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: 50

Answer

$f(x) =\frac{1}{x} $

Work Step by Step

$(f ∘ g)(x) = \frac{1}{x^2+3}$ $g(x)=x^2+3$ $(f ∘ g)(x) = f(g(x)) = f(x^2+3) = \frac{1}{x^2+3}$ Therefore lets choose $f(x) =\frac{1}{x} $ $(f ∘ g)(x) = f(g(x)) = \frac{1}{g(x)} =\frac{1}{x^2+3}$
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