A doubt about differentiability and increasing function

Suppose $$f:(a,b)\rightarrow (a,b)$$ is differentiable on $$(a,b)$$ and for an $$x_{0}$$ such that $$a , $$f'(x_{0}) >0$$ then is $$f$$ increasing in some neighborhood of $$x_{0}$$?

I have seen examples on this site on disproving this for the interval $$(0,1)$$ by taking the function $$x+2x^{2}\sin(\frac{1}{x})$$ when $$x\neq 0$$ and $$0$$ if $$x=0$$. But I have a doubt whether this would be true for $$x_{0}$$ being an interior point of the open interval $$(a,b)$$ . Can someone please clarify . I have tried to prove it using LMVT but since nothing is said about continuity on $$[a,b]$$ I am unable to proceed.

• "in some neighborhood of $(a,b)$" : what do you mean ? – TheSilverDoe Feb 22 at 10:39
• Oh no...typo...I meant in some neighborhood of $x_{0}$ . Thanks for noticing – Arghyadeep Chatterjee Feb 22 at 10:56
• Take the same counterexample $x+2x^{2}\sin(x)$ on $(-1,1)$. – TheSilverDoe Feb 22 at 10:59
• I think the result holds true when $f'$ is continuous at $x_0$. – cxh007 Feb 22 at 11:08
• @cxh007 math.stackexchange.com/questions/1603958/… see this . – Arghyadeep Chatterjee Feb 22 at 11:21

If we can assume that $$f'$$ is continuous at $$x_0$$, then by $$f'(x_0)>0$$ we know that there exists a closed interval $$[c,d]$$ satisfying $$x_0 \in [c,d]\subset (a,b)$$ and s.t. $$\forall t \in [c,d], f'(t)>0$$. Moreover $$f$$ is differentiable on $$(c,d)$$, $$f$$ is continuous at $$[c,d]$$. Then we apply LMVT: if $$x,y\in[c,d], x, then $$f(y)-f(x)=f'(\xi)(y-x)$$ where $$\xi \in [c,d]$$, and thus $$f'(\xi)>0, (y-x) >0$$, finally for $$x,y\in [c,d]$$, $$f(y)>f(x)$$ whenever $$y>x$$.
As noted, for counterexample just take $$(a,b)=(-1,1), x_0=0,f(x)=x+2x^2 \sin(\frac{1}{x})$$ when $$x \neq 0$$, $$f(x)=0$$ when $$x=0$$.