College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.6 - Applications and Models of Exponential Growth and Decay - Summary Exercises on Functions: Domains and Defining Equations - Page 463: 18

Answer

$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$

Work Step by Step

The denominator of the given function, $ f(x)=\dfrac{x^3-1}{x^2-1} ,$ cannot be $0$. Hence, \begin{array}{l}\require{cancel} x^2-1\ne0 \\\\ x^2\ne1 \\\\ x\ne\pm\sqrt{1} \\\\ x\ne\pm1 .\end{array} Hence, the domain is $ (-\infty,-1)\cup(-1,1)\cup(1,\infty) .$
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