The function given by y = f(x) shows the value of $5000 invested at 6% interest compounded continuously, x years after the money was originally invested.
(Round your answers to the nearest cent.)
Value of $5000 with Continuous Compounding at 6%
(a) Find the average amount earned per year between the Sth year and the 10th year.
To find the average amount earned per year between the 5th year and the 10th year, we need to calculate the value of the investment at these two time points and then find the difference.
For the 5th year:
x = 5 years
y = f(5) = $5000 * e^(0.06*5) = $5000 * e^(0.3) ≈ $5000 * 1.3499 ≈ $6749.79
For the 10th year:
x = 10 years
y = f(10) = $5000 * e^(0.06*10) = $5000 * e^(0.6) ≈ $5000 * 1.8221 ≈ $9110.49
The amount earned between the 5th and 10th year is:
$9110.49 - $6749.79 = $2360.70
The average amount earned per year between the 5th and 10th year is:
$2360.70 / 5 = $472.14
Therefore, the average amount earned per year between the 5th year and the 10th year is approximately $472.14.