College Algebra (10th Edition)

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

Chapter 6 - Section 6.2 - One-to-One Functions; Inverse Functions - 6.2 Assess Your Understanding - Page 419: 4

Answer

$\frac{x}{(1-x)}$

Work Step by Step

$\dfrac{\frac{1}{x}+1}{\frac{1}{x^2}-1}=\dfrac{\frac{x+1}{x}}{\frac{1-x^2}{x^2}}=\frac{x+1}{x}\frac{x^2}{1-x^2}=(x+1)\frac{x}{(1+x)(1-x)}=\frac{x}{(1-x)}$
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