College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 3 - Section 3.1 - Functions - 3.1 Assess Your Understanding: 78

Answer

$\displaystyle g(x)=\frac{x-1}{x+1}$

Work Step by Step

We are given: $f(x)=\displaystyle \frac{1}{x}$ and: $(\displaystyle \frac{f}{g})(x)=\frac{x+1}{x^{2}-x}$ Thus: $g(x)=\frac{f(x)}{(\displaystyle \frac{f}{g})(x)}$ $\displaystyle g(x)=\frac{\frac{1}{x}}{\frac{x+1}{x^{2}-x}}$ $\displaystyle g(x)=\frac{1*(x^2-x)}{x(x+1)}$ $\displaystyle g(x)=\frac{(x-1)x}{x(x+1)}$ $\displaystyle g(x)=\frac{x-1}{x+1}$
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