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: 25


$H(x)=\sqrt {x^2+2}=(g \circ h)(x)$

Work Step by Step

$h(x)=x^2+2$ $g(x)=\sqrt x$ $H(x)=\sqrt {x^2+2}$. So $H(x)=\sqrt {h(x)}=g(h(x))=(g \circ h)(x)$.
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