Chapter 3.3, Problem 22PPS

To determine

To find whether the given function is linear.

Answer to Problem 22PPS

The function is not a linear function.

Explanation of Solution

Given:

A table is given as

Concept Used:

A function is linear if the rate of change is constant.

  $\text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}$

Calculation:

Consider the given coordinates in the table.

$\left(-0.2,0.7\right),\left(0,0.4\right),\left(0.2,0.1\right),\left(0.4,0.3\right)$ , and $\left(0.6,0.6\right)$

In order to evaluate the rate of change

  $\begin{array}{c}\text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{0.4-0.7}{0-\left(-0.2\right)}\\ =\frac{-0.3}{0.2}\\ =-\frac{3}{2}\\ \text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{0.1-0.4}{0.2-0}\\ =\frac{-0.3}{+0.2}\\ =-\frac{3}{2}\end{array}$

And

  $\begin{array}{c}\text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{0.3-0.1}{0.4-\left(0.2\right)}\\ =\frac{0.2}{0.2}\\ =1\\ \text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{0.6-0.3}{0.6-0.4}\\ =\frac{0.3}{+0.2}\\ =\frac{3}{2}\end{array}$

Thus, rate of change is not constant.

Therefore, the function is not alinear function.