A doubt about differentiability and increasing function

Question

Suppose $f:(a,b)\rightarrow (a,b)$ is differentiable on $(a,b)$ and for an $x_{0}$ such that $a<x_{0}<b$ , $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.

Oh no...typo...I meant in some neighborhood of $x_{0}$ . Thanks for noticing — Mr. Gandalf Sauron, Feb 22 '21 at 10:56
I think the result holds true when $f'$ is continuous at $x_0$. — cxh007, Feb 22 '21 at 11:08
@cxh007 https://math.stackexchange.com/questions/1603958/sufficient-condition-to-show-f-is-monotonically-increasing-in-some-neighborhoo see this . — Mr. Gandalf Sauron, Feb 22 '21 at 11:21
@cxh007 if f is differentiable in (a,b) then it must be continuous at an interior point am I right? . Then the statement should be correct right? — Mr. Gandalf Sauron, Feb 22 '21 at 11:22
@cxh007 if $x_{0}$ is an interior point then can we work on a closed interval which is a subset of (a,b) and work with LMVT there? — Mr. Gandalf Sauron, Feb 22 '21 at 11:27
Yes, but in order to prove that $f$ is increasing, we may need that for every point $t$ in your closed interval $[c,d]$ there is $f'(t)>0.$ — cxh007, Feb 22 '21 at 11:42

cxh007 · Accepted Answer · 2021-02-22T11:48:58.927

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 $x,y\in[c,d], x<y$, 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 $x=0$.

A doubt about differentiability and increasing function

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