Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 2

Answer

a) $$0$$ b) $$\frac{e^2-1}{e^2+1}$$

Work Step by Step

To solve part a and b of this question, it is important to know that: $\tanh = \frac{\sinh x}{\cosh x} =\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}} =\frac{e^x-e^{-x}}{e^x+e^{-x}}$ a) $\tanh 0$ $=\frac{e^0-e^{-0}}{e^0+e^{-0}}=\frac{1-1}{1+1}=\frac{0}{2}=0$ b) $\tanh1$ $=\frac{e^1-e^{-1}}{e^1+e^{-1}}=\frac{e-\frac{1}{e}}{e+\frac{1}{e}} =\frac{e^2-1}{e^2+1}$
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.