Isolated positive derivative Is it possible that a function $f:\mathbb R\to\mathbb R$ satisfies $f'(p)&gt;0$ for some $p$ but for all $x\in\mathbb R\setminus\{p\}$, $f'(x)\le0$?

I made this question up myself, knowing basic calculus, and thought about it for a while, but couldn't come up with anything. Nevertheless I think it's an interesting problem.
If there is such a function, by the mean-value theorem, $f’(p)=\lim_{x\rightarrow p}\frac{f(x)-f(p)}{x-p}\leq 0$. This contradicts $f’(p)&gt;0$.
If $f'$ is continuous, then obviously not, that's basic calculus.
If you are willing to use distributions, you'll have to work on what $f'(p)&gt;0$ means, but then you can indeed provide an example, use the Heaviside function and $p=0$.