College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.1 - Composite Functions - 6.1 Assess Your Understanding - Page 409: 30

Answer

a) $f\circ g =\dfrac{x}{-2+3x}$ Domain: $\{x|x\ne-3,x\ne0,x\ne\frac{2}{3} \}$ b) $g\circ f = -2x-6$ Domain: $\{x|x\ne-3,x\ne0 \}$ c) $f\circ f =\dfrac{x+3}{3x+4}$ Domain: $\{x|x\ne-3,x\ne-\frac{4}{3} \}$ d) $g\circ g = x$ Domain: $\{x|x\ne0\}$

Work Step by Step

a) $f\circ g = f(g(x)) = $ $\dfrac{1}{-\frac{2}{x}+3}=$ $\dfrac{1}{\frac{-2}{x}+\frac{3x}{x}}=$ $\dfrac{1}{\frac{-2+3x}{x}}=$ $\dfrac{x}{-2+3x}$ Domain: $\{x|x\ne-3,x\ne0,x\ne\frac{2}{3} \}$ b) $g\circ f = g(f(x)) =$ $-\dfrac{2}{\frac{1}{x+3}}=$ $-\dfrac{2(x+3)}{1}=$ $-2x-6$ Domain: $\{x|x\ne-3,x\ne0 \}$ c) $f\circ f = f(f(x))=$ $\dfrac{1}{\frac{1}{x+3}+3}=$ $\dfrac{1}{\frac{1}{x+3}+\frac{3(x+3)}{x+3}}=$ $\dfrac{1}{\frac{1+3x+3}{x+3}}=$ $\dfrac{1(x+3)}{3x+4}=$ $\dfrac{x+3}{3x+4}$ Domain: $\{x|x\ne-3,x\ne-\frac{4}{3} \}$ d) $g\circ g = g(g(x))=$ $-\dfrac{2}{-\frac{2}{x}}=$ $-\dfrac{2x}{-2}=$ $x$ Domain: $\{x|x\ne0\}$
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