I was wondering when simplifying an expression AND function, would I have to include the "restriction/domain" following the simplified expression and function, such as when there is a hole or something that cancels out? (See below to understand what I mean.)
Question: Simplify $\frac{|x^3 + x|}{x}$, when x<0.
$\frac{|x^3 + x|}{x}$
= |x^2 + 1|(-x) / x
$= -|x^2 +1|$
$= -x^2 - 1, x ≠ 0$
What I am confused about/what I want to know:
Do I need to include the $x ≠ 0$ with the simplified expression, or just leave it as $-x^2 - 1$?
ALSO, if I were then to restate this expression as a function of x, f(x), would I then state the $x ≠ 0$, following the simplified function? For example, if I was told to graph the simplified function, would I label my answer as $f(x) = -x^2 - 1, x ≠ 0$?
(I ask this because I remember doing something similar when dealing with finding vertical asymptotes of a function. If they cancel out, you would have to indicate something after the answer, like x can not be equal to this value)
Let's think about what the function is. The absolute value function is defined as $$|x^3+x|=\begin{cases}x^3+x,\text{ if }x^3+x\ge 0\\ -x^3-x,\text{ if }x^3+x<0. \end{cases} $$
Here, the function is defined only for $x<0$. Notice that $x^3+x=x(x^2+1)$ is zero only when $x=0$. You can check that $x^3+x$ is negative to the left of $0$, so the function is actually
$$f(x)=\frac{-x^3-x}{x}$$
on its domain. Since $0$ is not in the domain, we may assume $x\neq 0$, and so
$$f(x)=-x^2-1.$$
Just to be clear, the following statements are correct:
(1) When $x<0$, the expression $\frac{|x^3+x|}{x}$ is equal to $-x^2-1$.
(2) When $x>0$, the expression $\frac{|x^3+x|}{x}$ is equal to $x^2+1$.
(3) When $x=0$, the expression $\frac{|x^3+x|}{x}$ is undefined.
Note that an expression is not a function, so if you want to be precise, we never talk about the "domain" or "range" of an expression.
Now for a function $f(x)=\frac{|x^3+x|}{x}$, you need to know the domain of $f$ in order to make any of these statements. Generally, if the domain is not given, we assume it to be the largest possible domain which makes sense. In this case, I would assume the domain is all real numbers except $0$. Thus, we could write
$$ f(x)=\begin{cases}-x^2-1,\text{ if }x<0\\ x^2+1,\text{ if }x>0. \end{cases} $$