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 477: 74


$(F\cdot G)(6)=0$ $(F\cdot G)(9)=2$

Work Step by Step

RECALL: $(F \cdot G)(x) = F(x) \cdot G(x)$ Using the rule above gives: $(F \cdot G)(6) = F(6) \cdot G(6)$ and $(F \cdot G)(9)=F(9)\cdot G(9)$ The graph shows that: $F(6)=0$; $F(9)=1$ $G(6)=3.5$; $G(9)=2$ Thus, using the values above give: $(F\cdot G)(6)=F(6) \cdot G(6)=0(3.5)=0$ $(F\cdot G)(9)=F(9) \cdot G(9) = 1(2)=2$
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