Chapter 7, Problem 34RE

To determine

To calculate: The inverse of the function $f\left(x\right)=\tan \left(2x+3\right)-1$ and domain of $f^{-1}$ and range of $f$ .

Answer to Problem 34RE

The inverse ofthe function $f\left(x\right)=\tan \left(2x+3\right)-1$ is $f^{-1}(y)=\frac{1}{2}\left[\left({\tan }^{-1}\left(y+1\right)\right)-3\right]$ . Domain of $f^{-1}$ and range of $f$ is $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\tan \left(2x+3\right)-1$ and its domain given as,

  $-\frac{3}{2}-\frac{\pi }{4}\le x\le -\frac{3}{2}+\frac{\pi }{4}$

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 $f\left(x\right)=\tan \theta +k$ is $\left(-\infty ,\infty \right)$ .

Where, $k$ is an integer and $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ .

Calculation:

Consider the function $f\left(x\right)=\tan \left(2x+3\right)-1$ ,

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

  $y=\tan \left(2x+3\right)-1$

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

  $x=f^{-1}\left(y\right)$

Apply it,

  $\begin{array}{c}\tan \left(2x+3\right)-1=y\\ \tan \left(2x+3\right)=y+1\\ 2x+3={\tan }^{-1}\left(y+1\right)\\ x=\frac{1}{2}\left[{\tan }^{-1}\left(y+1\right)-3\right]\end{array}$

Therefore, $f^{-1}$ is defined as $f^{-1}(y)=\frac{1}{2}\left[\left({\tan }^{-1}\left(y+1\right)\right)-3\right]$ .

Observe $f\left(x\right)=\tan \left(2x+3\right)-1$ is in the form of $f\left(x\right)=\tan \theta +k$ and $-1$ is an integer .

Also, $-\frac{\pi }{2}\le 2x+3\le \frac{\pi }{2}$ .

Recall that range of the function $f$ in the form $f\left(x\right)=\tan \theta +k$ is $\left(-\infty ,\infty \right)$ .

Where, $k$ is an integer.

Therefore, range of the function $f\left(x\right)=\tan \left(2x+3\right)-1$ is $\left(-\infty ,\infty \right)$ .

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

Thus we get,

Range of $f=$ Domain of $f^{-1}=$$\left(-\infty ,\infty \right)$ .