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

Question

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.

Answer 1

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.

In particular, I don't think you need the Mean Value Theorem for this, just the definition of a derivative.