Precalculus: Mathematics for Calculus, 7th Edition

$\left( -\infty, -2 \right] \cup \left[ 2,\infty \right)$
Using the properties of inequality, the given expression, $\dfrac{1}{2}|x|\ge1 ,$ is equivalent to \begin{array}{l}\require{cancel} 2\cdot\dfrac{1}{2}|x|\ge1\cdot2 \\\\ |x|\ge2 .\end{array} 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, $|x|\ge2 ,$ is \begin{array}{l}\require{cancel} x\ge2 ,\\\\\text{ OR }\\\\ x\le-2 .\end{array} In interval notation, the solution set is $\left( -\infty, -2 \right] \cup \left[ 2,\infty \right) .$