Chapter 1, Problem 31RE

Solution

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

The function $h\left(x\right)=5-e^x$ is bounded above in interval $\left(-\infty ,1.609\right]$ and bounded below in the interval $\left[1.609,\infty \right)$ .

Given data: The function is $h\left(x\right)=5-e^x$ .

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 $h\left(x\right)=5-e^x$ is shown in Fig. (1).

  

The function $h\left(x\right)=5-e^x$ is defined for all $x\in \left(-\infty ,\infty \right)$ .

From the graph, it is clear that in the interval $\left(-\infty ,1.609\right)$ the maximum value of the function $h\left(x\right)=5-e^x$ is 5. Therefore, the function $h\left(x\right)$ is bounded above in the interval $\left(-\infty ,1.609\right]$ .

In the interval $\left[1.609,\infty \right)$ , $0\le h\left(x\right)\le -\infty$ . Therefore, the function $h\left(x\right)$ is bounded below in the interval $\left[1.609,\infty \right)$ .

The function $h\left(x\right)=5-e^x$ is bounded above in interval $\left(-\infty ,1.609\right]$ and bounded below in the interval $\left[1.609,\infty \right)$ .