If $ f'(c) > 0 $, then there is an $ x $ such that $ f(x) > f(c) $. Here is the homework question that I have:

If $ f: [a,b] \to \Bbb{R} $ is differentiable at $ c $, where $ a < c < b $ and $ f^{\prime}(c) > 0 $, prove that there exists an $ x $ such that $ c < x < b $ and $ f(x) > f(c) $.

(I appreciate your help!)
 A: Hint: assume $f$ has a (global) maximum at $c$, conclude something about $f'(c)$ and reach a contradiction. 
A: Hint: You know that $\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=f^{\prime}(c)>0$, so consider what happens if $f(x)\le f(c)$ for $x>c$.
A: Depending on how many theorems we have at our disposal, this proof is unnecessary. I'll give it anyways just for fun though.
We have assumed that $$\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = l > 0$$
Therefore, for any $\varepsilon > 0$, there exists a neighborhood of $c$ where $$\left\lvert\frac{f(x)-f(c)}{x-c}-l\right\rvert<\varepsilon$$
for all $x$ in said neighborhood. What this means for such $x$ that are greater than $c$ (so that we can multiply by $x-c$ without altering the inequalities) is that: $$\quad \quad \,\,\,l-\varepsilon<\frac{f(x)-f(c)}{x-c}<\varepsilon + l \\
\begin{align} &\Rightarrow &(l-\varepsilon)(x-c)<f(x)-f(c) < (l+\varepsilon)(x-c) \end{align}$$
Take $\varepsilon = l/2$, and $x_{\varepsilon}$ close enough to $c$ (and greater than) to get $$\frac{l}{2}(x_{\varepsilon}-c) < f(x_{\varepsilon})-f(c)$$
and the left expression is greater than $0$, from which $$f(x_{\varepsilon}) > f(c)$$
As an extra, the last part of the proof would have worked for any positive $\varepsilon < l$, from which we in fact see that there is a whole open interval $(c,y)$ of points where $f(x) > f(c)$ (allowing $y = +\infty$).
