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The function is $$f(x)=\begin {cases} \log (x+1) & x\ge 0 \\ -\log(x+1) & x <0 \end {cases}$$

So $$f’(x)=\begin{cases} \frac{1}{x+1} & x\ge 0 \\ \frac{-1}{x+1} & x<0 \end {cases}$$

Now $f’(0) = 1 > 0$, so it should be strictly increasing

From the graph, it’s clear that the function is non monotonous at $x=0$. I also know that I am making a very trivial error, but I am confused about the cases when $x=0$

I am aware that the function is non differentiable at $x=0$, but discontinuous points are accepted when defining monotonicity

I have a very conceptual problem regarding this, so I chose a simple example.

If $f’(0)$ simply does not exist, then why do we consider discontinuous points?

Aditya
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  • When you define $f(x)$, you need to have the cases be $\log(x+1)$ if $\log(x+1)\geq 0$ and $-\log(x+1)$ if $\log(x+1)<0$. It doesn't make sense the way you have it defined now. – C Squared Sep 14 '20 at 04:00
  • @CSquared I think it can be defined in terms of $x$, since$ \log y <0$ for $y\in (0,1)$ – Aditya Sep 14 '20 at 05:44

1 Answers1

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Monotonicity does not depend on differentiability. A function can be monotone without being differentiable. This is because the definition of $f(x)$ to be monotone increasing, for example, is $\forall x_1, x_2 \in \mathbb R , \frac{ f(x_1)-f(x_2)}{x_1 - x_2} \ge 0$, where $x_1 \ne x_2$. This does not require differentiability at all.

However, if $f'(x) \ge 0$, then $f(x)$ is monotone increasing. (Remove all the equalities and you get the strict case.) This theorem enables us to determine whether a function is monotone, by inspecting its derivative. This method fails if this function has no derivative at all. So monotonicity is not affected, it is the link between the derivative and monotonicity that gets cut off.

Trebor
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  • So if basically, if $f’(a)$ does not exist, there is no mathematical way to tell if the function is monotone or not – Aditya Sep 14 '20 at 05:47
  • @Aditya Of course there is. Simply verify that $\frac{f(x_1)-f(x_2)}{x_1-x_2}\ge 0$, as said in the answer. In fact, you ought to have learnt this method before you even heard of derivatives. – Trebor Sep 14 '20 at 06:17
  • Right, so derivatives can’t be used. Out of curiosity, what would you plug in for $x_1$ and $x_2$? – Aditya Sep 14 '20 at 06:25
  • @Aditya I can't choose which. I have to plug in all possible values of $x_1, x_2$, and verify that inequality for all of them. – Trebor Sep 14 '20 at 06:29
  • Then how would you determine the function is not monotonous if graphing isn’t possible? – Aditya Sep 14 '20 at 08:47
  • It depends on what you know about $f(x)$. – Trebor Sep 14 '20 at 09:07
  • In the context of this question. How would you know that the function is not monotonous, assuming graphing is not allowed – Aditya Sep 14 '20 at 09:39