Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.1 Composite Functions - 4.1 Assess Your Understanding - Page 280: 40

Answer

(a) $ \frac{2}{3x+2}$, domain $\{x|x\ne0,-\frac{2}{3} \}$. (b) $ \frac{2x+6}{x}$, domain $\{x|x\ne-3,0 \}$. (c) $ \frac{x}{4x+9}$, domain $\{x|x\ne-3,-\frac{9}{4} \}$. (d) $ x$, domain $\{x|x\ne0 \}$.

Work Step by Step

Given $f(x)=\frac{x}{x+3}$ and $g(x)=\frac{2}{x}$, we have: (a) $f\circ g=\frac{\frac{2}{x}}{\frac{2}{x}+3}=\frac{2}{3x+2}$, domain $\{x|x\ne0,-\frac{2}{3} \}$. (b) $g\circ f=\frac{2}{\frac{x}{x+3}}=\frac{2x+6}{x}$, domain $\{x|x\ne-3,0 \}$. (c) $f\circ f=\frac{\frac{x}{x+3}}{\frac{x}{x+3}+3}=\frac{x}{4x+9}$, domain $\{x|x\ne-3,-\frac{9}{4} \}$. (d) $g\circ g=\frac{2}{\frac{2}{x}}=x$, domain $\{x|x\ne0 \}$.
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