Certainly! Let's tackle the problem step-by-step.
### Part (a): Evaluating [tex]\( P(0) \)[/tex]
The logistic growth function is given by:
[tex]\[ P(t) = \frac{3.194}{1 + 14.589 e^{-0.052 t}} \][/tex]
We need to evaluate [tex]\( P(0) \)[/tex]. This means we will substitute [tex]\( t = 0 \)[/tex] into the function:
[tex]\[ P(0) = \frac{3.194}{1 + 14.589 e^{-0.052 \cdot 0}} = \frac{3.194}{1 + 14.589 \cdot e^{0}} = \frac{3.194}{1 + 14.589 \cdot 1} = \frac{3.194}{1 + 14.589} = \frac{3.194}{15.589} \][/tex]
Calculating this fraction, we get:
[tex]\[ P(0) \approx 0.2049 \][/tex]
#### Interpretation
In the context of this problem, [tex]\( P(0) \)[/tex] represents the population of Los Angeles in the year 1900. Therefore, the population of Los Angeles in 1900 was approximately [tex]\( 0.2049 \)[/tex] million or approximately 204,900 people.
### Part (b): Predicting the Population on January 1, 2016
To find the population of Los Angeles on January 1, 2016, we first need to determine the value of [tex]\( t \)[/tex] in 2016. Since [tex]\( t \)[/tex] is the number of years since 1900, we calculate:
[tex]\[ t = 2016 - 1900 = 116 \][/tex]
Next, we substitute [tex]\( t = 116 \)[/tex] into the logistic growth function:
[tex]\[ P(116) = \frac{3.194}{1 + 14.589 e^{-0.052 \cdot 116}} \][/tex]
First, compute the exponent:
[tex]\[ -0.052 \cdot 116 = -6.032 \][/tex]
Now, compute the exponential term:
[tex]\[ e^{-6.032} \approx 0.0024 \][/tex]
Substituting back into the function:
[tex]\[ P(116) = \frac{3.194}{1 + 14.589 \cdot 0.0024} = \frac{3.194}{1 + 0.0358146} = \frac{3.194}{1.0358146} \][/tex]
Calculating this fraction:
[tex]\[ P(116) \approx 3.0859 \][/tex]
#### Conclusion
The predicted population of Los Angeles on January 1, 2016, is approximately [tex]\( 3.0859 \)[/tex] million or approximately 3,085,900 people.
This detailed, step-by-step approach shows how to evaluate and interpret the logistic growth function for both the initial year (1900) and a future year (2016).
