Answer:
a) 1296 bacteria per hour
b) 0 bacteria per hour
c) -1296 bacteria per hour
Step-by-step explanation:
We are given the following information in the question:
The size of the population at time t is given by:
[tex]b(t) = 9^6 + 6^4t-6^3t^2[/tex]
We differentiate the given function.
Thus, the growth rate is given by:
[tex]\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t[/tex]
a) Growth rates at t = 0 hours
[tex]\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}[/tex]
b) Growth rates at t = 3 hours
[tex]\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}[/tex]
c) Growth rates at t = 6 hours
[tex]\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}[/tex]
