## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 2 - Section 2.1 - Functions - Exercises: 25

#### Answer

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

#### Work Step by Step

For x = 2 $g(2) = \frac{1-2}{1+2}$ ... subtract numerator and add denominator $= \frac{-1}{3}$ ____________________ For x = -1 $g(-1) = \frac{1-(-1)}{1+(-1)}$ ... subtract numerator and add denominator $= \frac{2}{0} = undefined$ ____________________ For x = $\frac{1}{2}$ $g(\frac{1}{2}) = \frac{1-\frac{1}{2}}{1+\frac{1}{2}}$ ... convert whole numbers to fractions with common denominator $= \frac{\frac{2}{2}-\frac{1}{2}}{\frac{2}{2}+\frac{1}{2}}$...subtract numerator and add denominator $= \frac{\frac{1}{2}}{\frac{3}{2}}$...multiply top and bottom by 2 $=\frac{1}{3}$ ____________________ For x = a $g(a) = \frac{1-a}{1+a}$ ____________________ For x = a - 1 $g(a-1) = \frac{1-(a-1)}{1+(a-1)}$... simplify top and bottom $=\frac{1-a+1}{1+a-1}=\frac{-a+2}{a}$ ____________________ For x = $x^2 - 1$ $g( x^2 - 1) = \frac{1-( x^2 - 1)}{1+( x^2 - 1)}$... simplify top and bottom $=\frac{1- x^2 + 1}{1+ x^2 - 1}=\frac{-x^2+2}{x^2}$ ____________________

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.