Menu Value Theorem The population of a culture of cells grows according to the 100t function P(t) = t +1, where t? ? ? 0 is measured in weeks. a.? What is the average rate of change in the population over the interval [0, 8]? b. At what point of the interval [0, 8] is the instantaneous rate of change equal to the average rate of change?

Step-by-step solution Step 1 In this problem, the population of a culture of cells is given as P(t) = t +1, we have to find the average rate of change in the population over the interval [0,8]and we also have to find at what point of interval [0,8] is the instantaneous rate of change equal to the average rate of change. To find the requirement, we will be using the mean value theorem Mean Value Theorem: It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c such that f(b)f(a) f(c) = ba Step 2 Given the population of a culture of cells grows according to the function P(t) = 100t , t 0 …. 1) t +1 a. What is the average rate of change in the population over the interval [0, 8] Rate of change is nothing but the derivative. So let us find P (t. 100t We have P(t) = t +1 (t +1)100100t(1) P (t) = 2 (t +1) 100t + 100 100t 100 = (t+1) = (t +1) …. (2) Thus P(t)is differentiable in (0,8) Hence the requirement in mean value theorem is satisfied by P(t). Thus by using the mean value theorem stated in step 1, the average rate of chase in population is p(8)p(0) = 80 From (1) we can find the value of p(8),p(0)as follows 100(8) 800 100(0) 0 p(8) = 8+1 = 9 and p(0) = 0+1 = 1 Thus we get p(8)p(0) 8901 80 = 8 890 800 100 = 8 = 72 = 9 … (3) Thus the average rate of change in population is 100 cells per week. 9