Given the function: $f(x) = \sqrt[3]{x} \cdot \sqrt[5]{x-1}$, Wolfram claims the domain is $x\ge1$, the same of the $\sqrt[5]{x-1}$ part.

My question is: why isn't the domain $\mathbb{R}$? I've always considered the domain of odd roots to be $\mathbb{R}$, or ignored the root altogether when calculating the domain. The sources I checked confirmed that it is correct.

In this case, why must $x$ be $\ge0$ if, say, $\sqrt[5]{-2}$ exists?

I feel like I'm missing out something very basic here... Thanks in advance for the help.


1 Answer 1


You are absolutely right: the domain of the $n$th root function, for $n$ an odd positive integer, is the whole of $\mathbb R$. More generally, if a real number $a > 0$ can be expressed as a fraction $a=\frac{p}{q}$ with $q$ odd, then the domain of $x^a$ is the whole of $\mathbb R$.

But note that the set of such exponents $a$ is dense in $\mathbb R$: between any two real numbers, there is such a suitable exponent $a$ (in fact, an infinity of such). This means that the general problem of deciding, for a given exponent $a$, whether its domain is the whole of $\mathbb R$ is a difficult one (in fact, it is technically uncomputable). So Wolfram Alpha decided to ignore the problem, rather than checking for all the special cases (such as $a=\frac15$) that can be evaluated exactly as $\frac{p}{q}$ for odd $q$.

This was a reasonable design decision to make, but maybe one day they will get around to handling some of these special cases.


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