Chapter 1, Problem 30RE

Solution

To identify: The function as it is bounded above, bounded below, bounded or unbounded, using graphing utility.

The function $g\left(x\right)=\frac{6x}{x^2+1}$ is bounded in its domain $x\in \left(-\infty ,\infty \right)$ .

Given data: The function is $g\left(x\right)=\frac{6x}{x^2+1}$ .

Method/Formula used:

A real function $f\left(x\right)$ is said to be: (i) Bounded above, if there is a number K such that for all $x\in D\left(f\right)$ , $f\left(x\right)\le K$ ,(ii) Bounded below, if there is a number K such that for all $x\in D\left(f\right)$ , $f\left(x\right)\ge k$ , (iii) Bounded if the function is both bound above and bounded below, (iv) and not bounded if it is not bounded. If a tangent at point for finite x intersects y -axis then the function is bounded.

Calculations:

The graph of the function $g\left(x\right)=\frac{6x}{x^2+1}$ is shown in Fig. (1).

  

The function $g\left(x\right)=\frac{6x}{x^2+1}$ is defined for all $x\in \left(-\infty ,\infty \right)$ .

From the graph, it is clear that the graph of the function $g\left(x\right)=\frac{6x}{x^2+1}$ that when $x\to \infty$ , $g\left(x\right)\to 0$ , and when $x\to -\infty$ , $g\left(x\right)\to 0$ . Thus, the function $g\left(x\right)$ is bounded below between $x=-\infty$ and $x=0$

  $x\in \infty$ with its value at $x=0$ is 0. Also, the function $g\left(x\right)$ is bounded above between $x=0$ and $x=\infty$ .

The maximum value of the function in the interval $\left(-\infty ,0\right)$ is -3 at $x=-1$ and in the interval $\left(0,\infty \right)$ maximum value of the function is 3 at $x=1$ . Thus $\left|g\left(x\right)\right|\le 3,\text{for}\text{all}x\in \left(-\infty ,\infty \right)$ .

Hence, the function $g\left(x\right)=\frac{6x}{x^2+1}$ is bounded in its domain $x\in \left(-\infty ,\infty \right)$ .