Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Summary Exercises on Functions: Domains and Defining Equations - Exercises - Page 486: 7

Answer

$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.

Work Step by Step

The domain requirement for $f(x)=\frac{x^2+7}{x^2-9}$ is $x^2-9\ne0$ which gives $x\ne\pm3$, in interval notation $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$.
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