After t minutes of growth, a certain bacterial culture has a mass, in grams, of M(t) 􏰀=t^2.

To solve these questions, we need to use calculus concepts such as differentiation and integration. Don't worry if you're not familiar with these concepts; I will explain each step in detail.

a) To determine how much the bacterial culture grows during the time interval 3 < t < 3.01, we need to find the difference between the mass at time t = 3.01 and the mass at time t = 3.

M(t) = t^2

Let's calculate the mass at t = 3 and t = 3.01:

M(3) = (3)^2 = 9 grams
M(3.01) = (3.01)^2 = 9.0601 grams

Therefore, the bacterial culture grows by 9.0601 - 9 = 0.0601 grams during the time interval 3 < t < 3.01.

b) The average rate of growth during the time interval 3 < t < 3.01 can be found by dividing the change in mass by the change in time:

Average rate of growth = (Change in mass) / (Change in time)

The change in mass is 0.0601 grams (from part a), and the change in time is 3.01 - 3 = 0.01 minutes.

Average rate of growth = 0.0601 / 0.01 = 6.01 grams per minute

Therefore, the average rate of growth during the time interval 3 < t < 3.01 is 6.01 grams per minute.

c) To find the rate of growth when t = 3, we need to take the derivative of the mass function M(t) with respect to time t and then evaluate it at t = 3.

M(t) = t^2

To find the derivative of M(t), we can use the power rule of differentiation, which states that the derivative of t^n is n*t^(n-1).

dM(t)/dt = d(t^2)/dt = 2t

Now, let's evaluate the derivative at t = 3:

dM(t)/dt = 2t
dM(t)/dt at t = 3 = 2*3 = 6

Therefore, the rate of growth of the bacterial culture when t = 3 is 6 grams per minute.

I hope this explanation helps! Let me know if you have any further questions.