Chapter 3.5, Problem 20E

To determine

To sketch: a graph of given function in the same viewing window, to find which grows at a greater rate.

Answer to Problem 20E

The model $A={\left[1+\left(0.055/365\right)\right]}^{\left[365t\right]}$ grows at a higher rate.

Explanation of Solution

Given information:

If $1 is invested over a 10 years period, then the balance $A$ after $t$ years is given by either $A=1+0.06\left[t\right]$ or $A={\left[1+\left(0.055/365\right)\right]}^{\left[365t\right]}$ depending on whether the interest is simple interest at 6% or compound interest at $5\frac{1}{2}\%$ compounded daily.

Calculation:

To plot the graph of $A=1+0.06\left[t\right]$ , using the T1-83 graphing calculator follow the steps given below,

1. Enter the key $"Y="$ and then enter the function as shown below,

  

2. Press TRACE then the graph will appear on the screen as shown below,

  

From above graphs, the model $A={\left[1+\left(0.055/365\right)\right]}^{\left[365t\right]}$ grows at a higher rate.