Step 1 :The function given is \(f(t)=\frac{3,900,000}{1+400 e^{-0.75 t}}\).
Step 2 :To find the maximum rate of change, we need to find the derivative of this function, which represents the rate of change.
Step 3 :The derivative of \(f(t)\) is \(f'(t)\).
Step 4 :Using the quotient rule, we get \(f'(t) = \frac{(1+400 e^{-0.75 t}) * 0 - 3,900,000 * (0 - 400 * -0.75 * e^{-0.75 t})}{(1+400 e^{-0.75 t})^2}\).
Step 5 :Simplifying, we get \(f'(t) = \frac{3,900,000 * 300 * e^{-0.75 t}}{(1+400 e^{-0.75 t})^2}\).
Step 6 :To find the maximum rate of change, we need to find the maximum of \(f'(t)\). This occurs when the derivative of \(f'(t)\) is 0.
Step 7 :The derivative of \(f'(t)\) is \(f''(t)\).
Step 8 :Using the quotient rule again, we get \(f''(t) = \frac{(1+400 e^{-0.75 t})^2 * 0 - 2 * (1+400 e^{-0.75 t}) * (3,900,000 * 300 * e^{-0.75 t})}{(1+400 e^{-0.75 t})^4}\).
Step 9 :Simplifying, we get \(f''(t) = -\frac{2 * 3,900,000 * 300 * e^{-0.75 t} * (1+400 e^{-0.75 t})}{(1+400 e^{-0.75 t})^4}\).
Step 10 :Setting \(f''(t)\) equal to 0 and solving for \(t\), we get \(t = \ln(\frac{1}{400}) / -0.75\).
Step 11 :Evaluating this expression, we get \(t \approx 7.99\) weeks.
Step 12 :Substituting \(t = 7.99\) into \(f'(t)\), we get \(f'(7.99) \approx 1,000,000\) songs per week.
Step 13 :Substituting \(t = 7.99\) into \(f(t)\), we get \(f(7.99) \approx 3,900,000\) songs.
Step 14 :Therefore, the rate of change of the number of downloads is maximized after approximately 7.99 weeks, at which point the rate of change is approximately 1,000,000 songs per week and the song has been downloaded approximately 3,900,000 times.