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: 29

Answer

$f(3)=\sqrt{5} ,\\\\ f(2)=0 ,\\\\ f(1)=\text{does not exist} ,\\\\ f(-1)=\text{does not exist}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given function value in $ f(t)=\sqrt{5t-10} .$ $\bf{\text{Solution Details:}}$ If $ t=3 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{5t-10} \\\\ f(3)=\sqrt{5(3)-10} \\\\ f(3)=\sqrt{5} .\end{array} If $ t=2 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{5t-10} \\\\ f(2)=\sqrt{5(2)-10} \\\\ f(2)=0 .\end{array} If $ t=1 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{5t-10} \\\\ f(1)=\sqrt{5(1)-10} \\\\ f(1)=\sqrt{-5} \text{ (not a real number)} .\end{array} If $ t=-1 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{5t-10} \\\\ f(-1)=\sqrt{5(-1)-10} \\\\ f(1)=\sqrt{-15} \text{ (not a real number)} .\end{array} Hence, \begin{array}{l}\require{cancel} f(3)=\sqrt{5} ,\\\\ f(2)=0 ,\\\\ f(1)=\text{does not exist} ,\\\\ f(-1)=\text{does not exist} .\end{array}
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