Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises - Page 9: 7

Answer

$f(g(2))=2$ $g(f(-2)=-2$

Work Step by Step

We know that $g(2)=-2$ and $f(2)=f(-2)$ (because f is even): $f(g(2)=f(-2))=f(2)=2$ $g(f(-2)=g(f(2))=g(2)=-2$
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