Chapter 1.2, Problem 34E

a.

To determine

To find: the amount of interest in each year.

Answer to Problem 34E

  $\begin{array}{l}I\left(t_1\right)=\$400\\ I\left(t_2\right)=\$800\\ I\left(t_3\right)=\$1200\\ I\left(t_4\right)=\$1600\\ I\left(t_5\right)=\$2000\end{array}$

Explanation of Solution

Given information:

The equation for simple interest (I) is as follows:

  $\begin{array}{l}I=prt\\ \left(\text{where, }p\text{ is principal},r\text{ is rate of interest},\text{ and }t\text{ is time in years}\right)\end{array}$

  $\begin{array}{l}p=\$5000\\ r=8\%\\ t=5\text{ years}\end{array}$

Let the time for first five years be as follows:

  $\begin{array}{l}{t}_1=1\text{ year}\\ t_2=2\text{ year}\\ t_3=3\text{ year}\\ t_4=4\text{ year}\\ t_5=5\text{ year}\end{array}$

Calculation:

Put the values of p and r in the following equation to convert it into function of t as shown below.

  $\begin{array}{l}I\left(t\right)=prt\\ I\left(t\right)=\left(\$5000\right)\left(8\%\right)t\\ I\left(t\right)=\$400t\end{array}$

Substitute $t_1$ in the equation, to evaluate interest amount as shown below.

  $\begin{array}{c}I\left(t_1\right)=\$400t_1\\ =\$400\left(1\right)\\ =\$400\end{array}$

Substitute $t_2$ in the equation, to evaluate interest amount as shown below.

  $\begin{array}{c}I\left(t_2\right)=\$400t_2\\ =\$400\left(2\right)\\ =\$800\end{array}$

Substitute $t_3$ in the equation, to evaluate interest amount as shown below.

  $\begin{array}{c}I\left(t_3\right)=\$400t_3\\ =\$400\left(3\right)\\ =\$1200\end{array}$

Substitute $t_4$ in the equation, to evaluate interest amount as shown below.

  $\begin{array}{c}I\left(t_4\right)=\$400t_4\\ =\$400\left(4\right)\\ =\$1600\end{array}$

Substitute $t_5$ in the equation, to evaluate interest amount as shown below.

  $\begin{array}{c}I\left(t_5\right)=\$400t_5\\ =\$400\left(5\right)\\ =\$2000\end{array}$

Thus, the interest earned at $t_1,t_2,t_3,t_4,\text{and }{t}_5$ is $\$400,\$800,\$1200,\$1600$ and $\$2000$.

b.

To determine

To find: the domain and range of the relation.

Answer to Problem 34E

  $\begin{array}{l}\text{Domain}=\left\{1,2,3,4,5\right\}\\ \text{Range}=\left\{400,800,1200,1600,2000\right\}\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}I\left(t_1\right)=\$400t_1=1\text{ year}\\ I\left(t_2\right)=\$800t_2=2\text{ year}\\ I\left(t_3\right)=\$1200t_3=3\text{ year}\\ I\left(t_4\right)=\$1600t_4=4\text{ year}\\ I\left(t_5\right)=\$2000t_5=5\text{ year}\end{array}$

Let I be y and time (t) be x.

Calculation:

The relation for five (t,I) values will be as follows:

  $\left\{\left(1,400\right),\left(2,800\right),\left(3,1200\right),\left(4,1600\right),\left(5,2000\right)\right\}$

The domain is the set of all x or t values in order form the smallest to largest. Thus, the domain is

  $\left\{1,2,3,4,5\right\}$ .

The range is the set of all y or I values in order form the smallest to largest. Thus, the range is $\left\{400,800,1200,1600,2000\right\}$ .

c.

To determine

To justify: if this relation is a function or not.

Answer to Problem 34E

The relation is a function.

Explanation of Solution

Given information:

  $\left\{\left(1,400\right),\left(2,800\right),\left(3,1200\right),\left(4,1600\right),\left(5,2000\right)\right\}$

Calculation:

Since, there is only one value of y for a particular value of x. That is, no two ordered pairs have same first element. Thus, this relation is a function.