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 476: 32



Work Step by Step

RECALL: $(G\cdot G)(x) = G(x)\cdot G(x)$ Using the formula above gives: $(G\cdot G)(x) \\= G(x)\cdot G(x) \\=(5-x)(5-x) \\=5(5)-5(x)-x(5)-x(-x) \\=25-5x-5x+x^2 \\=x^2-10x+25$ To evaluate the given expression, substitute $6$ to $x$ in the equation above to obtain: $(G\cdot G)(x) \\= x^2-10x+25 \\=6^2-10(6)+25 \\=36-60+25 \\=-24+25 \\=1$
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