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: 89

Answer

$f(7)=2 ,\\\\ f(26)=3 ,\\\\ f(-9)=-2 ,\\\\ f(-65)=-4$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given function value in $ f(x)=\sqrt[3]{x+1} .$ $\bf{\text{Solution Details:}}$ If $ x=7 ,$ then \begin{array}{l}\require{cancel} f(x)=\sqrt[3]{x+1} \\\\ f(7)=\sqrt[3]{7+1} \\\\ f(7)=\sqrt[3]{8} \\\\ f(7)=2 .\end{array} If $ x=26 ,$ then \begin{array}{l}\require{cancel} f(x)=\sqrt[3]{x+1} \\\\ f(26)=\sqrt[3]{26+1} \\\\ f(26)=\sqrt[3]{27} \\\\ f(26)=3 .\end{array} If $ x=-9 ,$ then \begin{array}{l}\require{cancel} f(x)=\sqrt[3]{x+1} \\\\ f(-9)=\sqrt[3]{-9+1} \\\\ f(-9)=\sqrt[3]{-8} \\\\ f(-9)=-2 .\end{array} If $ x=-65 ,$ then \begin{array}{l}\require{cancel} f(x)=\sqrt[3]{x+1} \\\\ f(-65)=\sqrt[3]{-65+1} \\\\ f(-65)=\sqrt[3]{-64} \\\\ f(-65)=-4 .\end{array} Hence, \begin{array}{l}\require{cancel} f(7)=2 ,\\\\ f(26)=3 ,\\\\ f(-9)=-2 ,\\\\ f(-65)=-4 .\end{array}
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