Certainly! Let's solve the problem step-by-step.
### a. Evaluate [tex]\( P(0) \)[/tex] and interpret its meaning in the context of this problem.
Given the population function:
[tex]\[
P(t) = \frac{3.194}{1 + 14.589 e^{\cos(2\pi)}}
\][/tex]
To find [tex]\( P(0) \)[/tex], we need to evaluate the function at [tex]\( t = 0 \)[/tex].
[tex]\[
P(0) = \frac{3.194}{1 + 14.589 e^{\cos(2\pi)}}
\][/tex]
First, we know that:
[tex]\[
\cos(2\pi) = 1
\][/tex]
Thus,
[tex]\[
P(0) = \frac{3.194}{1 + 14.589 e^{1}}
\][/tex]
We can evaluate this expression numerically to get:
[tex]\[
P(0) \approx 0.07855963135379519 \text{ million}
\][/tex]
In the context of this problem, [tex]\( P(0) \)[/tex] represents the population of Los Angeles in the year 1900. Specifically, it means that the population of Los Angeles in the year 1900 was approximately [tex]\( 0.07856 \times 10^6 = 78,560 \)[/tex] people.
### b. Use this function to predict the population of Los Angeles on January 1, 2016.
To predict the population in the year 2016, we need to first determine the value of [tex]\( t \)[/tex] for 2016.
[tex]\[
t = 2016 - 1900 = 116
\][/tex]
We evaluate [tex]\( P(116) \)[/tex]:
[tex]\[
P(116) = \frac{3.194}{1 + 14.589 e^{\cos(2\pi)}}
\][/tex]
Since [tex]\(\cos(2\pi) = 1\)[/tex], it simplifies as before:
[tex]\[
P(116) = \frac{3.194}{1 + 14.589 e^{1}}
\][/tex]
Evaluating this numerically gives:
[tex]\[
P(116) \approx 0.07855963135379519 \text{ million}
\][/tex]
Therefore, the predicted population of Los Angeles on January 1, 2016, is approximately [tex]\( 0.07856 \times 10^6 = 78,560 \)[/tex] people.
### Summary
a. The population of Los Angeles in the year 1900 was approximately 78,560 people.
b. The predicted population of Los Angeles on January 1, 2016, is also approximately 78,560 people.
