Why is the domain of x raised to x (0,infinity)?

Question

My math teacher was explaining how to draw graphs of given functions. For $f(x)=x^x$ he put the domain as $(0,\infty)$. Why is this true (if it is)? For $x= -2$, $(-2)^{-2}$ is $1/4$, and so the function is defined at $x= -2$. Isn't it?

Answer 1

For negative numbers, the sign of the function changes with the values of $x$ and hence plotting a continuous graph is not possible.

For simplicity, consider $x = -1$, Then $f(-1) = (-1)^{-1} = -1$.

Whereas at $x = -2$, $f(-2) = (-2)^{-2} = 1/4$

Answer 2

Powers of the negatives are problematic. You can define rational powers using roots, and some fail to exist ($(-\frac14)^{-\frac14}$ ?). An irrational power of a negative is even a worse beast.

But you are right, you could extend the domain to a subset of the negative rationals.

Answer 3

As drhab suggested, this is a nice way to answer your question. But as additional comment, personally, I would attribute this as a convention (and a very good one). Technically, there is no supposed "right" or "wrong" here, but appending additional elements for which $x^x$ is defined in the reals is probably not that good of an idea.

Of course, $x^x$ is defined (in the real numbers) for some $x < 0$, but for most $x < 0$ (like $x = -1/2$ or $x = - \pi$ or $x = -3/4$), $x^x$ is not defined. Ideally, we would like our function $x^x$ to satisfy a number of nice properties on its domain, like continuity, differentiability, etc., but defining the domain of $x^x$ to be $[0, +\infty)$ and the union of all the nonpositive integers we would lose all of these properties.

Moreover, $f(x) = x^x$ is defined in the complex numbers without any of the sort of problems that real numbers face, so as Marc van Leeuwen said, why would we want to mix two different definitions that describe the same idea in such an unfruitful manner? It's better just to restrict $x^x$ to the domain $[0, +\infty)$.