College Algebra (11th Edition)

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

Chapter 2 - Review Exercises - Page 279: 115


$(g\circ f)(x)=x-2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the value of the given expression, $ \left( g\circ f \right)(x) ,$ 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)(x)=g(f(x)) .\end{array} Since $ f(x)=\sqrt{x-2} ,$ the equation above becomes \begin{array}{l}\require{cancel} (g\circ f)(x)=g(\sqrt{x-2}) .\end{array} Substituting $x$ with $\sqrt{x-2}$ in $g,$ the equation above becomes \begin{array}{l}\require{cancel} (g\circ f)(x)=(\sqrt{x-2})^2 \\\\ (g\circ f)(x)=x-2 .\end{array}
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