Chapter 1.3, Problem 52E

To determine

To show : the salary equation is given by $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ where $t$ is the time in years.

Explanation of Solution

Given information :

The initial salary is $\$36500$ and increase each year by $3.5\%$ for $4$ years.

Formula needed :

Salary increases per year and function is given by $y=initial{\left(1+rate\right)}^{time}$ where, rate in decimal and time in years.

Proof :

From initial to first year, the salary is $36500{\left(1.035\right)}^0$ .

From first to second year, the salary is $36500{\left(1.035\right)}^1$ .

From second to third year, the salary is $36500{\left(1.035\right)}^2$ .

From third to fourth year, the salary is $36500{\left(1.035\right)}^3$ .

Hence, for $t$ year, the salary is $36500{\left(1.035\right)}^{\mathrm{int}t}$

To determine

To graph : The salary function $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ .

Explanation of Solution

Given information :

The salary function is given by $y=36500{\left(1.035\right)}^{\mathrm{int}t}$ where $t$ is the time in years.

Graph :

So, the salary function for first four year is $y=\left\{\begin{array}{l}36500{\left(1.035\right)}^0,0\le t<1\\ 36500{\left(1.035\right)}^1,1\le t<2\\ 36500{\left(1.035\right)}^2,2\le t<3\\ 36500{\left(1.035\right)}^3,3\le t<4\end{array}\right.$ .

Thus, the graph is

  

Interpretation :

The function is continuous in the interval $\left(0,4\right)$ except at $x=1,2$ and $3$ .