Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Section 2.3 - The Algebra of Functions - Exercise Set - Page 133: 18


The answers are $(f+g)(x)=-7x+5$ $(f−g)(x)=-x-5$ $(fg)(x)=12x^2−20x$ $(\frac{f}{g})(x)=\frac{-4x}{-3x+5}$.

Work Step by Step

Given functions are $f(x)=-4x$ and $g(x)=−3x+5$ $(f+g)(x)=f(x)+g(x)$ Substitute values. $(f+g)(x)=-4x−3x+5$ $(f+g)(x)=-7x+5$ $(f−g)(x)=f(x)−g(x)$ Substitute values. $(f−g)(x)=-4x−(−3x+5)$ $(f−g)(x)=-4x+3x-5$ $(f−g)(x)=-x-5$ $(fg)(x)=f(x) \times g(x)$ Substitute values. $(fg)(x)=(-4x)×(−3x+5)$ $(fg)(x)=12x^2−20x$ $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$ Substitute values. $(\frac{f}{g})(x)=\frac{-4x}{-3x+5}$.
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