Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises: 79

Answer

If $f(x)$ and $g(x)$ are both even, $f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)+g(x)$ is neither even nor odd.

Work Step by Step

If $f(x)$ and $g(x)$ are both even, $f(x)=f(-x)$ (i) $g(x)=g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)+g(-x)$ (from i) and ii)) $=S(-x)$ $\therefore f(x)+g(x)$ is even. If $f(x)$ and $g(x)$ are both odd, $-f(x)=f(-x), f(x)=-f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=-f(-x)-g(-x)$ (from i) and ii)) $=-(f(-x)+g(-x))$ $=-S(x)$ $\because S(-x)=-S(x),$ $\therefore f(x)+g(x)$ is odd. If $f(x)$ is even and $g(x)$ is odd, $f(x)=f(-x)$ (i) $-g(x)=g(-x), g(x)=-g(-x)$ (ii) Let $S(x)=f(x)+g(x)$ $=f(-x)-g(-x)$ (from i) and ii)) $S(-x)=f(-x)+g(-x)$ $\ne \pm S(x)$ $\because S(-x)\ne \pm S(x)$ $\therefore f(x)+g(x)$ is neither even nor odd.
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