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: 57

Answer

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

Work Step by Step

We are given: $f(x)= \frac{x+2}{x^{2}-1}$ The domain of a function consists of all the values that $x$ is allowed to have. In this case, we can not have division by $0$: $x^{2}-1\neq 0$ $ x^{2}\neq 1$ $ x\neq\pm 1$ Therefore, the domain is: $(-\infty,\ -1)\cup (-1,1)\cup (1,\ \infty)$
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