Is it necessarily true that $x$ is forced to be nonnegative in the case of $\sqrt{x^3}$?

Question

I'm reading a textbook on Precalculus, and it makes the following claim:

If you have a single odd power under the square root, like $\sqrt{x^3}$, the variable is forced to be nonnegative, so absolute value is not necessary.

I'm familiar with the property that for any $a\in\mathbb R$, $\sqrt{a^2} = |a|$, but I don't see how it can even be used here to demonstrate $x$ is forced to be $\ge 0$. Can someone please explain what they're trying to demonstrate with this statement? I don't understand, because if $x=-1$, then $$\sqrt{(-1)^3} = \sqrt{-1} = i.$$

It just leaves me with a big fat "WHY?!" seeing that claim and coming up with this counterexample, but maybe I just don't understand what they're trying to claim. I see no other assumptions stated in the chapter prior to that, so I'm not sure what is even happening.

Answer 1

Because the real function $\sqrt{x}$ is generally understood as a function from $[0,+\infty)$ to $[0,+\infty)$. Hence, if $x<0$, as is $x=t^3$ for any $t<0$, then $\sqrt{x}$ is undefined.

In the context of complex analysis, this is much more nuanced. Any nonzero complex number has $n$ distinct $n$-th roots, so that in particular $x^{1/2}$ $\,''$$=$$''\,$$\sqrt{x}$ is now no longer singled-valued.

Taking your example, we can see that both $i^2=(-i)^2=-1$, so why would you prefer one over the other?

Answer 2

We want to treat $\sqrt{x^3}$ as a function

\begin{align} f:(D \subseteq \mathbb R) &\to \mathbb R \\ x &\to f(x)=\sqrt{x^3} \end{align}

and we ask what is the largest possible set that $D$ could be equal to?

If $x < 0$, then $x^3<0$ and $\sqrt{x^3}$ does not exists, or, at least, is not a real number. So we end up with

\begin{align} f:[0, \infty) &\to \mathbb R \\ x &\to \sqrt{x^3} \end{align}

Assuming that $x \ge 0$, we can simplify this to $\sqrt{x^3} = x\sqrt{x}$.

If we are talking about the function

\begin{align} g:(D \subseteq \mathbb R) &\to \mathbb R \\ x &\to g(x)=\sqrt{x^2} \end{align}

then, for all $x$, $x^2 \ge 0$ and $\sqrt{x^2}$ will always be a real number. So we end up with \begin{align} g:\mathbb R &\to \mathbb R \\ x &\to \sqrt{x^2} \end{align}

It turns out that this is true: $\sqrt{x^2} = |x|$.