Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.2 New Functions From Old - Exercises Set 0.2 - Page 25: 28


$$(f+g)(x)=\frac{1+2x^2}{x+x^3}$$ $$(f-g)(x)=-\frac{1}{x+x^3}$$ $$(fg)(x)=\frac{1}{1+x^2}$$ $$(f/g)(x)=\frac{x^2}{1+x^2}$$ $$D_{f+g}=D_{f-g}=D_{fg}=\mathbb{R}- \{0 \}$$ $$D_{f/g}=\mathbb{R}- \{0 \}$$

Work Step by Step

$$(f+g)(x)=f(x)+g(x)=\frac{x}{1+x^2}+\frac{1}{x}=\frac{1+2x^2}{x+x^3}$$ $$(f-g)(x)=f(x)-g(x)=\frac{x}{1+x^2}-\frac{1}{x}=-\frac{1}{x+x^3}$$ $$(fg)(x)=f(x)g(x)=(\frac{x}{1+x^2})(\frac{1}{x})=\frac{1}{1+x^2}$$ $$(f/g)(x)=f(x)/g(x)=(\frac{x}{1+x^2})/(\frac{1}{x})=\frac{x^2}{1+x^2}$$ The domains of $f+g$, $f-g$, and $fg$ are the intersection of the domains of $f$ and $g$, and the domain of $f/g$ is the intersection of the domains of $f$ and $g$ minus those points vanishing $g(x)$. So we have $$D_{f+g}=D_{f-g}=D_{fg}=\mathbb{R} \cap (\mathbb{R}- \{0 \} ) =\mathbb{R}- \{0 \}$$ $$ D_{f/g}=(\mathbb{R} \cap (\mathbb{R}- \{0 \} )-\varnothing =\mathbb{R}- \{0 \}.$$
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