Find real domain of a function results in $x \geq i$

Question

I have an equation of the form

$$f(x) = \sqrt{x^3 + x}$$

for which one needs to define the maximal domain, and image and domain are part of $\mathbb{R}$ (real numbers).

$$x^3 + x \geq 0 \implies x^2 \geq -1 \implies x \geq i$$

This seems a little confusing to me, since $i$ is an element of $\mathbb{C}$ and not $\mathbb{R}$.

What am I missing?

Can $x \geq i$ be a valid domain of the function given the above constraints?

Answer 1

The domain of $\sqrt{x^3+x}$ is defined for $$x^3+x=x\times(x^2+1)$$ Since $\forall x,x^2+1>0$, we need $x>0$. I.e., the domain of $f(x)$ is simply $$\{x\in\mathbb{R}:x\geq0\}$$

Answer 2

For real numbers, $x^2 + 1$ is always positive. That's not where the problem lies. The problem is the other factor in $x^3 + x = x (x^2 + 1)$.