Why is $x^x$ only defined for $x>0$ Consider the real function $x^x$.
I understand that $0^0$ is undefined so $x \neq 0$ but $x$ values like $-1$ and $-2$ have well defined function values (although curiously opposite sign). Why isn't the curve well defined for $x<0$?
Note: the domain that I would like to investigate this function for is x is an element of the real numbers
 A: Think about a number of the form $x=-(n+a)$, where $n\geq 1$ is a natural number
and ${0<a<1}$. Then
$$x^x=(-(n+a))^{-(n+a)}=(-(n+a))^{-n}(-(n+a))^{-a}$$
So if $x^x$ is a real number, then so is $(-(n+a))^{-a}$. But then, there are
very many choices of $a$ for which $(-(n+a))^{-a}$ is not real. For example, if $a=1/2$, then this number is not real. There are many other choices, too. 
That's why the "natural" domain of $x^x$ does not include negative numbers, although - for a selected subset of them, the values $x^x$ are real valued.
A: $$x^x=\begin{cases}
x^x\qquad\text{if } x>0 \\
(-1)^x(-x)^x \quad\text{if } x<0 
\end{cases}$$
$(-1)^x=\cos(\pi x)+i\,\sin(\pi x)$
$$x^x=\begin{cases}
x^x\qquad\text{if } x>0 \\
\left(\cos(\pi x)+i\,\sin(\pi x)\right)(-x)^x \quad\text{if } x<0 
\end{cases}$$
For $x<0$, one can draw separately the real part $\cos(\pi x)(-x)^x$ and the imaginary part $\sin(\pi x)(-x)^x$.
Of course, for $x=-n$ integer, the imaginary part is $0$. The real part is $\cos(\pi x)(-x)^x=\cos(-n\pi)(n)^{-n}=\frac{(-1)^n}{n^n}$. So, the result is real :
$$(-n)^{-n}=\frac{(-1)^n}{n^n}$$
A: Let's say $x<0$. Then make a change: $x=-y, y>0$ so that your function becomes: $$x^x=(-y)^{-y}=\frac{1}{(-y)^y}.$$
Well, if $y$ is an integer, it is real, but if $y$ is a rational like $\frac{1}{2}$ (or $\frac{2m+1}{2n}$), its value is a complex (unreal) number: $$\frac{1}{(-1/2)^{1/2}}=\frac{1}{(1/2)i}.$$
