Why $x^{\sqrt{2}}$ is a complex number when $x<0$?

Question

$f(x) = x^{\sqrt{2}}$ function is undefined when $x<0$, i.e its domain is $x\ge 0$. What is the reason for that?

Answer 1

I will answer to the title question, not to the question within the body.

If $x\in(-\infty,0)$, then $x$ has no real logarithm, since $e^y>0$ for every real number $y$. But, in general$$x^{\sqrt2}=\exp\bigl(\sqrt2\log(x)\bigr),$$and $x$ has infinitely many complex logarithms. For each one of them, you get a distinct value for $x^{\sqrt2}$.