Chapter 3.3, Problem 23PPS

To determine

To find whether the given function is linear.

Answer to Problem 23PPS

The function is 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.

$\left(\frac{1}{2},\frac{1}{2}\right),\left(\frac{3}{2},1\right),\left(\frac{5}{2},\frac{3}{2}\right),\left(\frac{7}{2},2\right)$ , and $\left(\frac{9}{2},\frac{5}{2}\right)$

In order to evaluate the rate of change

  $\begin{array}{c}\text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{1-\frac{1}{2}}{\frac{3}{2}-\frac{1}{2}}\\ =\frac{\frac{1}{2}}{1}\\ =\frac{1}{2}\\ \text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{\frac{3}{2}-1}{\frac{5}{2}-\frac{3}{2}}\\ =\frac{\frac{1}{2}}{1}\\ =\frac{1}{2}\end{array}$

And

  $\begin{array}{c}\text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{2-\frac{3}{2}}{\frac{7}{2}-\frac{5}{2}}\\ =\frac{\frac{1}{2}}{1}\\ =\frac{1}{2}\\ \text{Rate of change}=\frac{\text{change in }y\text{coordinate}}{\text{change in }x\text{coordinate}}\\ =\frac{\frac{5}{2}-2}{\frac{9}{2}-\frac{7}{2}}\\ =\frac{\frac{1}{2}}{1}\\ =\frac{1}{2}\end{array}$

Thus, rate of change is constant and it is $\frac{1}{2}$ .

Therefore, the function is a linear function.