What is the domain of $f(x)=x^x$? What is the domain of $f(x)=x^x$ ? 
I used Wolfram alpha where it is said that the domain is all positive real numbers. Isn't $(-1)^{(-1)} = -1$ ? Why does the domain not include negative real numbers as well?
I also checked graph and its visible for only $x>0$ . Can someone help me clarify this?
 A: Write:
$$y=x^x=e^{x\log x}$$  
If we want $y \in \mathbb{R}$ we must have $\log x \in \mathbb{R}$ and this is done only if $x> 0$
This is the usual definition for the function $y=f(x)=x^x$ for $x \in \mathbb{R}$, that gives $(0,+\infty)$ as the domain.

If we want $x\in \mathbb{Q}$ than we can define the function as:
$$
y=f(x)=x^x=\left( \frac{m}{n}\right)^{\frac{m}{n}} \iff y=\sqrt[n]{x^m} \iff y^n=\left(\frac{m}{n}\right)^m
$$
If we define $0^0=1$, this is a real number if $n=2k+1 \quad \forall k\in \mathbb{Z}$ so the domain of the function can be:
$$
\{q\in \mathbb{Q}|q=\frac{m}{2k+1}\quad , \quad  m,k \in \mathbb{Z} \}
$$
A: The expression $x^y$ can be assigned a reasonable meaning for all real $x$ and all rational numbers of the form $y=m/n$, where $m$ is even and $n$ is odd and positive. Thus $x^y=(x^m)^{1/n}$, interpreted as the unique real $n$th root of $x^m$ (define $0^0$ to be $1$). Since every real number can be arbitrarily well approximated by such "even/odd" rationals, by continuity, a synthetic definition of $x^y$ can be obtained for all real $x$ and $y$. For example, using this definition, a graph can be plotted for the relation $y^y=x^x$, which runs smoothly as a loop through all four quadrants (along with the obvious line $y=x$).
