Chapter 2.3, Problem 52E

(a)

To determine

To show: The equation that represent salary of a person is given by $y=36,500{\left(1.035\right)}^{\mathrm{int}t}$ .

Explanation of Solution

Given information: The initial salary of a person is $36,500 and a contract promises to increase salary each year for 4 years at 3.5% rate.

Proof:

The general formula for an amount is given by,

  $A=P{\left(1+r\right)}^t$

To find the first year salary $\left(0\le t<1\right)$ , substitute 0 for t , 0.035 for r and 36500 for P in the above formula.

  $\begin{array}{c}A=36500{\left(1+0.035\right)}^0\\ =36500\end{array}$

To find the second year salary $\left(1\le t<2\right)$ , substitute 1 for t , 0.035 for r and 36500 for P in the above formula.

  $\begin{array}{c}A=36500{\left(1+0.035\right)}^1\\ =36500\left(1.035\right)\end{array}$

To find the third year salary $\left(2\le t<3\right)$ , substitute 2 for t , 0.035 for r and 36500 for P in the above formula.

  $\begin{array}{c}A=36500{\left(1+0.035\right)}^2\\ =36500{\left(1.035\right)}^2\end{array}$

To find the fourth year salary $\left(3\le t<4\right)$ , substitute 3 for t , 0.035 for r and 36500 for P in the above formula.

  $\begin{array}{c}A=36500{\left(1+0.035\right)}^3\\ =36500{\left(1.035\right)}^3\end{array}$

So, the above function corresponds to $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ .

Hence, it is proved that the Salary increment is given by $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ .

(b)

To determine

To find: The graph of the function that represents salary and the values of t for which the function is continuous.

Answer to Problem 52E

The graph of the function $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ is shown in Figure (1) and it is continuous at all points of domain $\left[0,5\right)$ except $t=1,2,3,4$ .

Explanation of Solution

Given information: The initial salary of a person is $36,500 and a contract promises to increase salary each year for 4 years at 3.5% rate.

Calculation:

To draw the graph of the function $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ , follow the steps given below.

Press “ON” button to open the calculator. Press $\boxed{\text{Y}=}$ key and enter the right hand side of the equation $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ after the symbol ${\text{Y}}_{\text{1}}$ . Enter the keystrokes as given below:

  $\boxed{36500}\boxed{(}\boxed{1.035}\boxed{)}\boxed{\wedge }\boxed{\text{MATH}}\boxed{\Rightarrow }\boxed{5}\boxed{(}\boxed{\text{X}}\boxed{)}$

Set the window at $\left[0,5\right]$ by $\left[35000,45000\right]$ . Press $\boxed{\text{GRAPH}}$ key to produce the graph of the function as shown below.

  

Figure (1)

From the graph it can be seen that the function is not continuous at points $t=1,2,3,4$ .

Therefore, the graph of the function $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ is shown in Figure (1) and it is continuous at all points of domain $\left[0,5\right)$ except $t=1,2,3,4$ .