Chapter 1, Problem 25RE

Solution

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

The required intervals is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Given data: The given function is $f\left(x\right)=x^3/6$ .

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 $x_1,x_2\in \left(a,b\right)\text{or}\left[a,b\right],\text{or}\left[a,b\right)\text{or}\left(a,b\right]$ , $\left|f\left(x_1\right)\right|<\left|f\left(x_2\right)\right|$ when $x_1<x_2$ .

Calculation:

The graph of the function $f\left(x\right)=x^3/6$ is shown in Fig. (1

Consider two values of x , $x_1,x_2\in \left(0,\infty \right)$ such that $x_1<x_2$ , say $x_1=1,x_2=2$ , then $\begin{array}{c}f\left(x_1=1\right)={\left(1\right)}^3/6\\ =1/6\\ =0.167\end{array}$

And

  $\begin{array}{c}f\left(x_1=2\right)={\left(2\right)}^3/6\\ =8/6\\ =1.33\end{array}$

This shows that for $x_1<x_2\in \left(0,\infty \right)$ , $\left|f\left(x_1\right)\right|<\left|f\left(x_2\right)\right|$ , therefore, the function $f\left(x\right)=x^3/6$ is increasing in the interval.

Consider two values of x , $x_1,x_2\in \left(-\infty ,0\right)$ such that $x_1<x_2$ , say $x_1=-1,x_2=-2$ , then $\begin{array}{c}f\left(x_1=-1\right)={\left(-1\right)}^3/6\\ =-1/6\\ =-0.167\end{array}$

And

  $\begin{array}{c}f\left(x_1=-2\right)={\left(-2\right)}^3/6\\ =-8/6\\ =-1.33\end{array}$

  $-2<-1\Rightarrow \left\{f\left(-2\right)=-1.33\right\}<\left\{f\left(-1\right)=-0.167\right\}$

This shows that for $x_1<x_2\in \left(0,\infty \right)$ , $\left|f\left(x_1\right)\right|<\left|f\left(x_2\right)\right|$ , therefore, the function $f\left(x\right)=x^3/6$ is increasing in the interval $x_1,x_2\in \left(-\infty ,0\right)$ .

Note: Increase or decrease of a function is observed with respect to x -axis.

At $x=0,f\left(x\right)=0$ , the function $f\left(x\right)=x^3/6$ is neither increasing nor decreasing.

Thus, the function $f\left(x\right)=x^3/6$ is increasing in the interval $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .