Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 7 - Functions and Graphs - 7.4 The Algebra of Functions - 7.4 Exercise Set - Page 478: 76


$(F/G)(3)=2$ $(F/G)(7)=-\dfrac{1}{4}$

Work Step by Step

RECALL: $(F/G)(x) = \dfrac{F(x)}{G(x)}, G(x)\ne0$ Using the rule above gives: $(F/G)(3) = \dfrac{F(3)}{G(3)}$ and $(F/G)(7)=\dfrac{F(7)}{G(7)}$ The graph shows that: $F(7)=-1$; $F(3)=2$ $G(7)=4$; $G(3)=1$ Thus, using the values above give: $(F/G)(3)=\dfrac{F(3)}{G(3)}=\dfrac{2}{1}=2$ $(F/G)(7)=\dfrac{F(7)}{G(7)}=\dfrac{-1}{4}=-\dfrac{1}{4}$
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