the number of times a new pop song has been downloaded t weeks after its initial release is given by (…

the number of times a new pop song has been downloaded t weeks after its initial release is given by ( f(t)=\frac{4,000,000}{1 + 300e^{-0.45t}} ) a) after how many weeks is the rate of change of the number of downloads maximized? b) what is the rate of change of the number of downloads at the time found in part (a)? c) how many times has the song been downloaded at the time found in part (a)? a) the rate of change of the number of downloads is maximized after 12.68 weeks (do not round until the final answer. then round to two decimal places as needed.) b) the rate of change of the number of downloads is 450000 songs per week (round to the nearest whole number as needed.) c) the song has been downloaded ( square ) times (round to the nearest whole number as needed.)

the number of times a new pop song has been downloaded t weeks after its initial release is given by ( f(t)=\frac{4,000,000}{1 + 300e^{-0.45t}} ) a) after how many weeks is the rate of change of the number of downloads maximized? b) what is the rate of change of the number of downloads at the time found in part (a)? c) how many times has the song been downloaded at the time found in part (a)? a) the rate of change of the number of downloads is maximized after 12.68 weeks (do not round until the final answer. then round to two decimal places as needed.) b) the rate of change of the number of downloads is 450000 songs per week (round to the nearest whole number as needed.) c) the song has been downloaded ( square ) times (round to the nearest whole number as needed.)

Answer

Explanation:

Step1: Substitute (t = 12.68) into (f(t))

We have the function (f(t)=\frac{4000000}{1 + 300e^{-0.45t}}). Substitute (t = 12.68) into the exponent: (-0.45t=-0.45\times12.68=-5.706) Then calculate (e^{-5.706}\approx0.00334) Next, calculate (1+300e^{-5.706}=1 + 300\times0.00334=1+1.002 = 2.002)

Step2: Calculate (f(12.68))

Now, (f(12.68)=\frac{4000000}{2.002}\approx1998002)

Answer:

(1998002)