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: 34



Work Step by Step

We know that $g(x)=x^3$ and $F(x)=1/(x-3)$ and $f(x)=x^2-4$, so: $g(F(f(x))=g(F(x^2-4))=g(1/(x^2-4-3))=g(1/(x^2-7))=1/(x^2-7)^3$
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