# Prove that there is x, c < x < b, such that f(x) > f(c)

If $f : [a,b] \to R$ is differentiable at c, a < c < b and $f'(c) > 0$, prove that there is some x, c < x < b, such that $f(x) > f(c)$.

I'm not totally sure where to begin with this. Being that it is under the mean value section of my book, I would assume that that is relevant to the proof. My initial thought is that since the $f'(c) > 0$ we know the function is increasing at that point so you can peak a point x, c < x < b, so f(x) > f(c). I think you would need to know that the function is increasing from c to d for this to be the case though not just at the point c. So I'm not totally sure what to do from there.

Thank you anyone for the help.

Suppose otherwise, i.e., assume that for all $x \in (c,b)$, $f(x) \leq f(c)$. Then $$\frac{f(x) - f(c)}{x-c} \leq 0$$ for all $x \in (c,b)$. So as $x \rightarrow c^{+}$, the limit of $\frac{f(x) - f(c)}{x-c}$ will be $\leq 0$. This limit is $f'(c)$. So you have a contradiction.