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: 4

Answer

a) $$\frac{e^8-1}{2e^4}$$ b) $$\frac{15}{8}$$

Work Step by Step

Recall that: $$\sinh x =\frac{e^x-e^{-x}}{2}$$ a) $\sinh 4$ $=\frac{e^4-e^{-4}}{2} =\frac{e^4-\frac{1}{e^4}}{2} =\frac{\frac{e^8-1}{e^4}}{2} =\frac{e^8-1}{2e^4}$ b) $\sinh ({\ln4})$ $=\frac{e^{\ln 4}-e^{-({\ln 4})}}{2} =\frac{4-\frac{1}{4}}{2} =\frac{\frac{15}{4}}{2} =\frac{15}{8}$
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