## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 1

#### Answer

See the details below.

#### Work Step by Step

Since we have $$\sinh x= \frac{e^x-e^{-x}}{2}, \quad \cosh x= \frac{e^x-e^{-x}}{2}\cosh x= \frac{e^x+e^{-x}}{2}$$ Then we get $$\sinh -3= \frac{e^{-3}-e^{3}}{2}=-10.067, \quad \cosh x= \frac{e^{-3}+e^{3}}{2}\cosh x= 10.067$$ $$\sinh 0= \frac{e^{0}-e^{0}}{2}=0, \quad \cosh x= \frac{e^{0}+e^{0}}{2}\cosh x= 1$$ $$\sinh 5= \frac{e^{5}-e^{-5}}{2}=74.20, \quad \cosh 5= \frac{e^{5}+e^{-5}}{2}\cosh x= 74.02$$

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