Chapter 1, Problem 17RE

Solution

To find: The domain and range of a function.

The domain and the rage of the function $f\left(x\right)=\frac{x}{x^2-2x}$ are $\left(-\infty ,0\right)\cup \left(0,2\right)\cup \left(2,\infty \right)$ and $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ respectively.

Given information: The function is $f\left(x\right)=\frac{x}{x^2-2x}$ .

Method/Formula used:

Suppose $y=f\left(x\right)$ be a function $f:A\to B$ , then the set A is called the domain of function f and set B the co-domain. The range of the function is the set of points which are images of points of domain of function f . Thus, the range is the domain of the function $f^{-1}\left(x\right)$ . The figure below explains domain, co-domain and the range of f .

  

Calculations:

The given function $f:A\to B$ is $f\left(x\right)=\frac{x}{x^2-2x}$ .

(a) Domain:

Clearly, the function $f\left(x\right)=\frac{x}{x^2-2x}$ is defined for all real numbers x except when $x=0\text{and}x=2$ , therefore, the set $\left(-\infty ,\infty \right)-\left\{0,2\right\}$ or $\left(-\infty ,0\right)\cup \left(0,2\right)\cup \left(2,\infty \right)$ is the domain of function $f\left(x\right)=\frac{x}{x^2-2x}$ . This domain is shown in Fig. (1).

  

(b) Range:

Suppose $y=f\left(x\right)$ , then the set of all y for which there exists an ‘ x’ in the domain of the function of $f\left(x\right)$ or set of points y for which the inverse function $f^{-1}\left(y\right)=x$ is defined, is the range of $f\left(x\right)$ .

Now let, $y=\frac{x}{x\left(x-2\right)}$ . That implies $\frac{1}{y}=x-2$ or $x=\frac{1}{y}+2$ . Therefore,

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

The function $f^{-1}\left(y\right)$ is not defined for $y=0$ , therefore, 0 is not the image of any x . Hence, the range of function $f\left(x\right)$ is all real numbers except 0, i.e., the range of the function $f\left(x\right)$ is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ . This range is shown in Fig. (2).