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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 60 (page 451)

Find the inverse function $f^{- 1}$ of each function $f$ . Find the range of $f$ and the domain and range of $f^{- 1}$ .
$f (x) = 2 \cos (3 x + 2); - \frac{2}{3} 鈮� x 鈮� - \frac{2}{3} + \frac{蟺}{3}$ .

Short Answer

Expert verified

The inverse function is $f^{- 1} (x) = \frac{1}{3} [\cos^{- 1} (\frac{x}{2}) - 2]$ .

The range of the function $f (x)$ is $[- 2, 2]$ .

For the inverse function $f^{- 1} (x)$ , the domain is $[- 2, 2]$ and the range is $- \frac{2}{3} 鈮� x 鈮� - \frac{2}{3} + \frac{蟺}{3}$ .

Step by step solution

01

Step 1. Given Information.

The function is,

$f (x) = 2 \cos (3 x + 2); - \frac{2}{3} 鈮� x 鈮� - \frac{2}{3} + \frac{蟺}{3}$ .

02

Step 2. The range of the function.

$f (x) = 2 \cos (3 x + 2)$

Taking the equation as,

role="math" localid="1647627332881" $y = 2 \cos (3 x + 2) 鈬� \frac{y}{2} = \frac{2 \cos (3 x + 2)}{2} 鈬� \cos (3 x + 2) = \frac{y}{2}$

The cosine function has intervals, $- 1 鈮� \cos (x) 鈮� 1$ ,

$- 1 鈮� \cos (3 x + 2) 鈮� 1 - 1 鈮� \frac{y}{2} 鈮� 1 - 2 鈮� y 鈮� 2$

The interval notation of the range is $[- 2, 2]$ .

03

Step 3. The inverse function.

$f (x) = 2 \cos (3 x + 2)$

Writing it as,

$y = 2 \cos (3 x + 2)$

Replacing $x$ and $y$ ,

$x = 2 \cos (3 y + 2) 鈬� \cos (3 y + 2) = \frac{x}{2} 鈬� 3 y + 2 = \cos^{- 1} (\frac{x}{2}) 鈬� 3 y = \cos^{- 1} (\frac{x}{2}) - 2 鈬� y = \frac{1}{3} [\cos^{- 1} (\frac{x}{2}) - 2]$

The inverse function is $f^{- 1} (x) = \frac{1}{3} [\cos^{- 1} (\frac{x}{2}) - 2]$ .

04

Step 4. Domain and range of inverse function.

The range of $f^{- 1} (x)$ is $- \frac{2}{3} 鈮� x 鈮� - \frac{2}{3} + \frac{蟺}{3}$ .

The domain of $f^{- 1} (x)$ is $[- 2, 2]$ .

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