I'm trying to figure out the domain of the function $y=x^x$. When I graph it, it appears to be defined on $[0, \infty)$, but then when I plug in individual negative numbers, for some of them I get real numbers, such as $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$. However, for other numbers such as $x=2.2$, the result is imaginary.
On what real numbers is $y=x^x$ defined?
The standard domain for $x^x$, as a Real-valued function of a Real variable is $(0, \infty)$. There are values in the negative Real line where $x^x$ is defined, but there are no open sets in the negative Real line where $x^x$ is Real-valued.
Notice that for negative Real numbers, if $x=p/q$; $p$ a negative integer , q odd, then $x^x=(p/q)^{p/q}=((p/q)^p)^{1/q} $ is the $q-th$ root of a negative number, which exists(as a Real number) iff $q$ is odd. I think this is the entire set where $x^x$ is defined (As a Real number) on the negative Real axis. EDIT: As correctly pointed out by Dave Renfro, this works only for $p,q$ relatively-prime; otherwise , you run into the problem of having many representations for $p/q$, e.g., $1/3,2/6,...$ so that the function is defined for some representations but not for others.