Question

James is contemplating an investment opportunity represented by the function A(t)=P(1.06)^t, where P is the initial amount of the investment, and t is the time in years. If James invests $5000, what is the average rate of change in dollars per year (rounded to the nearest dollar) between years 15 and 20?

Answer 1

810.58 dollars per year

Explanation:

Ok, so we are given that P=5000, so that makes our function A(t)=5000(1.06)^t .

The average rate of change is really the slope of the line going through (15,y1) and (20,y2).

We can find the corresponding y values by plugging the x's there to A(t)=5000(1.06)^t.

So let's do that:

y1=A(15)=5000(1.06)^(15)=11982.79097

y2=A(20)=5000(1.06)^(20)=16035.67736

Now the slope of a line can be found by using the formula:

(y2-y1)/(x2_x1) or just lining up the points and subtracting vertically and remember y difference goes over x difference.

Like so:

( 15 , 11982.79097 )

- ( 20 , 16035.67736)

---------------------------------

-5 -4052.886391

So the average rate of change is whatever -4052.886391 divided by -5 is.....

which is approximately 810.58 dollars per year.

Answer 2

The average rate of change in dollars per year between years 15 and 20 is:

`m=\$810`

Explanation:

First we calculate the profit of the investment for the year 15

So

`P=5000\\t=15`

`A(15)=5000(1.06)^(15)`

`A(15)=11982.79`

Now we calculate the profit of the investment for the year 20

So

`P=5000\\t=20`

`A(20)=5000(1.06)^(20)`

`A(20)=16035.68`

Now the average rate of change m is defined as:

`m=(A(20) - A(15))/(20-15)`

Therefore:

`m=(16035.68-11982.79)/(20-15)`

`m=(4052.89)/(5)`

`m=\$810.58`