I know I can write the closed form of this interval using this formula:
$|x - \text{center}| ≤ \text{radius}$
if it was $[-3, 1]$ then number $-1$ would be the center and the radius is the distance from center to right and left side of the interval, which here is $2$.
then I can write it as: $|x + 1| ≤ 2$
but the problem is that it's closed on both sides. $\Rightarrow [-3, 1]$
if I write it as: $|x + 1| < 2$
then both sides will be open. $\Rightarrow (-3, 1)$
how can I show Its half-open form using an Absolute value inequality? or if it's not possible is there a way to do that using other methods?
my idea was to right $|x + 1| ≤ 2$ and add $x ≠ 1$ after it like this: $|x + 1| ≤ 2 x ≠ 1$
is this considered an Absolute value inequality?