When and why is $x^x$ undefined? I was learning in logarithmic differentiation how to differentiate a function like $f(x)=x^x$ so naturally i was curious about how the graph of the actual function looks like. I graphed it on this app and to my surprise the entire section from $(- \infty,0)$ is undefined. I do understand why we could have problems and abnormalities about $x=0$ but what about $x=-10$ shouldn't that just be $(-10)^{-10}$  which from my understanding is a very valid real number. So the question really begs itself why is the function undefined at $x<0$?
 A: Indeed, as you mentioned, something like $(-10)^{-10}$ is just $1/(-10)^{10}$.
The problem is that anywhere that we have rationals we will be very close to a value of $x$ which is of the form $-\frac{1}{2}p$.
In other words, note that $(-1/2)^{-1/2}$ is equal to $\frac{1}{(-1/2)^{1/2}}$ and we cannot take the square root of a negative number.
Likewise $$(-1/4)^{-1/4}=\frac{1}{(-1/4)^{1/4}}=\frac{1}{((-1/4)^{1/2})^{1/2}}$$ and again we have the square root of a negative number.
It turns out we have infinitely many of these cases along the negative real number line. Even though we technically can let the domain have some points, such as $-10$ or $-1$, it is easier if we just say the domain is $(0,\infty)$.
A: If $-1<x<0$ then $x^x$ translates into taking a root of a negative number, and that is undefined for real numbers.
For complex numbers, it is defined only when taking a branch, for example:
$$f(\frac{-1}{2})=\pm\sqrt{2}i$$
A function must have only one value.
The same logic applies also to any $x<0$
Regarding $x=0$:
$0^0$ is not defined even in the complex numbers domain, however
$$lim_{x\rightarrow 0^+}{x^x}=1$$
So for any practical use (e.g entropy) you can assume $0^0=1$
