College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Review Exercises - Page 788: 4


See below.

Work Step by Step

I plug in the respective values of $x$ into $f(x)=(-1)^{x+1}(\frac{1}{2})^x$ For $x=1$: $f(x)=(-1)^{1+1}(\frac{1}{2})^1=1\cdot\frac{1}{2}=\frac{1}{2}$ For $x=2$: $f(x)=(-1)^{2+1}(\frac{1}{2})^2=-1\cdot\frac{1}{4}=-\frac{1}{4}$ For $x=3$: $f(x)=(-1)^{3+1}(\frac{1}{2})^3=1\cdot\frac{1}{8}=\frac{1}{8}$ For $x=4$: $f(x)=(-1)^{4+1}(\frac{1}{2})^4=-1\cdot\frac{1}{16}=-\frac{1}{16}$
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