Chapter 5.2, Problem 60AYU

In Problems 51-62, 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 $f^{-1}$. (c) Graph $f$, $f^{-1}$, and $y=x$ on the same coordinate axes.

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

To determine

To find:

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

Answer to Problem 60AYU

Solution:

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

Explanation of Solution

Given:

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

Calculation:

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

Taking $f^{-1}$ on both sides,

$\Rightarrow$ We get, $f^{-1}\left(f\left(\frac{4}{x+2}\right)\right)=f^{-1}\left(y\right)$,

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

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

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

$\Rightarrow 4=yx+2y$

$\Rightarrow 4-2y=yx$

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

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

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

$\Rightarrow$ Rewriting this, we get,

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

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

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

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

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

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

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

Also,

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

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

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

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

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

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

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

$\Rightarrow$ 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 60AYU

Solution:

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

The range of $f=$ Domain of $f^{–1}=\left\{x\text{|}x\ne 0\right\}$.

Explanation of Solution

Given:

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

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

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

The range of $f\text{ =}$ Domain of $f^{\text{–1}}=\left\{x\text{|}x\ne 0\right\}$.

To determine

To find:

c. Graph $f$, $f^{-1}$, and $y\text{ = }x$ on the same coordinate axes.

Answer to Problem 60AYU

Solution:

c. The graph of $f$, inverse of $f$, and the line $y\text{ = }x$ in the same coordinate axes.

Explanation of Solution

Given:

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

Calculation:

c. The graph of $f$, inverse of $f$, and the line $y\text{ = }x$ in the same coordinate axes.