I need to find absolute maximum/minimum of this function (if this function has them), $$f(x) = x*\sqrt(|log(x)|)$$ domain of the function is: $f: ]0, +\infty[ \to R$ and the domain of its first derivative isn't defined in $x=e$, because when $x = e$ the sign function (i.e derivative of absolute value) isn't defined.Therefore, $$f': ]0, +\infty[\space (without\space e) \to R$$ the first derivative is (I've used product rule): $$1 * \sqrt(|log(x)|) + [(x * sign(log(x)) + 1/x) / (2 * \sqrt(|log(x)|))] $$ the only thing who can change the sign of the first derivative is the sign function, therefore I have: $(sign(log(x)) >= 0 \iff log(x) > 1$, so now I know that $f: [1, e[$ is increasing, and $f: ]e, +\infty[$ is increasing too.
$f: ]0, 1]$ is decreasing.
Therefore I have that 1 is a minimum, and the function doesn't have absolute max/absolute min. but this is wrong, because $e^-1/2$ must be a local maximum.
- EDIT: I think I got where the problem is, in the first derivative, as always.
The minimum of the function is $0$. Note that for any $x$ in $(0,\infty)$, we have $x>0$ and $\sqrt{|\ln x|}\geq 0$, so $f(x)=x\sqrt{|\ln x|}\geq 0\cdot 0=0$. It can be seen that $f$ actually attains the value $0$ at the input $1$, since $\ln(1)=0$, so $0$ is indeed the minimum.
This function does not have an absolute maximum because $\lim_{x\to\infty}f(x)=\infty$. However, as you found, it does have a local maximum at $1/\sqrt{e}$.