calculus problem?

Question

The number of yeast cells in a laboratory culture increases rapidly at first but levels off eventually. The population is modeled by the function below, where t is measured in hours. At time t = 0 the population is 20 cells and is increasing at a rate of 14 cells/hour.
n = f(t) = a/(1+be^(-0.8 t))
(a) Find the values of a and b.
a =
b =

(b) According to this model, in the long run the yeast population stabilizes at ? cells

Suppose that a population of bacteria triples every hour and starts with 300 bacteria.
(a) Find an expression for the number n of bacteria after t hours.
n(t) =

(b) Estimate the rate of growth of the bacteria population after 2.5 hours.
n'(2.5) =

Answer 1

1) 20 = a/(1+b) so a= 20+20b
df/dt = a*0.8*b/(1+be^-0.8t)^2
at t=0
14= 0.8ab/(a+b)^2 so
14=(16b+16b^2)/(1+b)^2
14+28b +14b^2 = 16b+16b^2
2b^2-12b-14=0
b^2-6b-7=0 b= ((6+-8))/2 = 7 or-1
The lasr one can´t be as gives a=0 so b= 7 and a = 160

lim f(t) t==> infinity = a = 160 cells

2) You must use another model
n= a * e^bt
300 =a
900 = 300 e^b so b = ln 3 and
n(t)= 300 *(3)^t
n´(t) = 300* ln 3 *(3)^t
n´(2.5) =300*ln3 *3^2.5=5138 cells/h