$f'(c) > 0$ implies increasing on $I$? I already know that if $f'$ is positive on $I$ then $f$ is increasing on $I$.
But, I wanna know what if we know just $f'(c)>0$ where $c\in I$, not on whole $I$.
Then is there an open interval $I$ which contains $c$ such that $f$ is increasing on $I$?
I think there might be, but I can not prove why.
Since we know only the fact $f'(c)>0$, I can compare the value of function at $x$ with $c$.
But we have to compare all value at $x,y\in I$ to guarantee the increase.
How can I prove this?
Thanks.
 A: If you do not assume continuity of $f'$ at $c$, this may not be true. For example, consider the function $f$ defined by $f(0)=0$ and $f(x)=\frac{x}2 +x^2\sin\frac{1}x$ if $x\neq 0$. Then $f$ is $\mathcal C^1$ outside $0$ and easily seen to be differentiable at $0$ with $f'(0)=\frac12>0$. However, $f$ cannot be increasing on any interval $(0,\delta)$. Indeed, if it were so we would have $f'(x)\geq 0$ on $(0,\delta)$. But if $x\neq 0$ then
$$f'(x)=\frac12+2x\sin\frac1x-\cos\frac1x\, ,$$
so $f'(x)\leq \frac12 +2x -\cos\frac1x $, and hence $f'(x)\leq \frac34-\cos\frac1x$ for all $x\in (0,\delta)$ if we assume (as we may) that $\delta\leq\frac18$. Since one can find $x>0$ arbitrarily close to $0$ such that $\cos\frac1x=1$, it follows that $f'$ cannot be $\geq 0$ on $(0,\delta)$.
A: The function
$$f(x) = \begin{cases} x^2\sin\frac{1}{x} + \frac{x}{2} & x \neq 0 \\ 0 & x = 0\end{cases}$$
is differentiable everywhere.  Its derivative is
$$f'(x) = \begin{cases}2x\sin\frac{1}{x} - \cos\frac{1}{x} + \frac{1}{2} & x \neq 0 \\ \frac{1}{2} & x = 0\end{cases}$$
This derivative is not continuous.  We have $f'(0) > 0$ but because of the cosine term one can see that $f$ is not increasing on any interval that contains $0$.
