Chapter 5.2, Problem 66AYU

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 $f^{-1}$.

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

To determine

To find:

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

Answer to Problem 66AYU

Solution:

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

Explanation of Solution

Given:

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

Calculation:

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

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

$\Rightarrow 3x+1=-yx$

$\Rightarrow 3x+xy=-1$

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

$\Rightarrow x=\frac{-1}{3+y}$

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

$\Rightarrow y=\frac{-1}{3+x}$

$\Rightarrow$ Rewriting this, we get,

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

Now,

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

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

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

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

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

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

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

Also,

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

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

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

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

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

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

$\Rightarrow$ 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 66AYU

Solution:

b. The domain of $f=$ Range of $f^{-1}=\left\{x\text{|}x\ne 0\right\}$.

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

Explanation of Solution

Given:

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

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

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

The domain of $f^{-1}\left(x\right)$ is $\left\{x\text{|}x\ne -3\right\}$.

Therefore, The domain of $f=$ Range of $f^{-1}=\left\{x\text{|}x\ne 0\right\}$.

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