Sufficient condition to show $f$ is monotonically increasing in some neighborhood I am curious if the following statement holds. 
Let $f:[a,b] \rightarrow \mathbb{R}$ be a continuous function differentiable on the open interval $(a,b)$. Then if $f'(c)>0$ for some $c \in (a,b)$, there exists a neighbourhood of $c$ in which $f$ is monotonically increasing. 
An ideal answer to this question would include either a proof or a counterexample. 
 A: As noted in other answers the results holds if $f\in C^1$. Consider the following counterexample:
$$f(x)=\begin{cases}
x+2x^2 \sin(\frac{1}{x})\quad \text{ if } x\neq 0 \\
0\qquad \quad \qquad \quad \quad \text{ if } x=0
\end{cases}$$
with derivative
$$f^\prime(x)=\begin{cases}
1+4x \sin(\frac{1}{x})-2\cos(\frac{1}{x}) \quad \text{ if } x\neq 0 \\
1\quad \qquad\qquad\qquad\quad \quad\qquad  \text{ if } x=0
\end{cases}$$
Note that $f^\prime(0)>0$ but every neighbourhood of $0$ has negative values (and positive values).
A: If you know $f'$ is continous, this is trivially true of course.
If it is not, that could be false.
A: This is true if $f \in C^1((a,b))$, i.e. $f'$ is continuous on $(a,b)$. To see this:
Suppose $f'(c)>0$, where $c \in (a,b)$ and $f \in C^1((a,b))$. Then $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that $|x-c|<\delta$ $\Rightarrow$ $|f'(x)-f'(c)|<\epsilon$. Take $\epsilon= \frac{1}{2} f'(c)>0$. Then $\exists$ $\delta>0$ with $|x-c|<\delta$ $\Rightarrow$ $0<\frac{1}{2}f'(c)<f'(x)$, so $f$ is monotone increasing $\forall$ $x \in (a,b)$ with $|x-c|<\delta$.
If $f \notin C^1((a,b))$ I see no reason for the claim to hold.
