Answer
$\left( -\infty, -2 \right] \cup \left[ 2,\infty \right)$
Work Step by Step
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)
.$
