Problem 76 - Beginning and Intermediate Algebra

Question

Approximate the point \((s)\) of intersection of the pair of equations. $$y=2.3 \ln (x+10.7), y=10 e^{-0.07 x^{2}}$$

Step-by-Step Solution

Answer

The approximate point of intersection (s) of the given functions can be found using the bisection method on the interval [0, 10] and a specified error value of 0.001. After performing the bisection method, we obtain the approximate value for x. Plugging this value into either \(f_1(x)\) or \(f_2(x)\) will give us the corresponding value of y. The approximate point of intersection is then given in the form (x, y).

Step 1: Define Functions and Interval

Define the given functions and the interval in which we want to look for the intersection point. Let's take the interval as [0, 10]. The functions are: \(f_1(x) = 2.3 \ln(x + 10.7)\) \(f_2(x) = 10 e^{-0.07x^2}\) We are looking for x in the interval [0, 10] such that \(f_1(x) = f_2(x)\).

Step 2: Define the Bisection Method Function

The bisection method function takes the lower and upper bounds of the interval as its arguments and returns the approximate point of intersection within a specified error bound. 1. Calculate the mid-point \(x_m\) of the interval [a, b]. 2. Check if \(|f_1(x_m) - f_2(x_m)|\) is less than some small error value (e.g., 0.001). If yes, return \(x_m\), else proceed. 3. If \(f_1(x_m) < f_2(x_m)\), update the interval to [\(x_m\), b] and go back to step 1. If \(f_1(x_m) > f_2(x_m)\), update the interval to [a, \(x_m\)] and go back to step 1.

Step 3: Perform Bisection Method

Using the bisection method function defined in Step 2, find the approximate value of x for the interval [0, 10] and a specified error value of 0.001. Keep iterating until the error condition is met.

Step 4: Calculate Intersection Point

Once we get the approximate value of x from Step 3, we can plug in the value in any of the given functions, \(f_1(x)\) or \(f_2(x)\), to calculate the corresponding value of y. This would give us the approximate point of intersection of the pair of equations.

Step 5: Present the Approximate Point of Intersection

Present the approximate point of intersection as calculated in Step 4 in the form (x, y).

Key Concepts

Intersection of Functions

When we talk about the intersection of functions, we are referring to the points in a graph where two or more functions have the same outputs. In other words, the values of the variables solve both equations simultaneously.
An intersection point involves finding a common x-value where the y-values are equal. In practical terms, this means that for two functions\( f_1(x) \) and \( f_2(x) \), the intersection occurs where \( f_1(x) = f_2(x) \).
This concept is crucial in various fields, such as physics and engineering, where systems must align or interact at specific points.
In the given problem, the focus is on finding this point of intersection for two particular functions within a defined interval using a method known as the bisection method.

Logarithmic Function

A logarithmic function is one of many forms of mathematical functions that involve logarithms. Logarithms are the inverses of exponential functions. They are particularly useful in solving problems related to multiplicative growth and decay.
The general form is given by \( y = \,a \ln(bx + c) \) where \( ln \) is the natural logarithm.

• The natural logarithm \( ln(x) \) uses the base \( e \), where \( e \approx 2.71828 \).
• Logarithms are handy for dealing with very large or very small numbers.

In our exercise, the function \( f_1(x) = 2.3 \ln(x + 10.7) \) plays a crucial role. The parameter \( 2.3 \) is a scaling factor, and \( 10.7 \) shifts the function horizontally.
This function reflects logarithmic growth, which is slower over time compared to linear or exponential growth. Understanding how to manipulate and interpret logarithmic functions is key in mathematical models, especially in compound interest or population growth.

Exponential Function

Exponential functions are known for their constant relative growth rate, with the general form \( y = a \, e^{(bx)} \). The base \( e \) is known as Euler's number and is approximately 2.71828.
In the problem, the function \( f_2(x) = 10 \, e^{-0.07x^2} \) represents an exponential decay.

• The coefficient 10 is a scaling factor that stretches or compresses the function vertically.
• The term \(-0.07x^2\) indicates that the growth factor is decaying, causing the function to decrease rapidly as \( x \) increases.

Exponential functions are pervasive in modeling real-world scenarios such as population dynamics, radioactive decay, and cooling processes.
It's vital to understand the behavior of exponential functions, as they can grow or shrink faster than any polynomial or logarithmic function, especially noticeable within a smaller interval like in our bisection analysis.

Error Bounds

In numerical methods like the bisection method, error bounds are a way to specify the precision needed in calculations. The goal is to find the root (or intersection) to within a certain accuracy level rather than an exact value, which can often be impossible due to irrational numbers.
Error bounds are defined as the difference between the actual value of x and the approximated value within the iteration steps.
In our exercise, the error bound was set to 0.001, meaning the solution should be accurate to three decimal places for the x-value at the intersection point.
Here's why error bounds are important:

• They decide when the algorithm should stop iterating.
• They ensure that calculations remain efficient without sacrificing accuracy.

By deciding on a suitable error bound, results can be obtained more quickly, allowing mathematical models to be both practical and reliable. This balance of precision and efficiency is crucial in computational applications and problem-solving scenarios.