University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 63



Work Step by Step

We have $$\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}$$ for $|x|\lt1$ Therefore, $$\tanh^{-1}\Big(-\frac{1}{2}\Big)=\frac{1}{2}\ln\frac{1-1/2}{1+1/2}$$ $$=\frac{1}{2}\ln (\frac{1}{2}\div\frac{3}{2})$$ $$=\frac{1}{2}\ln(\frac{1}{3})$$ $$=-\frac{\ln3}{2}$$
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