domain of $x^x$ What will be the domain of $f(x)=x^x$?
I have asked this to some teachers, they say that the domain is set of all nonnegative real numbers. It is true that there are infinite negative numbers for which $f(x)$ is not defined, but at the same time there are also infinite negative numbers (eg. $x=-3, -2/5,$ etc) for which $f(x)$ gives real numbers. So what exactly can be the domain of $f(x)= x^x$?
 A: It depends on the properties you want the function to have. At least for $x\in\mathbb Z$ you are right, as $x^x$ is well-defined for $x\in\mathbb Z,x<0$. When talking about $x\in\mathbb Q$, things get more difficult and I wouldn't argue that you can (easily) calculate $f(x)$ with $x=-\frac{2}{5}$. 
If you choose to include all $x\in\mathbb Z$ with $x<0$ in the domain of your function, your domain/function gets "messed up"; your domain is not an intervall anymore, it is not a differentiable function (it is however still continuous) etc. 
Now when we're talking about including some $x\in\mathbb Q$ for which $x^x$ is well-defined, things can get out of control. For $x=-\frac{1}{2}$ we would say that $$x^x=\left(-\frac{1}{2}\right)^{-\frac{1}{2}}=\frac{1}{\sqrt{-\frac{1}{2}}}$$ is undefined, however for $x=-\frac{2}{4}$ we have $$x^x=\left(-\frac{2}{4}\right)^{-\frac{2}{4}}=\frac{1}{\sqrt[4]{\left(-\frac{2}{4}\right)^2}}=\frac{1}{\sqrt[4]{\frac{4}{16}}}$$ which is well-defined. This means, that although we have $-\frac{1}{2}=-\frac{2}{4}$, that $f\left(-\frac{1}{2}\right)\neq f\left(-\frac{2}{4}\right)$ and thus $f$ is no longer a function.
For $x\in\mathbb R\setminus\mathbb Q$ we then finally need $x>0$, as for these $x$ we have $x^x=e^{x\ln(x)}$, which is only defined for $x>0$.
So although one could include all $x\in\mathbb Z$ with $x<0$, one normally chooses not to, to preserve those "nice" properties the function has when we restrict the domain to $\mathbb R_+^*$.
