Inequality query related to absolute value function

Question

I had a silly query in Inequalities and absolute value functions.

Suppose $| f(x) | > g(x)$

When $f(x)≥0$, then $f(x)> g(x)$. Let solutions of this be $S_1$.

When $f(x)<0$, then $-f(x)> g(x)$. Let solutions of this be $S_2$.

Final answer is $S_1 \cup S_2$

My query is:
let $g(x)<0$ be $S_3$.

Obviously, my final answer should be $S_1 \cup S_2 \cup S_3$.

But I have always seen the final answer boils down to $S_1 \cup S_2$.

Why is checking for $S_3$ not important?
Or am I making a mistake somewhere?

Answer 1

Every $x \in S_3$ is either in $S_1$ or in $S_2$, depending on the sign of $f(x)$. So $S_1 \cup S_2 \cup S_3 = S_1 \cup S_2$.