Chapter 11, Problem 10CR

To determine

$f^{-1}$, domain and range of $f$ and the domain and the range of $f^{-1}$, where the function $f\left(x\right)=\frac{5}{x+2}$ is a one-to-one.

Answer to Problem 10CR

Solution:

1)The inverse of $f\left(x\right)=\frac{5}{x+2}$ is $f^{-1}\left(x\right)=\frac{5}{x}-2$.

2) The domain of $f\left(x\right)$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$= range of $f^{-1}\left(x\right)$ and range of $f\left(x\right)=$ $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ = domain of $f^{-1}\left(x\right)$.

Explanation of Solution

Given information:

The one-to-one function $f\left(x\right)=\frac{5}{x+2}$ .

Explanation:

Step 1: Replace $f\left(x\right)$ with $y$.

Replace $f\left(x\right)$ with $y$ in $f\left(x\right)=\frac{5}{x+2}$, get $y=\frac{5}{x+2}$.

Step 2: Interchange the variables $x$ and $y$ in $y=\frac{5}{x+2}$.

By interchanging $x$ and $y$ in $y=\frac{5}{x+2}$

$x=\frac{5}{y+2}$.

Step 3: Solve for $y$ to obtain explicit form of $f^{-1}$.

Multiplying both sides by $y+2$,

$\Rightarrow x\left(y+2\right)=5$

Distributing left side,

$\Rightarrow xy+2x=5$

Subtracting $2x$ from both sides,

$\Rightarrow xy=5-2x$

Dividing both sides by $x$,

$\Rightarrow y=\frac{5-2x}{x}$

$\Rightarrow y=\frac{5}{x}-2$

Replace $y$ by $f^{-1}\left(x\right)$,

$\Rightarrow f^{-1}\left(x\right)=\frac{5}{x}-2$

Therefore, the inverse of $f\left(x\right)=\frac{5}{x+2}$ is $f^{-1}\left(x\right)=\frac{5}{x}-2$.

The function $f\left(x\right)=\frac{5}{x+2}$ is a rational function.

The rational function is undefined when its denominator is zero.

Hence the function is undefined when $x+2=0$, that is $x=-2$.

The domain of the function is set of all real numbers except $x=-2$.

That is domain of $f\left(x\right)$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$.

By using domain of $f\left(x\right)$ is range of $f^{-1}\left(x\right)$,

Range of $f^{-1}\left(x\right)$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$.

The function $f^{-1}\left(x\right)=\frac{5}{x}-2$ is a rational function.

The rational function is undefined where its denominator is zero.

Hence, $f^{-1}\left(x\right)$ is undefined when $x=0$.

The domain of $f^{-1}\left(x\right)$ is set of all real numbers except $x=0$.

That is domain of $f^{-1}\left(x\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$.

By using range of $f\left(x\right)$ is equal to domain of $f^{-1}\left(x\right)$,

The range of $f\left(x\right)=x^{\frac{2}{3}}-4$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$.

Hence, domain of $f\left(x\right)$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$= range of $f^{-1}\left(x\right)$ and range of $f\left(x\right)=$ $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ = domain of $f^{-1}\left(x\right)$.