Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 708: 34


$\sec^{-1}0$ does not exist

Work Step by Step

$y=\sec^{-1}x =$arcsec$x$ Domain: $(-\infty, -1]\cup[1, \infty)$ Range: $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ Quadrants (unit circle): I and II -------- 0 is not in the domain of $y=\sec^{-1}x,$ because 0 is not in the range of $\sec x$. (There is no y from $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi] $ such that $\displaystyle \sec y=\frac{1}{\cos y}=0) $ $\sec^{-1}0$ does not exist
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