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

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