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?

810.58 dollars per year

Step-by-step 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.

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

Step-by-step explanation:

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

So

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

So

Now the average rate of change m is defined as:

Therefore:

f is the midpoint of