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: 4

Answer

$\frac{1}{y^{6}+1}$

Work Step by Step

$f(x) = \frac{1}{x^{3}+1}$ Evaule $f(2)$ $f(2)= \frac{1}{2^{3}+1} = \frac{1}{9}$ Evaluate $f(y^{2})$ $f(y^{2})=\frac{1}{(y^{2})^{3}+1} = \frac{1}{y^{6}+1}$
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