Chapter 5.2, Problem 70AYU

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 $f^{-1}$.

$f(x)=\frac{-3x-4}{x-2}$

To determine

To find:

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

$f(x)=\frac{-3x-4}{x-2}$

Answer to Problem 70AYU

Solution:

a. $f^{-1}\left(x\right)=\frac{2x-4}{x+3}$

Explanation of Solution

Given:

i.e, A function $f$ is one-to-one if for all $a,b\in A$ and $a\ne b\Rightarrow f(a)\ne f(b)\Rightarrow x\ne y$.

Now, $f\left(a\right)=x\Rightarrow f^{-1}\left(f\left(a\right)\right)=f^{-1}\left(x\right)\Rightarrow a=f^{-1}\left(x\right)$.

Calculation:

a. Let $f\left(x\right)=\frac{-3x-4}{x-2}=y$.

$\Rightarrow$ i.e., $\frac{-3x-4}{x-2}=y$

$\Rightarrow -3x-4=y\left(x-2\right)$

$\Rightarrow -3x-4=xy-2y$

$\Rightarrow 2y-4=xy+3x$

$\Rightarrow 2y-4=x\left(y+3\right)$

$\Rightarrow x=\frac{2y-4}{y+3}$

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

$\Rightarrow y=\frac{2y-4}{y+3}$

$\Rightarrow$ Rewriting this, we get

$f^{-1}\left(x\right)=\frac{2x-4}{x+3}$

Now,

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

$\left(f.f^{-1}\right)\left(x\right)=f\left(\frac{2x-4}{x+3}\right)$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{-3\left(\frac{2x-4}{x+3}\right)-4}{\left(\frac{2x-4}{x+3}\right)-2}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{\frac{-6x+12-4x-12}{x+3}}{\frac{2x-4-2x-6}{x+3}}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{\frac{-10x}{x+3}}{\frac{-10}{x+3}}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{-10x}{-10}$

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

Also,

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

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=f^{-1}\left(\frac{-3x-4}{x-2}\right)$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{2\left(\frac{-3x-4}{x-2}\right)-4}{\left(\frac{-3x-4}{x-2}\right)+3}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\frac{-6x-8-4x+8}{x-2}}{\frac{-3x-4+3x-6}{x-2}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\frac{-10x}{x-2}}{\frac{-10}{x-2}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{-10x}{-10}$

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

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

To determine

To find:

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

$f(x)=\frac{-3x-4}{x-2}$

Answer to Problem 70AYU

Solution:

b. The domain of $f=\text{ Range of }{f}^{–1}=\left\{x\text{|}x\ne 2\right\}$.

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

Explanation of Solution

Given:

i.e, A function $f$ is one-to-one if for all $a,b\in A$ and $a\ne b\Rightarrow f(a)\ne f(b)\Rightarrow x\ne y$.

Now, $f\left(a\right)=x\Rightarrow f^{-1}\left(f\left(a\right)\right)=f^{-1}\left(x\right)\Rightarrow a=f^{-1}\left(x\right)$.

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 $f(x)$ cannot contain 0.

For this, $f\left(x\right)=\frac{-3x-4}{x-2}$ shows us that $x=2$ would not be an acceptable domain element, since it creates a zero denominator, setting $x-2=0$, $x=2$. Also exclude $2$ from the domain of $f(x)$. The domain of $f(x)$ is $\left\{x\text{|}x\ne 2\right\}$.

For this, $f^{-1}\left(x\right)=\frac{2x-4}{x+3}$ shows us that $x=-3$ would not be an acceptable domain element, since it creates a zero denominator, setting $x+3=0$, $x=-3$. 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 $f=\text{ Range of }{f}^{–1}=\left\{x\text{|}x\ne 2\right\}$.

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