Answer
$(0, \infty)$
Work Step by Step
Since $\sqrt{y}$ is not defined for $ y\lt 0$ and $ y^{-1}$ is not defined when $ y=0$, then the domain of $ g $ is $(0, \infty)$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 1 - Precalculus Review - 1.3 The Basic Classes of Functions - Exercises - Page 22: 11
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