Chapter 14, Problem 26CCSR

Solution

a.

To find: the inverse of the function $y=\frac{4}{x-1}$ .

The inverse of the given function $y=\frac{4}{x-1}$ is $f^{-1}\left(x\right)=\frac{4}{x}+1$ .

Given information:

The given function is $y=\frac{4}{x-1}$ .

Concept Used:

Step involved for inverse function:

• Switch $x$ and $y$ .
• Solve for $y$ .
• Replace $y$ with $f^{-1}\left(x\right)$ .

Calculation:

Switch $x$ and $y$ .

  $x=\frac{4}{y-1}$

Solve for $y$ .

  $\begin{array}{c}xy-x=4\\ xy=4+x\\ y=\frac{4+x}{x}\\ =\frac{4}{x}+1\end{array}$

Replace $y$ with $f^{-1}\left(x\right)$ .

  $f^{-1}\left(x\right)=\frac{4}{x}+1$

b.

To find: the value of $f\left(f^{-1}\left(x\right)\right)$ and $f^{-1}\left(f\left(x\right)\right)$ .

The value of $f\left(f^{-1}\left(x\right)\right)$ is $x$ .

The value of $f^{-1}\left(f\left(x\right)\right)$ is $x$ .

Given information:

The given function is $f\left(x\right)=\frac{4}{x-1}$ .

The inverse function is $f^{-1}\left(x\right)=\frac{4}{x}+1$ .

Concept Used:

The composition of functions $f\left(x\right)$ and $g\left(x\right)$ where $g\left(x\right)$ is acting first is represented by $f\left(g\left(x\right)\right)$ or $f\circ g\left(x\right)$ . It combines two or more functions to result in another function.Calculation:

First, evaluate the value of $f\left(f^{-1}\left(x\right)\right)$ .

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=f\left(\frac{4}{x}+1\right)\\ =\frac{4}{\frac{4}{x}+1-1}\\ =x\end{array}$

Now, evaluate the value of $f^{-1}\left(f\left(x\right)\right)$ .

  $\begin{array}{c}{f}^{-1}\left(f\left(x\right)\right)=f^{-1}\left(\frac{4}{x-1}\right)\\ =\frac{4}{\frac{4}{x-1}}+1\\ =x-1+1\\ =x\end{array}$

c.

To explain: the domain and range of $f$ and $f^{-1}$ relation.

The function $f^{-1}$ is the inverse of $f$ and their domain and range are reversed.

Given information:

The given function is $f\left(x\right)=\frac{4}{x-1}$ .

The inverse function is $f^{-1}\left(x\right)=\frac{4}{x}+1$ .

Concept Used:

The domain is the set of all the input values of a function and range is the possible output given by the function.

Calculation:

The vertical asymptote of the function $f\left(x\right)=\frac{4}{x-1}$ is $x=1$ .

The horizontal asymptote of the function $f\left(x\right)=\frac{4}{x-1}$ is $y=0$ .

So, the domain of the function $f\left(x\right)=\frac{4}{x-1}$ is a set of all real number except $x=1$ .

So, the range of the function $f\left(x\right)=\frac{4}{x-1}$ is a set of all real number except $y=0$ .

The vertical asymptote of the function $f^{-1}\left(x\right)=\frac{4}{x}+1$ is $x=0$ .

The horizontal asymptote of the function $f^{-1}\left(x\right)=\frac{4}{x}+1$ is $y=1$ .

So, the domain of the function $f^{-1}\left(x\right)=\frac{4}{x}+1$ is a set of all real number except $x=0$ .

So, the range of the function $f^{-1}\left(x\right)=\frac{4}{x}+1$ is a set of all real number except $y=1$ .

Therefore, the function $f^{-1}$ is the inverse of $f$ and their domain and range are reversed.