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 10 - Exponents and Radicals - 10.1 Radical Expressions and Functions - 10.1 Exercise Set - Page 635: 111



Work Step by Step

Since $(fg)(x)=f(x)g(x),$ with $f(x)=3x-1$ and $g(x)=\dfrac{1}{x},$ then \begin{array}{l}\require{cancel} (fg)(x)=f(x)g(x) \\\\ (fg)(x)=(3x-1)\left(\dfrac{1}{x}\right) \\\\ (fg)(x)=\dfrac{3x-1}{x} .\end{array}
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