Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.1 - The Algebra of Functions - Exercise Set - Page 843: 32


$f(x)=|x|$ $g(x)=x-1$

Work Step by Step

$h(x)=|x-1|$ Let $f(x)=|x|$ and $g(x)=x-1$. Then $(f \circ g)(x)=|(g(x)|=|x-1|=h(x)$.
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