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?
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.