Chapter 7, Problem 36RE

To determine

To calculate: The inverseof the function $f\left(x\right)=2\sin \left(-x+1\right)$ and domain of $f^{-1}$ and range of $f$ .

Answer to Problem 36RE

The inverse ofthe function $f\left(x\right)=2\sin \left(-x+1\right)$ is $f^{-1}\left(y\right)=1-\left({\sin }^{-1}\left(\frac{y}{2}\right)\right)$ . Domain of $f^{-1}$ and range of $f$ is $\left[-2,2\right]$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=2\sin \left(-x+1\right)$ and its domain given as,

  $1-\frac{\pi }{2}\le x\le 1+\frac{\pi }{2}$

Formula used:

When $y=f\left(x\right)$ , $f^{-1}$ implicitly defined as,

  $x=f^{-1}\left(y\right)$

Domain of $f^{-1}$ and range of $f$ are equal.

Range of the function $f$ in the form $f\left(x\right)=k\sin \theta$ is $\left[-k,k\right]$ .

Where, $k\in Z$ and $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ .

Calculation:

Consider the function $f\left(x\right)=2\sin \left(-x+1\right)$ ,

Let y is an element in the range of $f\left(x\right).$ That is

  $y=2\sin \left(-x+1\right)$

Recall that when $y=f\left(x\right)$ , $f^{-1}$ implicitly defined as,

  $x=f^{-1}\left(y\right)$

Apply it,

  $\begin{array}{c}2\sin \left(-x+1\right)=y\\ \sin \left(-x+1\right)=\frac{y}{2}\\ -x+1={\sin }^{-1}\left(\frac{y}{2}\right)\\ x=1-\left({\sin }^{-1}\left(\frac{y}{2}\right)\right)\end{array}$

Therefore, $f^{-1}$ is defined as $f^{-1}\left(y\right)=1-\left({\sin }^{-1}\left(\frac{y}{2}\right)\right)$ .

Observe $f\left(x\right)=2\sin \left(-x+1\right)$ is in the form of $f\left(x\right)=k\sin \theta$ and $2\in Z$ and $-\frac{\pi }{2}\le -x+1\le \frac{\pi }{2}$ .

Recall that, the range of the function $f$ in the form $f\left(x\right)=k\sin \theta$ is $\left[-k,k\right]$ .

Where, $k\in Z$ and $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

Therefore, range of the function $f\left(x\right)=2\sin \left(-x+1\right)$ is $\left[-2,2\right]$ .

Again recall that domain of $f^{-1}$ and range of $f$ are equal.

Thus we get,

Range of $f=$ Domain of $f^{-1}=$$\left[-2,2\right]$ .