Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.1 The Inverse Sine, Cosine, and Tangent Functions - 6.1 Assess Your Understanding - Page 474: 55

Answer

$f^{-1}(x)=\frac{1}{3}cos^{-1}(-\frac{x}{2})$ range of $f(x)$: $[-2,2]$, domain and range of $f^{-1}(x)$: $[-2,2]$ and $[0, \frac{\pi}{3}]$.

Work Step by Step

1. $f(x)=-2cos(3x) \Longrightarrow y=-2cos(3x) \Longrightarrow x=-2cos(3y) \Longrightarrow y=\frac{1}{3}cos^{-1}(-\frac{x}{2}) \Longrightarrow f^{-1}(x)=\frac{1}{3}cos^{-1}(-\frac{x}{2})$ 2. We can find the range of $f(x)$: $[-2,2]$, 3. We can find the domain and range of $f^{-1}(x)$: $[-2,2]$ and $[0, \frac{\pi}{3}]$.
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