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

$g(19)=2 ,\\\\ g(-13)=\text{does not exist} ,\\\\ g(1)=\text{does not exist} ,\\\\ g(84)=3$
$\bf{\text{Solution Outline:}}$ Substitute the given function value in $g(t)=\sqrt[4]{t-3} .$ $\bf{\text{Solution Details:}}$ If $t=19 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(19)=\sqrt[4]{19-3} \\\\ g(19)=\sqrt[4]{16} \\\\ g(19)=2 .\end{array} If $t=-13 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(-13)=\sqrt[4]{-13-3} \\\\ g(-13)=\sqrt[4]{-16} \text{ (not a real number)} .\end{array} If $t=1 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(1)=\sqrt[4]{1-3} \\\\ g(1)=\sqrt[4]{-2} \text{ (not a real number)} .\end{array} If $t=84 ,$ then \begin{array}{l}\require{cancel} g(t)=\sqrt[4]{t-3} \\\\ g(84)=\sqrt[4]{84-3} \\\\ g(84)=\sqrt[4]{81} \\\\ g(84)=3 .\end{array} Hence, \begin{array}{l}\require{cancel} g(19)=2 ,\\\\ g(-13)=\text{does not exist} ,\\\\ g(1)=\text{does not exist} ,\\\\ g(84)=3 .\end{array}