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: 51

Answer

$f(x) =x^2 $

Work Step by Step

$(f ∘ g)(x) = x^4+6x^2+20=(x^2+3)^2$ $g(x)=x^2+3$ $(f ∘ g)(x) = f(g(x)) = f(x^2+3) =(x^2+3)^2$ Therefore lets choose $f(x) =x^2 $ $(f ∘ g)(x) = f(g(x)) =(g(x))^2=(x^2+3)^2$
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