College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 192: 25


$g(2)=-\frac{1}{3}$ $g(-1)=$ undefined $g( \frac{1}{2})=\frac{1}{3}$ $g(a)=\frac{1-a}{1+a}$ $g(a-1)=\frac{2-a}{a}$ $g(x^{2}-1)=\frac{2-x^{2}}{x^{2}}$

Work Step by Step

We are given: $g(x)= \frac{1-x}{1+x}$ We evaluate: $g(2)=\frac{1-(2)}{1+(2)}=\frac{-1}{3}=-\frac{1}{3}$ $g(-1)=\frac{1-(-1)}{1+(-1)}=\frac{2}{0}=$ undefined $g( \frac{1}{2})=\frac{1-(\frac{1}{2})}{1+(\frac{1}{2})}=\frac{\frac{2}{2}-\frac{1}{2}}{\frac{2}{2}+\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{3}$ $g(a)=\frac{1-(a)}{1+(a)}=\frac{1-a}{1+a}$ $g(a-1)=\frac{1-(a-1)}{1+(a-1)}=\frac{1-a+1}{1+a-1}=\frac{2-a}{a}$ $g(x^{2}-1)=\frac{1-(x^{2}-1)}{1+(x^{2}-1)}=\frac{2-x^{2}}{x^{2}}$
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