Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 265: 73

Answer

For the function $arcsec~\frac{x}{2}$: The domain is $(-\infty, -2]\cup[2,\infty)$ The range is $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$ We can see a sketch of the graph of $~~arcsec~\frac{x}{2}~~$ below. Note that $~~y = \frac{\pi}{2}~~$ is a horizontal asymptote.
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Work Step by Step

Consider the function $sec~\frac{x}{2}$: The domain is all real numbers except $\pi+2\pi~n$, where $n$ is an integer The range is $(-\infty, -1]\cup[1,\infty)$ We can consider $~~sec~\frac{x}{2}~~$ with the domain restricted to $[0,\pi)\cup (\pi, 2\pi]$ Then for the function $arcsec~\frac{x}{2}$: The domain is $(-\infty, -2]\cup[2,\infty)$ The range is $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$ We can see a sketch of the graph of $~~arcsec~\frac{x}{2}~~$ below. Note that $~~y = \frac{\pi}{2}~~$ is a horizontal asymptote.
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