Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - 2.6 - Combinations of Functions: Composite Functions - 2.6 Exercises - Page 221: 73a

Answer

$g(-x)=g(x)$. So, $g(x)$ is even. $h(-x)=h(x)$. So, $h(x)$ is odd.

Work Step by Step

$g(x)=\frac{1}{2}[f(x)+f(-x)]$ $g(-x)=\frac{1}{2}[f(-x)+f(-(-x))]=\frac{1}{2}[f(-x)+f(x)]=g(x)$ $h(x)=\frac{1}{2}[f(x)-f(-x)]$ $h(-x)=\frac{1}{2}[f(-x)-f(-(-x))]=\frac{1}{2}[f(-x)-f(x)]=-\frac{1}{2}[f(x)-f(-x)]=-h(x)$
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