Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 11: 79

Answer

Given below.

Work Step by Step

We know that an even function equates $ f\left( -x \right)=f\left( x \right)$, while an odd function satisfies $ f\left( -x \right)=-f\left( x \right)$. If $f,g$ are even functions, then the sum is defined by $(f+g) (-x)= f(-x)+g(-x)=[f(x)] + [g(x)]= (f+g) (x)$. That is, the sum is an even function. Similarly; if $f,g$ are odd functions, then the sum is defined by $(f+g) (-x)= f(-x)+g(-x)=[-f(x)] + [-g(x)]= -[f(x)+g(x)]= -(f+g) (x)$. That is, the sum is an odd function.
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