College Algebra (11th Edition)

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

Chapter 2 - Section 2.8 - Function Operations and Composition - 2.8 Exercises - Page 266: 5

Answer

$(fg)(x)=-38$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To evaluate the given expression, $ (fg)(4) ,$ given \begin{array}{l}\require{cancel} f(x)=x^2+3 \\ g(x)=-2x+6 ,\end{array} use the definition of the appropriate function operation. Then substitute $x$ with $ 4 .$ $\bf{\text{Solution Details:}}$ Since $(fg)(x)=f(x)g(x),$ then \begin{array}{l}\require{cancel} (fg)(x)=(x^2+3)(-2x+6) \\\\ (fg)(x)=x^2(-2x)+x^2(6)+3(-2x)+3(6) \\\\ (fg)(x)=-2x^3+6x^2-6x+18 .\end{array} Substituting $x$ with $ 4 ,$ then \begin{array}{l}\require{cancel} (fg)(x)=-2(4)^3+6(4)^2-6(4)+18 \\\\ (fg)(x)=-2(64)+6(16)-6(4)+18 \\\\ (fg)(x)=-128+96-24+18 \\\\ (fg)(x)=-38 .\end{array}
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