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I've been taught that the domain of the function $f(x) = x^{1/x}$ is $x > 0$ assuming the function to be from $\mathbb{R} \to \mathbb{R}$. But $f(-1) = (-1)^{(-1)} = -1$, so why does the domain not include $x = -1$?

Learning
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3 Answers3

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In fact the domain of the function as defined in the question does include not only $-1$, but the reciprocals of all negative integers (e.g. you can define $(-1/2)^{-2}$); depending on how you define exponentiation you can even include (the dense set of) all negative rational numbers whose lowest-term representations have an odd numerator. However, the function is discontinuous at negative arguments, so mathematicians tend to limit their analysis to positive ones, possibly also zero if they use a limiting value.

Oscar Lanzi
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    This is correct. There is nothing wrong with defining it at $-1$, but you can't do calculus/analysis stuff with it "over there." – Randall Jul 06 '21 at 16:11
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Generally, in calculus, you consider expressions of the form $$f(x)^{g(x)}$$ as an abbreviation for $$\mathrm e^{g(x)\ln f(x)},$$ which is only defined when $f(x)>0$. (Do you see why both expressions are equal in that case?)

You are correct to say that this does not tell the whole truth, but the missing cases are generally not interesting from the point of view of calculus.

Zuy
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  • How do you reconcile this with $h(x)=x^3$? Do you really think of this as $h(x)=e^{3\ln x}$ so that $h$ is not definited for $x=1/2$? – Randall Jul 06 '21 at 16:01
  • @Randall Good point. I'd say that one shall use my suggested rewriting only in cases where $g(x)$ is non-constant. – Zuy Jul 06 '21 at 16:08
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A few things:

  • The notation "$f:X\to Y$" means that the domain of $f$ is $X$ and its codomain is $Y$. So it does not make sense to say "the domain of the function $f:\Bbb{R}\mapsto\Bbb{R}$ given by $f(x)=x^{1/x}$ is $x>0$".
  • Formally, a function is set of input-out pairs, and to specify that set, you have to state the domain. In practice, mathematicians are a little sloppy, and the domain is often assumed to be the set of real numbers that make sense when you plug them into a formula, unless otherwise specified. This set is sometimes called the "natural domain" of $f$.
  • With the above in mind, we can try to find the "natural domain" of $f(x)=x^{1/x}$. For $x>0$, the definition of $x^y$ is typically $\exp(y\log x)$. This definition works makes sense for irrational exponents as well as rational ones. However, if $x$ is negative, then there are instances where $x^{1/x}$ according to the "schoolbook" definition of exponentiation (e.g. when $x=-1$ as you have mentioned in your question).
  • Nevertheless, the set of negative $x$ for which $x^{1/x}$ makes sense is only countable, meaning that negative exponents are extremely ugly to work with in analysis. Because of this, some authors try to avoid this mess entirely by only defining $x^y$ for $x>0$. (This is particularly common if the exponent is variable rather than constant.) The only time where negative bases ever come up is when dealing with polynomials such as $x^2$, and perhaps with odd power roots as well.
  • The moral of the story is that the author about how exactly $f$ is being defined: right now it is not clear.
Joe
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