Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 1 - Functions and Their Graphs - Section 1.1 Functions - 1.1 Assess Your Understanding - Page 54: 57

Answer

The domain is {$x|x\gt9$}.

Work Step by Step

Requirement (1) For a square root to be defined, the expression inside the radial sign must be greater than or equal to $0$. The domain is defined by applying the requirement: $x-9\geq0$ $x-9+9\ge 0+9$ $x\geq 9$ Requirement (2) For a fraction to be defined, the denominator can not be equal to $0$. Thus, $\sqrt{x-9}\ne0$ $(\sqrt{x-9})^2\ne 0^2$ $x-9\ne0$ $x-9+9\ne0+9$ $x\ne9$ The two requirements can be applied together, to define the domain: $x\geq9$ and $x\ne9$. Therefore, the domain is {$x|x\gt9$}.
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