Chapter 1, Problem 32RE

Solution

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

The function $k\left(x\right)=1000+\frac{x}{1000}$ is bounded.

Given data: The function is $k\left(x\right)=1000+\frac{x}{1000}$ .

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 $k\left(x\right)=1000+\frac{x}{1000}$ is shown in Fig. (1).

  

The function $k\left(x\right)=1000+\frac{x}{1000}$ is defined for all $x\in \left(-\infty ,\infty \right)$ .

From the graph, it is clear that in the interval $\left(-\infty ,\infty \right)$ the function $k\left(x\right)=1000+\frac{x}{1000}$ is a constant function.

In interval $\left[0,1000\right]$ , the function $k\left(x\right)\le 1000$ , therefore function $k\left(x\right)$ is bounded above.

In interval $\left[0,-1000\right]$ , the function $k\left(x\right)\le 1000$ , therefore function $k\left(x\right)$ is bounded above.

In interval $\left[1000,\infty \right)$ , the function $1000\le k\left(x\right)$ , therefore function is $k\left(x\right)$ bounded below. Thus the function $k\left(x\right)=1000+\frac{x}{1000}$ is bounded above and bounded below.

Therefore, the function $k\left(x\right)=1000+\frac{x}{1000}$ is bounded.