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: 1


a) 0 b) 1

Work Step by Step

a) Since, $\sinh(x)$ = $\dfrac{e^x-e^-x}{2}$ Thus, $\sinh(0)=\dfrac{(e^0)-(e^-0)}{2}=\dfrac{1-1}{2}=0$ b) Since, $\cosh(x)$ = $\dfrac{e^x+e^{-x}}{2}$ Thus, $\cosh(0)$= $\dfrac{(e^0)+(e^-0)}{2}$=$\dfrac{1+1}{2}=1$
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