Chapter 14, Problem 10GR

Solution

To find:The inverse $f^{-1}$ of a function $f$ and domain and range of function $f^{-1}$ .

The domain and range of $f\left(x\right)=5/x$ are $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ , and the domain and range of $f^{-1}$ are $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ respectively.

Given data: The function $f\left(x\right)=\frac{5}{x}$ .

Method/Formula used:

If $y=f\left(x\right)$ is a function then $f^{-1}\left(y\right)=x$ is inverse function. The domain of f is the values of x for which $f\left(x\right)$ is defined. The range of $f\left(x\right)$ is the domain of $f^{-1}$ . The inverse exist iff

  $f^{-1}\text{o}f=x$

Calculations:

The given function $f\left(x\right)=\frac{5}{x}$ can be written as:

  $y=\frac{5}{x}$

Domain of f :

The set of values of x for which y is defined is the domain D of function f . Since, functiony is defined for all $x\in \left(-\infty ,\infty \right)$ except when $x=0$ , therefore, the interval $\left(-\infty ,\infty \right)-\left\{0\right\}$ or $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ is the domain of the function $f\left(x\right)=5/x$ .

Range of f :

The value of function is $f\left(x\right)<0$ or $f\left(x\right)>0$ because $x=\pm \infty$ is not point of the domain. Therefore, the range of $f\left(x\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Inverse function$f^{-1}$ :

  $\begin{array}{l}y=\frac{5}{x}\\ \Rightarrow x=\frac{5}{y}\leftarrow x=f^{-1}\left(y\right)\\ \Rightarrow f^{-1}\left(y\right)=\frac{5}{y}\end{array}$

Replace y by x . Thus, the inverse function $f^{-1}\left(x\right)=\frac{5}{x}$ .

Domain of$f^{-1}$:

The function $f^{-1}\left(x\right)=\frac{5}{x}$ is defined for all values of x , except $x=0$ therefore, $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ , is the domain of $f^{-1}$ .

Range of$f^{-1}$:

The range of $f^{-1}\left(x\right)=\frac{5}{x}$ equals all real numbers that are in the domain of $f\left(x\right)$ . Therefore, $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ is the range of $f^{-1}$ .