Increasing function on small interval given positive derivative Suppose that $f'(0)>0$. Does it imply that there exists a $\delta > 0 $ such that $f$ is increasing on $[0,\delta]$?
I think this is false and I've been trying to think of a counter example. I was thinking using the function
$f(x) = x^2sin(1/x)$ if $x \neq 0 $ and $f(x) = 0 $ if $x=0$, but I'm not sure if this example works?
I'd appreciate any help!

I came across this problem as well. The solution I found dealt with decreasing functions, but the problem is equivalent to the increasing case (just take $-f$ as defined below).
Consider the function $f$ on $[ 0 , 2/\pi ]$:
$$
f(x)=\begin{cases}
x^2\sin^2 \left( \frac 1x \right) -\frac 12 x&, \text{if } x\neq 0 \\
-\frac 12 x &, \text{if } x = 0
\end{cases}
$$
Its derivative is:
$$
f'(x) = \begin{cases}
2x\sin^2 \left(\frac{1}{x} \right) - \sin \left( \frac{2}{x} \right) - \frac 12 &, \text{if } x\neq 0 \\
-\frac 12 &, \text{if } x = 0
\end{cases}
$$
After doing some fiddling with the identities, it is possible to find a sequence $\langle a_n \rangle_{n=1}^\infty$ such that $f'(x)>0$. Letting $\langle a_n \rangle_{n=1}^\infty=\left\langle \left( \frac{5\pi}{8} + n\pi \right)^{-1} \right\rangle_{n=1}^\infty$, then $a_n \to 0 $ as $n \to \infty$. Therefore, in any open interval $[0, \delta]$ for positive $\delta$ one can find an element of this sequence contained within it.
Consider:
\begin{align*}
f'(a_n) &= 2a_n\sin^2 \left(\frac{1}{a_n} \right) - \sin \left( \frac{2}{a_n} \right) - \frac 12 \\
& = \frac{2}{\frac{5\pi}{8} + n\pi}\sin^2 \left(\frac{5\pi}{8} + n\pi \right) - \sin \left( \frac{5\pi}{4} + 2n\pi \right) - \frac 12\\
& = \frac{2}{\frac{5\pi}{8} + n\pi}\sin^2 \left(\frac{5\pi}{8} \right) -  \left( \frac{-1}{\sqrt{2}} \right) - \frac 12 \\
& = \frac{2}{\frac{5\pi}{8} + n\pi}\sin^2 \left(\frac{5\pi}{8} \right) +  \frac{\sqrt{2}-1}{2} \\
&>0
\end{align*}
But we know that a differentiable function $f$ is decreasing (strictly or not) on an interval $I$ only if $\forall x \in I: f'(x) \le 0$. But since for any interval $[0,\delta]$, we can find an $n$ such that $f'(a_n) > 0$, we see that for no $\delta > 0$ is $f$ decreasing on $[0,\delta]$, just as you had conjectured.
