Consider the function $f(x) = x^x$.
Wolfram alpha tells me that this function’s domain is $x : x>0$, $x \in \mathbb{R}$. I can’t see why it cannot be defined for a number like $(-2)$. I mean $(-2)^{-2}=0.25$, the same Wolfram Alpha told me. I realize that fractional powers for negative numbers may cause problems, but it could be defined for integers. Thanks for any help.
For most purposes (including the hotly debated question of what to make of $0^0$) you can consider any instances of exponentiation $x^y$ (where $x,y$ can stand for expressions) to stand for on of two quite disparate definitions that happen to coincide on the intersection of their domains:
If $y$ designates an integer, then $x^y$ is defined algebraically; recursively by $x^0=1$ and $x^{n+1}=xx^n$ for the case $y\geq0$, and provided $x$ is invertible by $x^{-n}=(x^{-1})^n$ for $y<0$.
In other cases one must assume that $x$ is real and positive, and $x^y$ stands for $\exp(y\ln x)$ (note that I did not write $e^{y\ln x}$, which would result in a circular definition). Here $\exp$ is a perfectly defined function $\Bbb C\to\Bbb C$, so one can allow $y$ to be any complex number (or one could even amuse oneself by taking square matrices for $y$), but $x$ must be restricted to avoid ambiguity of $\ln x$. One can of course extend the definition by making choices for $\ln x$, but using a bare $x^y$ for such cases would be confusing, and also one must be aware that many properties of exponentiation will start to fail.
Given this, the real function $x\to x^x$ can only be defined using the second variant, which justifies taking the domain to be the (strictly) positive reals. One could extend to domain to contain the non-positive integers as well (using the first definition), but mixing the two definitions of $x^y$ in a single usage is generally not a very fruitful idea.
Even if you take $x=0$, $f(x)=0^0 $ is an indeterminate form. $f(x)$ is defined for all positive values of $x$