I need help with what this question is asking.
Define $f$ by: $$f(x) = \begin{cases}x^2\sin\frac{1}{x}, & \mbox{if }x\neq 0 \\ 0 & \mbox{if }x=0\end{cases}$$ Let $g(x) = x + 2f(x)$. Show that $g'(0) > 0$ but that $g$ is not monotonic in any open interval about $0$.
Okay so I understand how to show $g'(0) > 0$ but concerning the following part of the question, showing it is not monotonic about $0$, is it asking given some interval $(a,b)$ where $0\in (a,b)$ we need to show that the interval $(a,b)$ is not monotonic. Or other words, prove that at some point the function does not consistently increase or decrease, or I can find two values $x_1,x_2$ such that $f(x_1) = f(x_2)$ but $x_1 \neq n_2$. If that is the case would that involve somehow finding a value in the derivative equal to $g'(0)$? Any help would be greatly appreciated!
The mean value theorem tells you that $\forall x,y\in (a,b)\exists c\in(a,b)$ such that $f'(c)(y-x)=f(y)-f(x)$. But the other direction is true too. If you find the $c$, the $x,y$ exist. Thus if $f'(c)=0$ then $f(x)=f(y)$. Now look at $g'(x)$ to find suitable points. You'll note that there is the additional requirement that $x,y$ need to be on opposite sides of $0$, but there's a simple argument for why that isn't a concern. A hint for this part would be to look at the graph of the function.