Chapter 1, Problem 29RE

Solution

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

The function $f\left(x\right)=x+\sin x$ is an unbounded function.

Given data: The function is $f\left(x\right)=x+\sin x$

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.

Calculations:

The graph of the function $f\left(x\right)=x+\sin x$ is shown in Fig. (1).

  

The function $f\left(x\right)=x+\sin x$ is defined for all $x\in \left(-\infty ,\infty \right)$ .

From the graph, it is clear that the graph of the function $f\left(x\right)=x+\sin x$ extends along straight line $y=x$ or $f\left(x\right)=x$ on both sides of origin. Thus, there does not exist k or K such that $k\le f\left(x\right)\text{or}f\left(x\right)\le K$ in the domain $\left(-\infty ,\infty \right)$ of the function $f\left(x\right)$ . Therefore, neither the function $f\left(x\right)=x+\sin x$ is bounded below nor the bounded above.

Hence, the function $f\left(x\right)=x+\sin x$ is an unbounded function.