Answer
False.
Work Step by Step
$f(x)$ is not defined for real numbers x that yield 0 in the denominator.
So, f is not defined for $ x=2\quad$AND $\quad x=-2.\\\\$
The statement is false.
Calculus with Applications (10th Edition)
by Lial, Margaret L.; Greenwell, Raymond N.; Ritchey, Nathan P.
Published by
Pearson
ISBN 10:
0321749006
ISBN 13:
978-0-32174-900-0
Chapter 2 - Nonlinear Functions - Chapter Review - Review Exercises - Page 113: 8
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