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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 $f(x)= x^x$?

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    $\begingroup$ to define $x^x$ over the reals you Need $x>0$ $\endgroup$
    – Dr. Sonnhard Graubner
    Nov 29, 2015 at 14:35
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    $\begingroup$ @Dr.SonnhardGraubner Why? $(-1)^{-1} = \frac{1}{-1} = -1$. $\endgroup$
    – layman
    Nov 29, 2015 at 14:37
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    $\begingroup$ If we define exponentiation as $a^b=e^{b\ln a}$, then $x^x$ is only defined on positive reals. $\endgroup$
    – Element118
    Nov 29, 2015 at 14:38
  • $\begingroup$ yes this is true, but the maximal range is $x>0$ e.g. $(-\sqrt{2})^{-\sqrt{2}}$ is not defined $\endgroup$
    – Dr. Sonnhard Graubner
    Nov 29, 2015 at 14:39
  • $\begingroup$ @Dr.SonnhardGraubner but if we have real $f(a)$ for any $a$ then $a$ must belong to domain of $f$, isn't it? $\endgroup$
    – Potemkin Metro Card
    Nov 29, 2015 at 14:42

1 Answer 1

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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 $\mathbb R_+^*$.

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  • $\begingroup$ Does that mean the domain of the function is not strictly defined? $\endgroup$
    – Potemkin Metro Card
    Nov 29, 2015 at 14:54
  • $\begingroup$ The domain of a function is always strictly defined. Just saying $f(x)=x^2$ does not give you a function, one always has to specify what the domain and codomain are. I might choose to look at $f:(0,\infty)\rightarrow (0,\infty),~f(x)=x^2$ because it fits the environment I'm working in e.g. $x$ is the speed of a car in kilometres per hour and $f(x)$ is the braking distance of the car at this speed. As the car won't have a negative speed, I choose to only have $x>0$ in my domain, although one could define $f:\mathbb R\rightarrow\mathbb R,~f(x)=x^2$ without any problems. $\endgroup$
    – Hirshy
    Nov 29, 2015 at 14:59
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    $\begingroup$ The domain is the set of all positive reals and the negative rationals with an odd denominator. $\endgroup$
    – user348749
    Aug 13, 2016 at 7:41

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