## Precalculus: Mathematics for Calculus, 7th Edition

$-5$ $,$ $-1$ $,$ $\sqrt{5}$ $,$ $3$ and $5$ satisfy the inequality
$\dfrac{1}{x}\le\dfrac{1}{2}$ $;$ $S=\{-5,-1,0,\frac{2}{3},\frac{5}{6},1,\sqrt{5},3,5\}$ $x=-5$ $\dfrac{1}{-5}\le\dfrac{1}{2}$ $-\dfrac{1}{5}\le\dfrac{1}{2}$ True $x=-1$ $\dfrac{1}{-1}\le\dfrac{1}{2}$ $-1\le\dfrac{1}{2}$ True $x=0$ $\dfrac{1}{0}\le\dfrac{1}{2}$ Left side undefined. False $x=\dfrac{2}{3}$ $\dfrac{1}{(\frac{2}{3})}\le\dfrac{1}{2}$ $\dfrac{3}{2}\le\dfrac{1}{2}$ False $x=\frac{5}{6}$ $\dfrac{1}{(\frac{5}{6})}\le\dfrac{1}{2}$ $\dfrac{6}{5}\le\dfrac{1}{2}$ False $x=1$ $\dfrac{1}{1}\le\dfrac{1}{2}$ $1\le\dfrac{1}{2}$ False $x=\sqrt{5}$ $\dfrac{1}{\sqrt{5}}\le\dfrac{1}{2}$ $0.4472\le0.5$ True $x=3$ $\dfrac{1}{3}\le\dfrac{1}{2}$ True $x=5$ $\dfrac{1}{5}\le\dfrac{1}{2}$ True