University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 47


$sec^{-1}{0}$ is undefined. $\sin^{-1}{\sqrt {2}}$ is undefined.

Work Step by Step

a) Since $\sec{x}={\frac{1}{cos{x}}}$, $\sec x$ can not be zero and thus $sec^{-1}{0}$ is undefined. b) The value of $\sin^{-1}{\sqrt {2}}$ is not defined as the value of $\sqrt{2}=1.414$, which is greater than 1 and the range of sin is between -1 and 1.
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