Problem 381 - Precalculus

Question

For the following exercises, use the logistic growth model \(f(x)=\frac{150}{1+8 e^{-2 x}}\) Graph the model.

Step-by-Step Solution

Answer

Graph the function using points and approach the horizontal asymptote at \( y = 150 \).

Step 1: Understanding the Logistic Growth Model

The logistic growth model formula is given as \( f(x) = \frac{L}{1 + b e^{-kx}} \), where \( L \) represents the carrying capacity, \( b \) is a constant representing the growth factor, and \( k \) is the growth rate. For the given function \( f(x)=\frac{150}{1+8 e^{-2 x}} \), \( L = 150 \), \( b = 8 \), and \( k = 2 \).

Step 2: Identify Key Features of the Graph

To graph the function, first identify key features: asymptotes, intercepts, and behavior as \( x \to \pm \infty \). The horizontal asymptote is \( y = L = 150 \), because as \( x \to \infty \), \( e^{-2x} \to 0 \) and \( f(x) \to L \). As \( x \to -\infty \), \( e^{-2x} \to \infty \) making \( f(x) \approx 0 \). The y-intercept is found by setting \( x = 0 \): \( f(0) = \frac{150}{1 + 8} = \frac{150}{9} \approx 16.67 \).

Step 3: Plot the Graph Using Calculated Points

Calculate important points such as the y-intercept and points of inflection to plot accurately. \( f(0) \approx 16.67 \) is a point on the graph. As \( x \) increases, compute additional points such as \( f(1) \) and \( f(2) \): \( f(1) = \frac{150}{1 + 8e^{-2}} \approx 37.5 \) and \( f(2) = \frac{150}{1 + 8e^{-4}} \approx 82.5 \). These values will help plot the curve showing the slow start, rapid growth in the middle, and leveling off towards the carrying capacity.

Step 4: Draw the Curve on the Graph

Using the plotted points and asymptote, draw a smooth curve starting from near \( y = 0 \) at \( x \to -\infty \), passing through calculated points like \( (0, 16.67) \), \( (1, 37.5) \), and \( (2, 82.5) \), and approaching \( y = 150 \) as \( x \to \infty \). Make sure the curve exhibits logistic growth characteristics: slow initial increase, rapid growth toward the middle, and then plateauing.

Key Concepts

Graphing Functions

When we talk about graphing functions, especially logistic growth models, we dive into the art of visual representation. With the logistic function, the aim is to visualize how the population (or variable in question) evolves over time. Let's break down how to approach graphing these models.

Firstly, identify key features of the function before plotting any points. For the logistic growth function given as \( f(x) = \frac{150}{1+8 e^{-2 x}} \), the horizontal asymptote at \( y = 150 \) is vital. Why is this important? Because it represents the carrying capacity, indicating the maximum value the function will approach but not exceed.

Next, pinpoint the y-intercept by substituting \( x = 0 \) into the function. This gives us an initial position of the curve on the graph. For \( f(x) \), the y-intercept is approximately 16.67, marking where the curve intersects the y-axis.

When graphing, remember these key steps:

• Determine any limits or asymptotes as \( x \to \pm \infty \).
• Identify critical points such as the y-intercept.
• Plot calculated points to define the curve's shape, ensuring the graph accurately shows the slow start and swift growth before leveling.

With these preparations, you'll be poised to sketch a reliable graph that exemplifies the logistic growth model's features.

Carrying Capacity

Carrying capacity is a crucial aspect of the logistic growth model. It represents the maximum population size that the environment can sustain indefinitely. In the logistic formula \( f(x) = \frac{L}{1 + b e^{-kx}} \), \( L \) is the carrying capacity. In our example, this value is 150.

Understanding carrying capacity is vital because it explains why resources limit growth. Initially, the population grows rapidly due to limited competition for resources. However, as resources become scarcer, growth slows and eventually stabilizes.

The carrying capacity affects how the function behaves at large \( x \) values. As \( x \to \infty \), the logistic function \( f(x) \to L \). This means the population approximates the carrying capacity over time. Hence, the horizontal asymptote, \( y = L \), indicates this stable state.

In terms of graphing:

• Expect the curve to level off as it nears \( y = 150 \).
• Understand that \( L \) limits how high the function can rise.
• Recognize that in nature, this mirrors how ecosystems have a limit to the number of individuals they can support.

With this in mind, interpreting and predicting growth using logistic models becomes more intuitive.

Exponential Growth

Exponential growth is a concept where a quantity increases at a consistent rate over time. In the initial phases of population growth, it seems unrestricted and rapid. The logistic growth model juxtaposes this with a more realistic scenario by incorporating limitations, modeled after real-world ecosystems.

With the function \( f(x) = \frac{150}{1+8 e^{-2 x}} \), the term \( e^{-2x} \) represents exponential decay as \( x \to \infty \). Initially, the population grows exponentially when resources are abundant, and interaction between individuals is low.

However, due to this model's design, such exponential growth transitions into a plateau as it approaches the carrying capacity. At this point, exponential growth's initial rapid rise gives way to logistic growth behavior characterized by:

• An initial exponential-like increase when \( x \) is negative or small, demonstrating quick growth.
• A slowdown in growth nearing the midpoint.
• A transition to a stable maximum as it reaches carrying capacity.

This shift highlights why exponential growth is unsustainable without carrying capacity limitations, making the logistic model a favored choice in ecological and population studies.