Answer
$r\in(-\infty,-3]\cup[1,\infty)$
Work Step by Step
Use properties from the table 'Absolute Values and Intervals".
Applying Property $9$:
$|x|\geq a\quad\Leftrightarrow\quad x\geq a\ \ $or$\ \ x\leq-a, \quad $ for a = positive number
We have
$\begin{array}{ll}
\frac{r+1}{2}\geq 1 & /\times 2\\
r+1\geq 2 & /-1\\
r\geq 1 &
\end{array}$ $\ \ $
or
$\ \ $ $\begin{array}{ll}
\frac{r+1}{2}\leq-1 & /\times 2\\
r+1\leq-2 & /-1\\
r\leq-3 &
\end{array}$
In interval form, we can write this result as
$r\in(-\infty,-3]\cup[1,\infty)$
