What would be the further steps for the case like this: I am finding the first derivative of a function: $f(x) = \ln(1+x^2)$ So the procedure would then be:
- $f'(x) = \frac{2x}{1+x^2}$.
- $f'(x) = 0 \implies \frac{2x}{1+x^2}$ , where $1+x^2 \neq 0$.
- Next step would be to show points at which $f'(x) \neq 0$, so I would have $1+x^2 = 0$ which results in $x^2 = -1$, which does not have any solutions, since the root is always positive.
Am I doing something wrong?
Graph shows clearly that at $y = 0$ at $x = 0$, but how to show it in calculations?