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: 33



Work Step by Step

We know that $F(x)=1/(x-3)$, so: $F(F(x))=\frac{1}{\frac{1}{x-3}-3}=\frac{1}{\frac{1}{x-3}-\frac{3(x-3)}{(x-3)}}=\frac{1}{\frac{1-3x+9}{x-3}}=\frac{1}{\frac{10-3x}{x-3}}=\frac{x-3}{10-3x}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.