Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 431: 75

Answer

See below.

Work Step by Step

$f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}$ $f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}+0$ $f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}+[\frac{f(-x)}{2}-\frac{f(-x)}{2}]$ $f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(x)}{2}+\frac{f(-x)}{2}-\frac{f(-x)}{2}$ ... addition of real numbers is commutative $f(x)=\displaystyle \frac{f(x)}{2}+\frac{f(-x)}{2}+\frac{f(x)}{2}-\frac{f(-x)}{2}$ ... and associative $f(x)=[\displaystyle \frac{f(x)}{2}+\frac{f(-x)}{2}]+[\frac{f(x)}{2}-\frac{f(-x)}{2}]$ $f(x)=\displaystyle \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$ ... which we needed to show $F_{1}(x)=\displaystyle \frac{f(x)+f(-x)}{2}\qquad $ is even because $F_{1}(-x)=\displaystyle \frac{f(-x)+f(x)}{2}=\frac{f(x)+f(-x)}{2}=F_{1}(x)$ $F_{1}(-x)=F_{1}(x)$ $F_{2}(x)=\displaystyle \frac{f(x)-f(-x)}{2}\qquad $ is odd because $F_{2}(-x)=\displaystyle \frac{f(-x)-f(x)}{2}=\frac{-[f(x)-f(-x)]}{2}=-F_{2}(x)$ $F_{2}(-x)=F_{2}(x)$
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