First you have to change each one into a decimal.
10% = 0.10
1/9 = 0.11
So the answer has to be between 0.10 and 0.11.
A is more than 0.11.
B is more than 0.11.
C is between 0.10 and 0.11.
D is less than 0.10.
The answer is C. 0.108.
Answer:
P = 6200 / (1 + 5.2e^(0.0013t))
increases the fastest
Step-by-step explanation:
dP/dt = 0.0013 P (1 − P/6200)
Separate the variables.
dP / [P (1 − P/6200)] = 0.0013 dt
Multiply the left side by 6200 / 6200.
6200 dP / [P (6200 − P)] = 0.0013 dt
Factor P from the denominator.
6200 dP / [P² (6200/P − 1)] = 0.0013 dt
(6200/P²) dP / (6200/P − 1) = 0.0013 dt
Integrate.
ln│6200/P − 1│= 0.0013t + C
Solve for P.
6200/P − 1 = Ce^(0.0013t)
6200/P = 1 + Ce^(0.0013t)
P = 6200 / (1 + Ce^(0.0013t))
At t = 0, P = 1000.
1000 = 6200 / (1 + C)
1 + C = 6.2
C = 5.2
P = 6200 / (1 + 5.2e^(0.0013t))
You need to change the exponent from negative to positive.
The inflection points are where the population increases the fastest.
Two main facts are needed here:
1. The logarithm 
, regardless of the base of the logarithm, exists for 
.
2. The square root 
exists for 
.
(in both cases we're assuming real-valued functions only)
By (2) we know that 
exists if 
, or 
.
By (1), we know that 
exists if 
, or 
. But as long as the square root exists, it will always be positive, so this condition will always be met.
Ultimately, then, we only require , so the function has domain 
.
To determine the range, we need to know that, in their respective domains, and increase monotonically without bound. We also know that 
at minimum, at which point the square root term vanishes, so the least value the function takes on is 
. Then its range would be 
.
