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: 54

Answer

$f(x) =(x-3)^\frac{1}{3} +3$

Work Step by Step

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