0$ so that the inequality can be transformed as
$$
f(x)-f(c)>ε·(x-c)>0
$$
which gives the contradiction.

On the other hand, $x<c$ results in $(x-c)<0$ and thus the transformed inequality
$$
f(x)-f(c)<ε·(x-c)<0
$$
which is not a contradiction at all since $f(c)$ was assumed to be the maximum.
Neighbourhoods in the usual topology of the real numbers extend on both sides of the point of which they are a neighbourhood.

The neighbourhood $V\_\delta$ must contain elements $x$ that are greater than $c$ and these elements lead to a contradiction.

So the contradiction follows from the established fact that such a neighbourhood has at least one $x>c.$ The contradiction does not prove that (any) $x$ is less than $c,$ but that no such neighbourhood exists.