Problem 20 - Calculus Early Transcendentals

Question

Use a graphing utility to determine the number of times the curves intersect and then apply Newton’s Method, where needed, to approximate the \(x\)-coordinates of all intersections. \(y=e^{-x}\) and \(y=\ln x\)

Step-by-Step Solution

Answer

The curves intersect once at approximately \( x = 0.6518 \).

Step 1: Graph the Functions

Start by graphing the functions \( y = e^{-x} \) and \( y = \ln x \) using a graphing utility. Observe the graph to visually determine where the curves intersect. The graph confirms that the curves intersect in the first quadrant, indicating there is one intersection point.

Step 2: Set Up Intersection Equation

To find the intersection points algebraically, equate the two functions: \( e^{-x} = \ln x \). This equation does not have a simple algebraic solution, so we need to use numerical methods to find where \( f(x) = e^{-x} - \ln x = 0 \).

Step 3: Choose an Initial Guess for Newton's Method

From the graph, estimate an initial guess for where the intersection occurs. Let's choose an initial \( x_0 = 0.5 \), which looks like a reasonable starting point based on the graph.

Step 4: Apply Newton's Method

Newton's Method formula is \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). First, find the derivative \( f'(x) = -e^{-x} - \frac{1}{x} \). Using the initial guess \( x_0 = 0.5 \), compute subsequent values: - \( x_1 = 0.5 - \frac{e^{-0.5} - \ln(0.5)}{-e^{-0.5} - \frac{1}{0.5}} \) - Continue iterating until \( |f(x_n)| ightarrow 0 \) to find \( x \approx 0.6518 \).

Step 5: Verify Intersection Points

Check the accuracy by substituting \( x \approx 0.6518 \) back into the original functions to ensure they produce approximately the same \( y \)-value, confirming the intersection point.

Key Concepts

Understanding Graphing Utilities

Graphing utilities are powerful tools used to visualize mathematical functions and their intersections. These tools assist in generating the visual representation of equations, like curves or lines, and allow us to perceive their behavior over a particular range. In this exercise, the goal is to use a graphing utility to graph the functions \( y = e^{-x} \) and \( y = \ln x \). The plots of these two functions can help us locate their intersection points visually.

Why are graphing utilities so crucial? They simplify the exploration of complicated mathematical concepts by providing:

• Immediate visual feedback, which helps in estimating values such as points of intersection.
• Insights into the behavior of functions over defined intervals.
• Enhanced understanding of theoretical math concepts through tangible visuals.

In this task, the graph displayed both curves, affirming that they intersect once in the first quadrant. This visual information guides the entire problem-solving process.

Applying Newton's Method

Newton's Method is a common numerical technique used to find successively better approximations to the roots (or zeroes) of a real-valued function. Here, it is employed to approximate the intersection point of the functions \( y = e^{-x} \) and \( y = \ln x \). This intersection signifies the solution to the equation \( e^{-x} = \ln x \), which doesn't lend itself to simple algebraic manipulation.

To use Newton’s Method, follow these steps:

• Start with an initial guess for the solution, \( x_0 \), based on insights from the graph.
• Apply the iteration formula:
  \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
  where \( f(x) = e^{-x} - \ln x \) and \( f'(x) = -e^{-x} - \frac{1}{x} \).
• Repeat the iteration until the function value \( |f(x_n)| \) approaches zero, indicating convergence to an accurate approximation of the intersection \(x\)-coordinate.

Through these iterative calculations, you'll find that \( x \approx 0.6518 \), refining our understanding of where the curves precisely meet.

Finding the Intersection of Curves

The intersection of curves refers to the points where two or more curves share the same coordinates on a graph. These points are solutions to equations formed by equating the functions defining the curves. In our problem, these are the \(x\)-values where \( y = e^{-x} \) and \( y = \ln x \) are equal.

To find the exact points of intersection, we use methods like:

• Graphical interpretation through graphing utilities to visually identify overlap or point of tangency.
• Algebraic or numerical methods like Newton’s Method to calculate precise intersection values when algebraic solutions are complex or unavailable.

Here, the finite graph gave the initial clue, while Newton’s Method pinpointed the intersection at \( x \approx 0.6518 \), verifying this by plugging back the \(x\)-value into the functions to check if they produce the same \(y\)-value ensures accuracy.
Indeed, intersections signify consensus between functions at a specific point, a critical aspect when determining solutions to equations graphically.