Chapter 5.2, Problem 65AYU

In Problems 63-74, the function $f$ is one-to-one. (a) Find its inverse function $f^{-1}$ and check your answer. (b) Find the domain and the range of $f$ and $f^{-1}$.

$f(x)=\frac{2x}{3x-1}$

To determine

To find:

a. Find its inverse function $f^{-1}$ and check your answer.

Answer to Problem 65AYU

Solution:

a. $f^{-1}\left(x\right)=\frac{x}{3x-\text{ 2}}$

Explanation of Solution

Given:

$f(x)=\frac{2x}{3x-\text{ 1}}$

Calculation:

a. Let $f\left(x\right)=\frac{2x}{3x-\text{ 1}}=y$.

$\Rightarrow$ i.e., $\frac{2x}{3x-1}=y$

$\Rightarrow 2x=y\left(3x-1\right)$

$\Rightarrow 2x=3xy-y$

$\Rightarrow y=3xy-2x$

$\Rightarrow y=x\left(3y-2\right)$

$\Rightarrow x=\frac{y}{3y-2}$

$\Rightarrow$ Change $x$ to $y$ and $y$ to $x$.

$\Rightarrow y=\frac{x}{3x-2}$

$\Rightarrow$ Rewriting this, we get,

$\Rightarrow$ $f^{-1}\left(x\right)=\frac{x}{3x-2}$

Now,

$\Rightarrow$ $\left(f\cdot f^{-1}\right)\left(x\right)=f(f^{-1}\left(x\right))$

$\Rightarrow$ $\left(f\cdot f^{-1}\right)\left(x\right)=f\left(\frac{x}{3x-2}\right)$

$\Rightarrow \left(f\cdot f^{-1}\right)\left(x\right)=\frac{2\left(\frac{x}{3x-2}\right)}{3\left(\frac{x}{3x-2}\right)-1}$

$\Rightarrow \left(f\cdot f^{-1}\right)\left(x\right)=\frac{\frac{2x}{3x-2}}{\frac{3x-3x+2}{3x-2}}$

$\Rightarrow \left(f\cdot f^{-1}\right)\left(x\right)=\frac{\frac{2x}{3x-2}}{\frac{2}{3x-2}}$

$\Rightarrow \left(f\cdot f^{-1}\right)\left(x\right)=\frac{2x}{2}$

$\Rightarrow \left(f\cdot f^{-1}\right)\left(x\right)=x$

Also,

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)$

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=f^{-1}\left(\frac{2x}{3x-1}\right)$

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=\frac{\frac{2x}{3x-1}}{3\left(\frac{2x}{3x-1}\right)-2}$

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=\frac{\frac{2x}{3x-1}}{\frac{6x-6x+2}{3x-1}}$

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=\frac{\frac{2x}{3x-1}}{\frac{6x-6x+2}{3x-1}}$

$\Rightarrow$ $\left(f^{-1}\cdot f\right)\left(x\right)=\frac{\frac{2x}{3x-1}}{\frac{2}{3x-1}}$

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=\frac{2x}{2}$

$\Rightarrow \left(f^{-1}\cdot f\right)\left(x\right)=x$

$\Rightarrow$ Here we verified that $\left(f\cdot f^{-1}\right)\left(x\right)=x=\left(f^{-1}\cdot f\right)\left(x\right)$.

To determine

To find:

b. Find the domain and the range of $f$ and $f^{-1}$.

Answer to Problem 65AYU

Solution:

b. The domain of $f=$ Range of $f^{-1}=\left\{x\text{|}x\ne \frac{\text{1}}{3}\right\}$.

The range of $f=$ Domain of $f^{–1}=\left\{x\text{|}x\ne \frac{2}{3}\right\}$.

Explanation of Solution

Given:

$f(x)=\frac{2x}{3x-\text{ 1}}$

Calculation:

b. When working with rational functions, domain elements must not create division by zero. Any $x$ not in the domain of $f$ must be excluded. So, we know that the domain of $f\left(x\right)$ cannot contain 0.

For this, $f\left(x\right)=\frac{2x}{3x-1}$ shows us that $x=\frac{1}{3}$ would not be an acceptable domain element, since it creates a zero denominator, setting $3x-1=0$, $x=\frac{1}{3}$. Also exclude $\frac{1}{3}$ from the domain of $f\left(x\right)$. The domain of $f\left(x\right)$ is $\left\{x\text{|}x\ne \frac{1}{3}\right\}$.

For this, $f^{-1}\left(x\right)=\frac{x}{3x-2}$ shows us that $x=\frac{2}{3}$ would not be an acceptable domain element, since it creates a zero denominator, setting $3x-2=0$, $x=\frac{2}{3}$. Also exclude $\frac{2}{3}$ from the domain of $f^{-1}\left(x\right)$. The domain of $f^{-1}\left(x\right)$ is $\left\{x\text{|}x\ne \frac{2}{3}\right\}$.

Therefore, The domain of $f=$ Range of $f^{-1}=\left\{x\text{|}x\ne \frac{\text{1}}{3}\right\}$.

The range of $f=$ Domain of $f^{–1}=\left\{x\text{|}x\ne \frac{2}{3}\right\}$.