Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 489: 3


a) $\dfrac{13}{5}$; b) $\dfrac{e^{10}+1}{2e^5}$

Work Step by Step

Apply definition of hyperbolic function for $\cosh x$ a) $\cosh ({\ln{5}})=\dfrac{e^{(\ln 5)}+e^{-(\ln 5)}}{2}$ or,$=\dfrac{5+\frac{1}{5}}{2}$ or, $=\dfrac{\dfrac{26}{5}}{2}$ or, $=\dfrac{13}{5}$ b) $\cosh 5=\dfrac{e^{(5)}+e^{-(5)}}{2}$ Thus, $=\dfrac{e^5+\dfrac{1}{e^5}}{2}$ or, $=\dfrac{\dfrac{e^{10}+1}{e^5}}{2}$ $=\dfrac{e^{10}+1}{2e^5}$
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