Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.1 Radical Expressions and Functions - 10.1 Exercise Set: 30

Answer

$g(-6)=\sqrt{11} ,\\\\ g(3)=\text{does not exist} ,\\\\ g(6)=\sqrt{11} ,\\\\ g(13)=12$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given function value in $ g(x)=\sqrt{x^2-25} .$ $\bf{\text{Solution Details:}}$ If $ x=-6 ,$ then \begin{array}{l}\require{cancel} g(x)=\sqrt{x^2-25} \\\\ g(-6)=\sqrt{(-6)^2-25} \\\\ g(-6)=\sqrt{36-25} \\\\ g(-6)=\sqrt{11} .\end{array} If $ x=3 ,$ then \begin{array}{l}\require{cancel} g(x)=\sqrt{x^2-25} \\\\ g(3)=\sqrt{(3)^2-25} \\\\ g(3)=\sqrt{9-25} \\\\ g(3)=\sqrt{-16} \text{ (not a real number)} .\end{array} If $ x=6 ,$ then \begin{array}{l}\require{cancel} g(x)=\sqrt{x^2-25} \\\\ g(6)=\sqrt{(6)^2-25} \\\\ g(6)=\sqrt{36-25} \\\\ g(6)=\sqrt{11} .\end{array} If $ x=13 ,$ then \begin{array}{l}\require{cancel} g(x)=\sqrt{x^2-25} \\\\ g(13)=\sqrt{(13)^2-25} \\\\ g(13)=\sqrt{169-25} \\\\ g(13)=\sqrt{144} \\\\ g(13)=12 .\end{array} Hence, \begin{array}{l}\require{cancel} g(-6)=\sqrt{11} ,\\\\ g(3)=\text{does not exist} ,\\\\ g(6)=\sqrt{11} ,\\\\ g(13)=12 .\end{array}
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