Chapter 5.2, Problem 68AYU

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 $f^{-1}$.

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

To determine

To find:

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

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

Answer to Problem 68AYU

Solution:

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

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

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

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

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

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

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

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

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

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

$\Rightarrow$ Rewriting this, we get

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

Now,

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

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

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

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

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{\frac{11x}{2-x}}{\frac{11}{2-x}}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{11x}{11}$

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

Also,

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

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

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

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\frac{8x-12+3x+12}{x+4}}{\frac{2x+8-2x+3}{x+4}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\frac{11x}{x+4}}{\frac{11}{x+4}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{11x}{11}$

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

$\Rightarrow$ 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 $f^{-}{}^1$.

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

Answer to Problem 68AYU

Solution:

b. The domain of $f=\text{ Range of }{f}^{–1}=\left\{x\text{|}x\ne -4\right\}$.

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

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

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

The domain of $f^{-1}\left(x\right)$ is $\left\{x\text{|}x\ne \text{ 2}\right\}$.

Therefore, The domain of $f=\text{ Range of }{f}^{–1}=\left\{x\text{|}x\ne -4\right\}$.

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