Answer
$(-\infty,-3]\cup[3,\infty)$.
Work Step by Step
The domain requirement for $f(x)=\sqrt {x^2-9}$ is $x^2-9\geq0$ which gives $x\geq3$ or $x\leq-3$, in interval notation $(-\infty,-3]\cup[3,\infty)$.
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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: 6
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