Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, use this scenario: The population \(P\) of an endangered species habitat for wolves is modeled by the function \(P(x)=\frac{558}{1+54.8 e^{-0.462 x}},\) where \(x\) is given in years. How many wolves will the habitat have after 3 years?

Step-by-Step Solution

Answer

38 wolves.

Step 1: Understanding the Problem

The problem provides a function \(P(x)\) which models the population of wolves in a habitat over time \(x\). Our task is to determine the population after 3 years, or \(x = 3\).

Step 2: Substitute the Given Value

We begin solving for \(P(3)\) by substituting \(x = 3\) into the equation: \[P(3) = \frac{558}{1 + 54.8 e^{-0.462 \times 3}}.\]

Step 3: Calculate the Exponent

First, calculate the exponent \(-0.462 \times 3\) which simplifies to \(-1.386\).

Step 4: Evaluate the Exponential Expression

Next, find \(e^{-1.386}\) using a calculator, which approximately equals \(0.250\).

Step 5: Update the Function with the Exponential Value

Substitute this result back into the population function: \[P(3) = \frac{558}{1 + 54.8 \times 0.250}.\]

Step 6: Simplify the Denominator

Now, calculate \(54.8 \times 0.250\) which equals \(13.7\), and add it to 1, giving \(1 + 13.7 = 14.7\).

Step 7: Final Calculation for Population

Finally, compute the population using the simplified denominator: \[P(3) = \frac{558}{14.7} \approx 37.959.\]

Step 8: Round to Nearest Whole Number

Since the population must be a whole number, round \(37.959\) to \(38\).

Key Concepts

Exponential Functions

Exponential functions are powerful mathematical tools used to model situations where change occurs at a constant rate. Unlike linear functions, where changes occur by addition or subtraction, exponential functions involve multiplication or division. This is why they can depict rapid growth or decay. In our exercise, the population of wolves is modeled using an exponential function:

• Exponential functions commonly take the form of \(f(x) = a \cdot e^{bx}\).
• The base \(e\) is an irrational number approximately equal to 2.718, known as Euler's number.
• The function provided models population changes over time, making it a suitable choice for population studies.

When dealing with exponential functions, understanding the base and exponent is key. The exponent determines how the function grows or shrinks. If negative, the function models decay, as seen in this wolf population model.

Endangered Species

Endangered species, like the wolves in the exercise, are plants or animals at risk of becoming extinct. Population modeling for these species is critical for conservation efforts. By knowing the population trends, environmentalists and policymakers can strategize better to help protect these species.

• Endangered species face threats like habitat loss, climate change, and poaching.
• Such species require careful monitoring and protective measures for recovery.
• Mathematical models provide insights into breeding programs and the status of these populations over time.

This exercise's scenario demonstrates how exponential functions can predict the population's future, aiding in planning necessary interventions.

Mathematical Modeling

Mathematical modeling involves creating equations and functions to represent real-world situations, making them easier to analyze. In our case, the population of wolves is modeled using the function \( P(x) = \frac{558}{1 + 54.8e^{-0.462x}} \). Models are essential because they:

• Simplify complex biological, chemical, or physical processes into understandable equations.
• Help predict future outcomes based on current data and trends.
• Are tools for decision-making and policy development.

Understanding the variables in a model is crucial. In the population function, \(x\) represents time in years. Such clarity is vital for interpreting results correctly and applying them effectively in policy and conservation planning.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations that represent values. The function used for modeling the wolf population is rich in algebraic expression complexity:

• Includes constants like 558, which may represent an initial population carrying capacity.
• Contains the variable \(x\) for time, directly influencing population change.
• Utilizes operations like exponentiation and multiplication to form the model.

Understanding each component in the expression ensures clear problem-solving steps. Simplifying these expressions accurately is necessary for determining the population values effectively. Whether calculating the exponential part or simplifying the denominator, each step uses algebraic knowledge to reach a precise result.