Chapter 6.3, Problem 19E

To determine

To find: the nature of the graph formed by the points, $(t,d)=\left(t,f(t)\right)=\left\{\left(1,0.7\right),\left(2,1.4\right),(3,2.1),(4,2.1),(5,2.8)\right\}$ . And verify whether it represent linear or exponential function.

Answer to Problem 19E

the data cannot be represent by a linear nor a exponential function.

  

Explanation of Solution

Given information :

Given points $(t,d)=\left(t,f(t)\right)=\left\{\left(1,0.7\right),\left(2,1.4\right),(3,2.1),(4,2.1),(5,2.8)\right\}$ , where $t$ represents the independent variable time and $d=f(t)$ is the distance covered by bus.

Calculation:

Given points of the graph are, $(t,d)=\left(t,f(t)\right)=\left\{\left(1,0.7\right),\left(2,1.4\right),(3,2.1),(4,2.1),(5,2.8)\right\}$ .

The common ratios are,

  $\frac{1.4}{0.7}=2,\frac{2.1}{1.4}=1.5,\frac{2.1}{2.1}=1,\frac{2.8}{2.1}=1.3$

So there is no fixed common ratio for each term. Hence the graph is not representing exponential function.

Therefore the difference is,

  $1.4-0.7=0.7,2.1-1.4=0.7,2.1-2.1=0,2.8-2.1=0.7$

So the common difference between CD is different from all other only. So if the slope of CD is equal as any one of the slope of AB, BC and DE, then it will ;lie on same line.

  $\begin{array}{l}\text{Slope of }AB=\frac{1.4-0.7}{2-1}=0.7\\ \text{Slope of }CD=\frac{2.1-2.1}{4-3}=0\end{array}$

Since the slope are not same therefore all points are not in same line and the graph is not a linear function.

Hence the data cannot be represent by a linear nor a exponential function.

Graph: