Calculus: Early Transcendentals 8th Edition

$|1-2x^{2}|=(1-2x^{2}), x€(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ And $|1-2x^{2}|=-(1-2x^{2}), x€(-∞,-\frac{1}{\sqrt 2})∪x€( \frac{1}{\sqrt 2},∞)$
For $x =0 \ or \ x> 0,| x| = x$ For $x< 0,| x| = -x$ Now, apply above definition for the expression $|1-2x^{2}|$. $|1-2x^{2}|>0$ if $x€(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ And $|1-2x^{2}|<0$ if $x€(-∞,-\frac{1}{\sqrt 2})$$∪$$x€( \frac{1}{\sqrt 2},∞)$ Therefore, $|1-2x^{2}|=(1-2x^{2}),x€(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ And $|1-2x^{2}|=-(1-2x^{2}), x€(-∞,-\frac{1}{\sqrt 2})∪x€( \frac{1}{\sqrt 2},∞)$