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My teacher proposed this exercise:

Let $f$ be a differentiable function on $(a, b)$ except in $x_1, x_2 \in (a, b)$. Suppose that $f$ is continuous on $[a, b]$. Prove the next statement: if $f'(x) > 0$ for all points of $(a, b)$ where $f$ is differentiable then $f$ is increasing on $[a, b]$. Is the converse true?

My attempt:

I applied the MVT for intervals $[a, x_1), (x_1, x_2), (x_2, b]$ taking two arbitrary numbers in each interval, checking the conditions for MVT and concluding that $f$ is increasing in each interval. But how I conclude that $f$ is increasing on $[a, b]$? I think I have to prove that $f(x) \le f(y)$ for every $x \in [a, x_1)$ and $y \in (x_1, x_2)$, the same way for the other interval, but I am not sure how to do it.

As the exercise is proposed, I consider the converse is not true. (I add the definition of increasing function given by the teacher. $f$ is increasing in $I$ if: $x \lt y \implies f(x) \lt f(y)$). So, taking $f(x) = k \in \Bbb R$ we have a increasing function but $f'(x) = 0$ for all $x \in \Bbb R$. I am reasoning in the right way?

I would really appreciate your help. And if anyone can provide me another example of the converse, or, off course, correct me.

Jhon C
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    Try starting with a lemma: if $f$ is differentiable at a point with a derivative greater than zero, then there must be some neighborhood of that point on which $f$ is increasing. – user3716267 Feb 06 '23 at 21:24
  • $f$ is still continuous at $x_1$, so you can replace $f(x_1)$ with its limit as you approach from either direction. You should be able to argue then that if you pick a point to the left, it's less than this limit, and if you pick a point to the right, it's greater, and then transitivity allows you to compare the two. – ZKe Feb 06 '23 at 21:27
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    @JhonC, the function $f(x)=k$ is not increasing, but you are right when you say that the converse of the statement is not true, indeed a counterexample is $f(x)=x^3$ which is increasing on $[-1,1]$ but $f’(0)=0$. – Angelo Feb 06 '23 at 21:33
  • @user3716267: the statement in your comment is not True. – Paramanand Singh Feb 06 '23 at 22:44
  • Split into $[a, x_1],[x_1,x_2],[x_2,b]$ and apply mvt to conclude that $f$ is strictly increasing on each on these subintervals and hence on $[a, b] $. Why did you choose open / semi open intervals in your approach? – Paramanand Singh Feb 06 '23 at 22:47
  • See related https://math.stackexchange.com/q/1845927/72031 – Paramanand Singh Feb 06 '23 at 22:52
  • @ParamanandSingh We have that $f$ is continuous, as well, from the problem statement. Do you have a counterexample? – user3716267 Feb 07 '23 at 02:34
  • @user3716267: I misinterpreted your last comment (thinking it to be coming from asker). A counter example is $f(x) =x^2\sin(1/x)+(x/2),x\neq 0,f(0)=0$. $f$ is continuous and differentiable everywhere with $f'(0)=1/2$ and yet there is no interval around $0$ where $f$ is increasing. – Paramanand Singh Feb 09 '23 at 16:01

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