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: 3

Answer

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

Work Step by Step

Remember that: $$\cosh x=\frac{e^x+e^{-x}}{2}$$ a) $\cosh ({\ln{5}})$ $=\frac{e^{(\ln 5)}+e^{-(\ln 5)}}{2} =\frac{5+\frac{1}{5}}{2} =\frac{\frac{26}{5}}{2} =\frac{26}{10} =\frac{13}{5}$ b) $\cosh 5$ $=\frac{e^{(5)}+e^{-(5)}}{2} =\frac{e^5+\frac{1}{e^5}}{2} =\frac{\frac{e^{10}+1}{e^5}}{2} =\frac{e^{10}+1}{2e^5}$
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