Praxis Core Academic Skills for Educators: Mathematics (5733)Algebra and FunctionsHard
A biologist is tracking the population of a certain species of fish in a lake. The population, P(t), after t years, is modeled by the function P(t) = 1500 / (1 + 4e^(-0.5t)). What is the limiting value of the fish population as time approaches infinity?
- A300
- B1500
- C0
- D6000
Show answer & explanationAnswer & explanation
Correct answer: B. 1500
This is a logistic growth model. As time t approaches infinity (t → ∞), the exponential term e^(-0.5t) approaches e^(-∞), which is 0. So, the denominator approaches 1 + 4(0) = 1. Therefore, the population P(t) approaches 1500 / 1 = 1500. The limiting value is the carrying capacity, which is the numerator of the logistic function.
Why the other options are wrong
- A. This is the initial population P(0) = 1500 / (1+4) = 300, not the limiting value.
- C. This would imply the population dies out, which is not what the model suggests.
- D. This would be 1500 * 4, which is not how the logistic function works.
Logistic Growth Model Limiting Value
For a logistic growth model in the form P(t) = K / (1 + Ae^(-Bt)), the limiting value (or carrying capacity) of the population as time approaches infinity is K, the numerator of the fraction.
- Represents a population that grows exponentially at first, then slows down as it approaches a maximum limit.
- The 'K' value is the carrying capacity of the environment.
- As t approaches infinity, the exponential term e^(-Bt) approaches 0.
Memory trick: The Ceiling is the 'K'!