Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Problems - Page 76: 6

Answer

$$ g(x)=\left|x^{2}-1\right| -\left|x^{2}-4\right| $$ $$ \left|x^{2}-1\right|=\left\{\begin{array}{ll}{x^{2}-1} & {\text { if }|x| \geq 1} \\ {1-x^{2}} & {\text { if }|x|<1}\end{array} \text { and }\left|x^{2}-4\right|=\left\{\begin{array}{ll}{x^{2}-4} & {\text { if }|x| \geq 2} \\ {4-x^{2}} & {\text { if }|x|<2}\end{array}\right.\right. $$ So for $$ 0 \leq|x|<1, g(x)=1-x^{2}-\left(4-x^{2}\right)=-3 $$ for $$ 1 \leq|x|<2, g(x)=x^{2}-1-\left(4-x^{2}\right)=2 x^{2}-5 $$ and for $$ |x| \geq 2, g(x)=x^{2}-1-\left(x^{2}-4\right)=3 $$

Work Step by Step

$$ g(x)=\left|x^{2}-1\right| -\left|x^{2}-4\right| $$ $$ \left|x^{2}-1\right|=\left\{\begin{array}{ll}{x^{2}-1} & {\text { if }|x| \geq 1} \\ {1-x^{2}} & {\text { if }|x|<1}\end{array} \text { and }\left|x^{2}-4\right|=\left\{\begin{array}{ll}{x^{2}-4} & {\text { if }|x| \geq 2} \\ {4-x^{2}} & {\text { if }|x|<2}\end{array}\right.\right. $$ So for $$ 0 \leq|x|<1, g(x)=1-x^{2}-\left(4-x^{2}\right)=-3 $$ for $$ 1 \leq|x|<2, g(x)=x^{2}-1-\left(4-x^{2}\right)=2 x^{2}-5 $$ and for $$ |x| \geq 2, g(x)=x^{2}-1-\left(x^{2}-4\right)=3 $$
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