Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises - Page 94: 49

Answer

On the graph of $y = tan(2~sin~x)$ on the interval $[-\pi, \pi]$, we can see that the four asymptotes are approximately at the points $x = \pm 0.9$ and $x = \pm 2.24$ The four asymptotes are: $x = -\pi + sin^{-1}(\frac{\pi}{4})$ $x = -sin^{-1}(\frac{\pi}{4})$ $x = sin^{-1}(\frac{\pi}{4})$ $x = \pi - sin^{-1}(\frac{\pi}{4})$

Work Step by Step

On the graph of $y = tan(2~sin~x)$ on the interval $[-\pi, \pi]$, we can see that the four asymptotes are approximately at the points $x = \pm 0.9$ and $x = \pm 2.24$ We can find the exact equations of these asymptotes: $y = tan(2~sin~x)$ $y = \frac{sin(2~sin~x)}{cos(2~sin~x)}$ There will be an asymptote when $cos(2~sin~x) = 0$ $2~sin~x = -\frac{\pi}{2},\frac{\pi}{2}$ $sin~x = -\frac{\pi}{4},\frac{\pi}{4}$ If $sin~x = -\frac{\pi}{4}$: $x = sin^{-1}(-\frac{\pi}{4}) = -sin^{-1}(\frac{\pi}{4})$ $x = -\pi + sin^{-1}(\frac{\pi}{4})$ If $sin~x = \frac{\pi}{4}$: $x = sin^{-1}(\frac{\pi}{4})$ $x = \pi - sin^{-1}(\frac{\pi}{4})$
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