College Algebra (11th Edition)

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

Chapter 2 - Review Exercises - Page 279: 116

Answer

$(f\circ g)(-6)=\sqrt{34}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the value of the given expression, $ \left( f\circ g \right)(-6) ,$ 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} (f\circ g)(-6)=f(g(-6)) .\end{array} Substituting $x$ with $-6$ in $g(x)=x^2,$ then \begin{array}{l}\require{cancel} g(-6)=(-6)^2 \\\\ g(-6)=36 .\end{array} Using $g(-6)=36,$ the equation of the function composition above becomes \begin{array}{l}\require{cancel} (f\circ g)(-6)=f(g(-6)) \\\\ (f\circ g)(-6)=f(36) .\end{array} Substituting $x$ with $36$ in $f(x)=\sqrt{x-2},$ then \begin{array}{l}\require{cancel} f(36)=\sqrt{36-2} \\\\ f(36)=\sqrt{34} .\end{array} Using $ f(36)=\sqrt{34} ,$ the equation of the function composition above becomes \begin{array}{l}\require{cancel} (f\circ g)(-6)=f(g(-6)) \\\\ (f\circ g)(-6)=f(36) \\\\ (f\circ g)(-6)=\sqrt{34} .\end{array}
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