Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Section 1.3 - New Functions from Old Functions - 1.3 Exercises - Page 45: 66

Answer

$h$ is an odd function if $f$ is an odd function. $h$ is an even function if $f$ is an even function. Therefore, $h$ is not always an odd function.

Work Step by Step

It is given that $g(x)$ is an odd function. Then $g(-x) = -g(x)$ Suppose that $f(x)$ is an odd function. Then $f(-x) = -f(x)$ We can consider $h(-x)$: $h(-x) = f \circ g(-x)$ $= f \circ -g(x)$ $= -[f \circ g(x)]$ $= -h(x)$ $h$ is an odd function if $f$ is an odd function. Suppose that $f(x)$ is an even function. Then $f(-x) = f(x)$ We can consider $h(-x)$: $h(-x) = f \circ g(-x)$ $= f \circ -g(x)$ $= f \circ g(x)$ $= h(x)$ $h$ is an even function if $f$ is an even function. Therefore, $h$ is not always an odd function.
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