## Precalculus: Mathematics for Calculus, 7th Edition

$\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]$
For any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$. (The symbol $\lt$ may be replaced with $\le$.) Using the concept above, the solutions to the given inequality, $|16x|\le8 ,$ is \begin{array}{l}\require{cancel} -8\le 16x\le8 \\\\ -\dfrac{8}{16}\le x\le\dfrac{8}{16} \\\\ -\dfrac{1}{2}\le x\le\dfrac{1}{2} .\end{array} In interval notation, the solution set is $\left[ -\dfrac{1}{2},\dfrac{1}{2} \right] .$