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: 43


$f ∘ G =\frac{1}{|x-2|}$ Therefore the domain is $x\ne2$

Work Step by Step

$f(x) = |x|$ $G(x) = \frac{1}{(x-2)}$ $f ∘ G = f(G(x)) = f( \frac{1}{(x-2)}) = | \frac{1}{(x-2)}|=\frac{1}{|x-2|}$ Therefore the domain is $|x-2|\ne0$ $x-2\ne 0$ $x\ne2$
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