Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.3 Inverse, Exponential, and Logarithmic Functions - 1.3 Exercises - Page 36: 28


$f^{-1}(x)=\sqrt (\frac{2}{x}-1)$

Work Step by Step

$f(x)=\frac{2}{x^{2}+1}$ $y=\frac{2}{x^{2}+1}$ ${x^{2}+1}=\frac{2}{y}$ ${x^{2}}=\frac{2}{y}-1$ $x=\sqrt (\frac{2}{y}-1)$ $f^{-1}(x)=\sqrt (\frac{2}{x}-1)$
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.