Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Review Exercises - Page 318: 67


a) $(fg)(x)=36x^3-9x^2$ b) $(fg)(-1)=-45$

Work Step by Step

a) Using $(fg)(x)=f(x)g(x),$ then with $f(x)=12x^2-3x$ and $g(x)=3x$, \begin{array}{l}\require{cancel} (fg)(x)=(12x^2-3x)(3x) .\end{array} Using $(b+c)a=ab+ac$ or the Distributive Property, the expression above is equivalent to \begin{array}{l}\require{cancel} (fg)(x)=12x^2(3x)-3x(3x) \\\\ (fg)(x)=36x^3-9x^2 .\end{array} b) Replacing $x=-1$ in $(fg)(x)=36x^3-9x^2,$ then $(fg)(-1)$ evaluates to \begin{array}{l}\require{cancel} (fg)(-1)=36(-1)^3-9(-1)^2 \\\\ (fg)(-1)=36(-1)-9(1) \\\\ (fg)(-1)=-36-9 \\\\ (fg)(-1)=-45 .\end{array}
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