## Precalculus: Mathematics for Calculus, 7th Edition

$\left( -\infty, -2 \right] \cup \left[ 0,\infty \right)$
For any $a\gt0$, $|x|\gt a$ implies $x\gt a \text{ OR } x\lt -a$. (The symbol $\gt$ may be replaced with $\ge$.) Using the concept above, the solutions to the given inequality, $|x+1|\ge1 ,$ is \begin{array}{l}\require{cancel} x+1\ge1 \\\\ x\ge1-1 \\\\ x\ge0 ,\\\\\text{ OR }\\\\ x+1\le-1 \\\\ x\le-1-1 \\\\ x\le-2 .\end{array} In interval notation, the solution set is $\left( -\infty, -2 \right] \cup \left[ 0,\infty \right) .$