Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises: 79

Answer

If $f$ and $g$ are both even functions, $f+g$ is even. If $f$ and $g$ are both odd functions, $f+g$ is odd. If $f$ is even and $g$ is odd, $f+g$ can be even, odd, or neither of them.

Work Step by Step

Assume $f$ and $g$ are both even functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x).$$Thus, $f+g$ is even. Assume $f$ and $g$ are both odd functions, so we have $$(f+g)(-x)=f(-x)+g(-x)=-f(x)+(-g(x))=-(f+g)(x).$$Thus, $f+g$ is odd. Assume $f$ is even and $g$ is odd, so we have $$(f+g)(-x)=f(-x)+g(-x)=f(x)+(-g(x))=(f-g)(x).$$In this case, $f+g$ is even,if $g=0$, and $f+g$ is odd, if $f=0$, but if $g \neq 0$ then $f+g$ is neither even nor odd (For example, take $f=x^2$ and $g=x$; then $f+g=x^2+x$ is neither even or odd since $(f+g)(-x)=x^2-x$ and $(f+g)(-x) \neq (f+g)(x)$ and $(f+g)(-x) \neq -(f+g)(x)$.
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