$a$ - point of strict local minima. Does there exist an interval, that contains $a$, that $f'(x) < 0$ for every $x<a$ in that interval, and $f'(x)>0$ for every $x > a$ in that interval?
I tried to get an example, where statement is wrong. I tried something like $x^2\sin(\frac{1}{x}) $ with some modifications.
How to prove or disprove it?
No. Put$$f(x)=\begin{cases}2x^2+x^2\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}$$Then$$f'(x)=\begin{cases}2x\left(2+\sin\left(\frac1x\right)\right)-\cos\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}$$So, $f$ has an absolute minimum at $0$, but it is not true that $f'(x)<0$ in some interval $(-a,0)$ and that $f'(x)>0$ in some interval $(0,a)$, with $a>0$.