Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 266: 6

Answer

(a) $sinh~1 = \frac{e^2-1}{2e}$ (b) $sinh^{-1}~1 = ln(1+\sqrt{2}) = 0.8813$

Work Step by Step

(a) $sinh~x = \frac{e^x-e^{-x}}{2}$ $sinh~1 = \frac{e^1-e^{-1}}{2}$ $sinh~1 = \frac{e-\frac{1}{e}}{2}$ $sinh~1 = \frac{\frac{e^2-1}{e}}{2}$ $sinh~1 = \frac{e^2-1}{2e}$ (b) $sinh^{-1}~x = ln(x+\sqrt{x^2+1})$ $sinh^{-1}~1 = ln(1+\sqrt{1^2+1})$ $sinh^{-1}~1 = ln(1+\sqrt{2})$ $sinh^{-1}~1 = 0.8813$
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