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: 92

Answer

$f(0)=1 ,\\\\ f(15)=2 ,\\\\ f(-82)=\text{does not exist} ,\\\\ f(80)=3$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given function value in $ f(t)=\sqrt[4]{t+1} .$ $\bf{\text{Solution Details:}}$ If $ t=0 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt[4]{t+1} \\\\ f(0)=\sqrt[4]{0+1} \\\\ f(0)=\sqrt[4]{1} \\\\ f(0)=1 .\end{array} If $ t=15 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt[4]{t+1} \\\\ f(15)=\sqrt[4]{15+1} \\\\ f(15)=\sqrt[4]{16} \\\\ f(15)=2 .\end{array} If $ t=-82 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt[4]{t+1} \\\\ f(-82)=\sqrt[4]{-82+1} \\\\ f(-82)=\sqrt[4]{-81} \text{ (not a real number)} .\end{array} If $ t=80 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt[4]{t+1} \\\\ f(80)=\sqrt[4]{80+1} \\\\ f(80)=\sqrt[4]{81} \\\\ f(80)=3 .\end{array} Hence, \begin{array}{l}\require{cancel} f(0)=1 ,\\\\ f(15)=2 ,\\\\ f(-82)=\text{does not exist} ,\\\\ f(80)=3 .\end{array}
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