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: 80

Answer

1) If both $f$ and $g$ are even $fg$ is even; 2) If both $f$ and $g$ are odd $fg$ is even; 3) If $f$ is even and $g$ is odd $fg$ is odd.

Work Step by Step

The new function of some argument $x$ is the product: $$h(x)=f(x)g(x).$$ To say when it is odd and when it is even we have to check what happens when we substitute $x$ for $-x$. Check all the combinations: 1) $f$ is even and $g$ is even. Then $f(-x)=f(x)$ and $g(-x)=g(x)$ so we have: $$h(-x)=f(-x)g(-x)=f(x)g(x)=h(x)$$ so we have $h(-x)=h(x)$ and we see that $h$ is even. 2) $f$ is odd and $g$ is odd. Then $f(-x)=-f(x)$ and $g(-x)=-g(x)$ so we have: $$h(-x)=f(-x)g(-x)=(-f(x))\cdot(-g(x))=f(x)g(x)=h(x)$$ so we have $h(-x)=-h(x)$ and we see that $h$ is even. 3) $f$ is even and $g$ is odd. Then $f(-x)=f(x)$ and $g(-x)=-g(x)$ so we have: $$h(-x)=f(-x)g(-x)=f(x)\cdot(-g(x))=-h(x)$$ so we have $h(-x)=-h(x)$ and we see that $h$ is odd.
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