College Algebra (11th Edition)

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

Chapter 2 - Review Exercises - Page 279: 105

Answer

$(f+g)(-4)=68$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the expression, $ (f+g)(-4) ,$ given that \begin{array}{l}\require{cancel} f(x)=3x^2-4 \text{ and } \\g(x)=x^2-3x-4 ,\end{array} use the definition of the appropriate function operation. Then substitute $x$ with $-4.$ $\bf{\text{Solution Details:}}$ Using $(f+g)(x)=f(x)+g(x),$ then \begin{array}{l}\require{cancel} (f+g)(x)=(3x^2-4)+(x^2-3x-4) \\\\ (f+g)(x)=3x^2-4+x^2-3x-4 \\\\ (f+g)(x)=(3x^2+x^2)-3x+(-4-4) \\\\ (f+g)(x)=4x^2-3x-8 .\end{array} Sustituting $x$ with $-4,$ then \begin{array}{l}\require{cancel} (f+g)(-4)=4(-4)^2-3(-4)-8 \\\\ (f+g)(-4)=4(16)-3(-4)-8 \\\\ (f+g)(-4)=64+12-8 \\\\ (f+g)(-4)=68 .\end{array}
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