Domain of $\log\lvert x^3+1\rvert$

Question

This is a really simple question I think, but I'm looking for justification/clarification as well. I have a function \begin{align} y\left(x\right)=\log\left|x^3+1\right|,\tag{1} \end{align} state this function's domain.

I understand that $\left|x^3+1\right|$ is $0$ when $x=-1$, but does this form a discontinuity in the domain of $y\left(x\right)$? My answer was that $x\in\left(-\infty,\infty\right)$, although we see that the function goes to $-\infty$ when \begin{align} \lim_{n\rightarrow -1^+}\log\left|x^3+1\right|=-\infty,\tag{2}\\ \lim_{n\rightarrow -1^-}\log\left|x^3+1\right|=-\infty,\tag{3} \end{align} but does this necessitate that we have a domain of \begin{align} \left(-\infty,-1\right)\cup\left(-1,\infty\right)?\tag{4} \end{align} And I apologize as I did not know which "tag" to add to this question since "domain" is not one of them.

Answer 1

In general, the domain of $y=log(g(x))$ is whenever $g(x)>0$. For your problem, $|x^3+1|>0$ whenever $x \neq -1$. Hence the domain is precisely what you've proposed as (4).

Answer 2

The domain is everywhere where the function is defined. Sometimes there are ways to extend the domain of a function, even without destroying continuity, by specifiying what the function should be for particular points where it is not yet defined, but for which the limit does exist. Now for the actual question, $\log(x)$ is only defined for $x>0$ so the domain is $\mathbb{R}\backslash \{-1\}$, as you found.