College Algebra (10th Edition)

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

Chapter 6 - Review Exercises - Page 499: 5

Answer

$f\circ g =\sqrt{3+3x+3x^2}$ Domain: $\{x|\Re\}$ $g\circ f = 1+\sqrt{3x}+x$ Domain: $\{x|x\geq0 \}$ c) $f\circ f = \sqrt[4]{3x}$ Domain: $\{x|x\geq0 \}$ d) $g\circ g =3+3x+4x^2+2x^3+x^4$ Domain: $\{x|\Re\}$

Work Step by Step

$f\circ g = f(g(x)) = $ $\sqrt{3(1+x+x^2)}=$ $\sqrt{3+3x+3x^2}$ Domain: $\{x|\Re\}$ $g\circ f = g(f(x)) =$ $1+\sqrt{3x}+\sqrt{x^2}=$ $1+\sqrt{3x}+x$ Domain: $\{x|x\geq0 \}$ c) $f\circ f = f(f(x))=$ $\sqrt{\sqrt{3x}}=$ $\sqrt[4]{3x}$ Domain: $\{x|x\geq0 \}$ d) $g\circ g = g(g(x))=$ $1+(1+x+x^2)+(1+x+x^2)^2=$ $2+x+x^2+1+x+x^2+x+x^2+x^3+x^2+x^3+x^4=$ $3+3x+4x^2+2x^3+x^4$ Domain: $\{x|\Re\}$
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