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: 20

Answer

$e^{x}$ = 5 sinh x = $\frac{12}{5}$ cosh x = $\frac{13}{5}$

Work Step by Step

We know that $\frac{1 + tanh x}{1 - tanh x}$ = $e^{2x}$ Substitute tanh x = $\frac{12}{13}$ , To get $\frac{1 + \frac{12}{13}}{1 - \frac{12}{13}}$ = $e^{2x}$ $\frac{13 + 12}{13 - 12}$ = $e^{2x}$ 25 = $e^{2x}$ $e^{x}$ = 5 Thus, we know: sinh = $\frac{e^{x}-e^{-x}}{2}$ = $\frac{5 - \frac{1}{5}}{2}$ = $\frac{\frac{25}{5}-\frac{1}{5}}{2}$ = $\frac{12}{5}$ cosh x = $\frac{e^{x} + e^{-x}}{2}$ = $\frac{5 + \frac{1}{5}}{2}$ = $\frac{\frac{25}{5} + \frac{1}{5}}{2}$ = $\frac{13}{5}$
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