Answer
False.
Work Step by Step
This statement is not necessarily true since $g(x)= \frac{1}{\sqrt{f(x)}}$ is not defined for $x$ for which $f(x) \le 0$.
The domain of $g$ consists of all those real numbers $x$ for which $f(x)>0$.
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 0 - Before Calculus - 0.1 Functions - Exercises Set 0.1 - Page 13: 22
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