Suppose $f$ is differentiable on $\mathbb{R}$ and $f'(0)$ is positive (a) Must there be an interval around $0$ on which the derivative remains positive?
(b) Is there an open interval around $0$ on which $f$ is increasing?
For (a) I believe the answer is yes because $f$ is differentiable and the derivative $f'(0)$ exists.
Since $f'(x) = lim_{x \rightarrow 0} \dfrac{f(x)-0}{x-0}>0$,
then
$lim_{x \rightarrow 0-} \dfrac{f(x)-0}{x-0}>0$ and $lim_{x \rightarrow 0+} \dfrac{f(x)-0}{x-0}>0$
Am I on the right track here?
As far as (b) is concerned, I thought I would use the Mean Value Theorem and define the interval around $c>0$, but I know I can't start with point $c$ and from there find an interval $(a,b)$, so maybe there's another way, like Darboux's Theorem?
Thanks in advance!
 A: The counter-example from user detnvvp's answer shows that the answer to both your queries is NO. I want to address some concerns over your approach in this answer.
You start off with $$f'(x) = \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} > 0$$ I think there is a typo and the correct equation is $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} > 0$$ And then you show that the corresponding one sided limits are also positive. From that you get two things (which you should have tried to work out):
1) There is an $h > 0$ such that $f(x) > f(0)$ for all $x \in (0, h)$.
2) There is a $k > 0$ such that $f(x) < f(0)$ for all $x \in (-k, 0)$.
Such a behavior is expressed by saying that $f$ is strictly increasing at point $0$. But note that this is not the same as saying that $f$ is increasing in some interval containing $0$. Precisely we can't combine 1), 2) above to say that $f$ is strictly increasing in $(-k, h)$.
Thus you should not expect the answer of $(b)$ to be affirmative.
Regarding $(a)$ note that continuous functions have sign preserving property i.e. if $f$ is continuous at $a$ and $f(a) \neq 0$ then $f$ has same sign as that of $f(a)$ in some interval containing $a$.
So the answer to $(a)$ is affirmative if $f'$ is continuous at $0$. But we are not given this so we should not expect an affirmative answer for $(a)$.
A: (b) is wrong: to see that, consider the function $$f(x)=\left\{\begin{array}{c l}x^2\sin\frac{1}{x}+\frac{1}{2}x, & x\neq 0 \\ 0, & x=0\end{array}\right..$$ Then $f'(0)=\frac{1}{2}>0$, but $f$ is not increasing on a neighborhood of $0$. To see that, consider the derivative at any $x>0$: $$f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}+\frac{1}{2},$$ which changes sign infinitely often as $x\to 0^+$.
