## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 265: 69

#### Answer

For the function $sec^{-1}~x$: The domain is $(-\infty, -1]\cup[1,\infty)$ The range is $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$ We can see a sketch of the graph of $~~sec^{-1}~x~~$ below. Note there is a horizontal asymptote at $y = \frac{\pi}{2}$ #### Work Step by Step

Consider the function $sec~x$: The domain is all real numbers except $\frac{\pi}{2}+\pi~n$, where $n$ is an integer The range is $(-\infty, -1]\cup[1,\infty)$ We can consider the function $~~sec^{-1}~x~~$ as the inverse function of $~~sec~x~~$ by considering the domain of $~~sec~x~~$ restricted to $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$ Then for the function $sec^{-1}~x$: The domain is $(-\infty, -1]\cup[1,\infty)$ The range is $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$ We can see a sketch of the graph of $~~sec^{-1}~x~~$ below. Note there is a horizontal asymptote at $y = \frac{\pi}{2}$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.