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!
 A: 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.
A: If $f$ is any function at all such that
$$ x \leq f(x) \leq x + x^2$$
then $f'(0) = 1$ will hold. Note $\frac{d}{dx} x |_{x=0} = \frac{d}{dx} (x+x^2) |_{x=0} = 1$, so this will follow from the squeeze theorem for limits. If you try drawing functions trapped between $x$ and $x+x^2$, you will quickly convince yourself they do not need to be increasing near zero. Now just come up with an explicit example, perhaps along the lines of the one you discussed above. 
