## Thomas' Calculus 13th Edition

$x\in(-3,-2)\cup(2,3)$
The square root function is defined for nonnegative real numbers, and is an increasing function. This means that if a and b are nonnegative, $a \lt b\Rightarrow\sqrt{a} \lt \sqrt{b}$ An alternate definition of the absolute value is $|x|=\sqrt{x^{2}}$. All parts of the compound inequality sign are nonnegative, so we may take the square root of all parts, with the direction of inequality being unchanged: $\sqrt{4} \lt \sqrt{x^{2}} \lt \sqrt{9}$ $2 \lt |x| \lt 3$ Now we have two inequalities: $\left[\begin{array}{lll} & 2 \lt |x| & \\ & \swarrow\searrow & \\ x \lt -2 & ..or.. & x \gt 2 \end{array}\right]\qquad$and$\qquad \left[\begin{array}{l} |x| \lt 3\\ -3 \lt x \lt 3\\ \end{array}\right]$ $x$ is in the intersection of $(-3,3)$ with $(-\infty,-2)\cup(2,\infty)$ From the image below, this solution set in interval form is $x\in(-3,-2)\cup(2,3)$