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: 32

Answer

a) $f\circ g =\dfrac{2}{2+3x}$ Domain: $\{x|x\ne-3,x\ne-\frac{3}{2},x\ne0 \}$ b) $g\circ f = \dfrac{2x+6}{x}$ Domain: $\{x|x\ne-3,x\ne0 \}$ c) $f\circ f = \dfrac{x}{4x+9}$ Domain: $\{x|x\ne-3, x=\ne-\frac{9}{4} \}$ d) $g\circ g = x$ Domain: $\{x|x\ne0\}$

Work Step by Step

a) $f\circ g = f(g(x)) = $ $\dfrac{\frac{2}{x}}{\frac{2}{x}+3}=$ $\dfrac{\frac{2}{x}}{\frac{2}{x}+\frac{3x}{x}}=$ $\dfrac{\frac{2}{x}}{\frac{2+3x}{x}}=$ $\dfrac{2(x)}{x(2+3x)}$ $\dfrac{2}{2+3x}$ Domain: $\{x|x\ne-3,x\ne-\frac{3}{2},x\ne0 \}$ b) $g\circ f = g(f(x)) =$ $\dfrac{2}{\frac{x}{x+3}}=$ $\dfrac{2(x+3)}{x}=$ $\dfrac{2x+6}{x}$ Domain: $\{x|x\ne-3,x\ne0 \}$ c) $f\circ f = f(f(x))=$ $\dfrac{\frac{x}{x+3}}{\frac{x}{x+3}+3}=$ $\dfrac{\frac{x}{x+3}}{\frac{x}{x+3}+\frac{3(x+3)}{x+3}}=$ $\dfrac{\frac{x}{x+3}}{\frac{x+3x+9}{x+3}}=$ $\dfrac{x(x+3)}{(x+3)(4x+9)}=$ $\dfrac{x}{4x+9}$ Domain: $\{x|x\ne-3, x=\ne-\frac{9}{4} \}$ d) $g\circ g = g(g(x))=$ $\dfrac{2}{\frac{2}{x}}=$ $\dfrac{2x}{2}=$ $x$ Domain: $\{x|x\ne0\}$
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