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

$g(0)=1 ,\\\\ g(-62)=5 ,\\\\ g(-13)=3 ,\\\\ g(63)=-5$
$\bf{\text{Solution Outline:}}$ Substitute the given function value in $g(x)=-\sqrt[3]{2x-1} .$ $\bf{\text{Solution Details:}}$ If $x=0 ,$ then \begin{array}{l}\require{cancel} g(x)=-\sqrt[3]{2x-1} \\\\ g(0)=-\sqrt[3]{2(0)-1} \\\\ g(0)=-\sqrt[3]{-1} \\\\ g(0)=1 .\end{array} If $x=-62 ,$ then \begin{array}{l}\require{cancel} g(x)=-\sqrt[3]{2x-1} \\\\ g(-62)=-\sqrt[3]{2(-62)-1} \\\\ g(-62)=-\sqrt[3]{-125} \\\\ g(-62)=-(-5) \\\\ g(-62)=5 .\end{array} If $x=-13 ,$ then \begin{array}{l}\require{cancel} g(x)=-\sqrt[3]{2x-1} \\\\ g(-13)=-\sqrt[3]{2(-13)-1} \\\\ g(-13)=-\sqrt[3]{-27} \\\\ g(-13)=-(-3) \\\\ g(-13)=3 .\end{array} If $x=63 ,$ then \begin{array}{l}\require{cancel} g(x)=-\sqrt[3]{2x-1} \\\\ g(63)=-\sqrt[3]{2(63)-1} \\\\ g(63)=-\sqrt[3]{125} \\\\ g(63)=-5 .\end{array} Hence, \begin{array}{l}\require{cancel} g(0)=1 ,\\\\ g(-62)=5 ,\\\\ g(-13)=3 ,\\\\ g(63)=-5 .\end{array}