How to determine different absolute value equation cases?

Question

This is a question from this post.

From: $$ |3x|=\left\{ \begin{align} 3x & \text{ , if }x\geq 0 \\ -3x & \text{ , if }x <0 \end{align} \right\} $$ $$ |4x+1|=\left\{ \begin{align} 4x+1 & \text{ , if }x\geq \frac{-1}{4} \\ -(4x+1) & \text{ , if }x <\frac{-1}{4} \end{align} \right\} $$

How do you determine the different cases to check?

How do you get these as the cases: $x<\frac{-1}{4}$, $\frac{-1}{4}\leq x <0$, and $x\geq 0$?

Answer 1

If you want to solve $|4x+1|\lt |3x|$ in the link, then note that $$4x+1=0\iff x=-\frac 14,\ \ \ \ 3x=0\iff x=0.$$

Then, we can separate it into three cases :

For $x\lt -\frac 14$, then $|4x+1|=-(4x+1)$ and $|x|=-x$.

For $-\frac 14\le x\lt 0$, then $|4x+1|=4x+1$ and $|x|=-x$.

For $x\ge 0$, then $|4x+1|=4x+1$ and $|x|=x$.

Answer 2

Hint: Consider that:$$ x<−\frac{1}{4} \iff 4x< -1 \iff 4x +1 < 0$$ $$ x\ge-\frac{1}{4} \iff 4x\ge -1 \iff 4x +1 \ge 0$$

Answer 3

Well, we know that $$ |x|=\left\{ \begin{align} x & \text{ , if }x\geq 0 \\ -x & \text{ , if }x <0 \end{align} \right\}.$$

Then, it must also be true that $$ |f(x)|=\left\{ \begin{align} f(x) & \text{ , if }f(x) \geq 0 \\ -f(x) & \text{ , if }f(x) <0 \end{align} \right\}.$$

Substituting in for $f(x)$ and solving for the variable $x$ produces the desired result.