Answer
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$
Work Step by Step
The domain requirement for this function is that the denominator can not be zero; we have $x^2-9\ne0$. Thus, $x\ne\pm3$ or $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 6 - Cumulative Review Exercises - Page 801: 13
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