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 - Page 264: 26

Answer

$cosh^{-1}~x = ln(x+\sqrt{x^2-1})~~~~~~~$ for $~~x \geq 1$

Work Step by Step

Let $y = cosh^{-1}~x$ Then: $x = cosh~y = \frac{e^y+e^{-y}}{2}$ $e^y-2x+e^{-y} = 0$ $e^{2y}-2xe^y+1 = 0$ We can use the quadratic formula: $e^y = \frac{2x\pm\sqrt{4x^2-4}}{2}$ $e^y = x\pm\sqrt{x^2-1}$ This will lead to two solutions, but by convention, we can use $x+\sqrt{x^2-1}$ as the solution. $e^y = x+\sqrt{x^2-1}$ $y = ln(x+\sqrt{x^2-1})$ $cosh^{-1}~x = ln(x+\sqrt{x^2-1})$ Note that $x\geq 1$ because of the term $\sqrt{x^2-1}$ and because it is required that $x+\sqrt{x^2-1} \gt 0$
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