Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX A - Numbers, Inequalities, and Absolute Values - A Exercises: 12

Answer

$|1-2x^{2}|=(1-2x^{2}), x€(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ And $|1-2x^{2}|=-(1-2x^{2}), x€(-∞,-\frac{1}{\sqrt 2})∪x€( \frac{1}{\sqrt 2},∞)$

Work Step by Step

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 $|1-2x^{2}|=-(1-2x^{2}), x€(-∞,-\frac{1}{\sqrt 2})∪x€( \frac{1}{\sqrt 2},∞)$
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