Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.1 Properties of Functions - 2.1 Exercises: 23

Answer

$[3,+\infty)$

Work Step by Step

The domain is the set of all possible values of x for which a value f(x) exists. $(x-3)^{1/2}=\sqrt{x-3}$ produces a real number, that is, f(x) is defined only when the radicand $x-3 $ is non-negative. So $x-3 \geq 0$ $x \geq 3,$ In interval notation, the domain is $[3,+\infty)$
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