## Thomas' Calculus 13th Edition

$x\in(-1,3)$
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 $|x|=\sqrt{x^{2}}$. Both sides of the inequality sign are nonnegative, so we may take the square root of both sides, with the direction of inequality being unchanged: $\sqrt{(x-1)^{2}} \lt \sqrt{4}$ $|x-1| \lt 2$ $-2 \lt |x-1| \lt 2$ add 1 to each side: $-1 \lt |x| \lt 3$ This solution set in interval form is $x\in(-1,3)$