College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Review Exercises - Page 279: 117

Answer

$(g\circ f)(3)=1$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the value of the given expression, $ \left( g\circ f \right)(3) ,$ given that \begin{array}{l}\require{cancel} f(x)=\sqrt{x-2} \\g(x)=x^2 ,\end{array} use the definition of function composition. $\bf{\text{Solution Details:}}$ Using $(f\circ g)(x)=f(g(x)),$ then \begin{array}{l}\require{cancel} (g\circ f)(3)=g(f(3)) .\end{array} Substituting $x$ with $3$ in $f(x)=\sqrt{x-2},$ then \begin{array}{l}\require{cancel} f(3)=\sqrt{3-2} \\\\ f(3)=\sqrt{1} \\\\ f(3)=1 .\end{array} Using $f(3)=1,$ the equation of the function composition above becomes \begin{array}{l}\require{cancel} (g\circ f)(3)=g(f(3)) \\\\ (g\circ f)(3)=g(1) .\end{array} Substituting $x$ with $1$ in $g(x)=x^2,$ then \begin{array}{l}\require{cancel} g(1)=1^2 \\\\ g(1)=1 .\end{array} Using $g(1)=1,$ the equation of the function composition above becomes \begin{array}{l}\require{cancel} (g\circ f)(3)=g(f(3)) \\\\ (g\circ f)(3)=g(1) \\\\ (g\circ f)(3)=1 .\end{array}
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