Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises - Page 10: 44

Answer

$f ∘ g ∘ G =|(\frac{1}{(x-2)})^2 - 4|$ Therefore the domain is $x\ne2$

Work Step by Step

$f(x) = |x|$ $G(x) = \frac{1}{(x-2)}$ $g(x) = x^2-4$ $f ∘ g ∘ G = f(g(G(x))) = f( g(\frac{1}{(x-2)})) = f((\frac{1}{(x-2)})^2 - 4)=|(\frac{1}{(x-2)})^2 - 4|$ Therefore the domain is $x-2\ne0$ $x\ne2$
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