Chapter 1, Problem 27RE

Solution

To find: The interval in which the function increases, using graphing utility.

The required intervals is $\left(-\infty ,1\right)\cup \left(1,\infty \right)$ .

Given data: The given function is $y=\frac{x}{1-x^2}$ .

Method/Formula used:

An interval $\left(a,b\right)\text{or}\left[a,b\right],\text{or}\left[a,b\right)\text{or}\left(a,b\right]$ of the domain of function $f\left(x\right)$ is said to be interval of increasing, if for two numbers $x_1,x_2\in \left(a,b\right)\text{or}\left[a,b\right],\text{or}\left[a,b\right)\text{or}\left(a,b\right]$ , $f\left(x_1\right)<f\left(x_2\right)$ , when $x_1<x_2$ .

Calculation:

Let $y=\frac{x}{1-x^2}$ or $f\left(x\right)=\frac{x}{1-x^2}$ . The graph of the function $f\left(x\right)=\frac{x}{1-x^2}$ is shown in Fig. (1).

  

From the graph, it is clear that when $x\to -\infty$ , $f\left(x\right)\to 0$ , and when $x\to \infty$ , $f\left(x\right)\to 0$ , therefore minimum value of the function $f\left(x\right)$ occurs at $x=\pm \infty$ .

Also, when $x\to -1$ , $f\left(x\right)\to \pm \infty$ , and when $x\to 1$

  $f\left(x\right)\to \pm \infty$ , that shows , the function $f\left(x\right)$ is minimum at $x=\pm \infty$ and maximum at $x=\pm 1$ .

Therefore, the function $f\left(x\right)=\frac{x}{1-x^2}$ is an increasing function in the interval

  $\left(-\infty ,1\right)\cup \left(1,\infty \right)$ ,