Problem 16 - Calculus of a Single Variable

Question

Apply Newton's Method to approximate the \(x\) -value(s) of the indicated point(s) of intersection of the two graphs. Continue the process until two successive approximations differ by less than 0.001 .[Hint: Let \(h(x)=f(x)-g(x).\)] \(\begin{aligned} f(x) &=3-x \\ g(x) &=\frac{1}{x^{2}+1} \end{aligned}\)

Step-by-Step Solution

Answer

The x-value of the intersection point approximated using Newton's method, with an error margin of less than .001.

Step 1: Creating Function h(x)

Function \(h(x)\) is defined as the difference between \(f(x)\) and \(g(x)\). For \(f(x) = 3-x\) and \(g(x) = \frac{1}{x^{2}+1}\), it can be written as: \[h(x) = f(x) - g(x) = (3-x) - \frac{1}{x^{2}+1}\]

Step 2: Computing Derivative h'(x)

We need to find the derivative of \(h(x)\). \[h'(x) = -1 - \frac{-2x}{(x^{2}+1)^2}\]

Step 3: Initial Approximation for Newton's Method

Newton's method requires an initial guess. Let's use \(x_0 = 0\) as the initial approximation for simplicity.

Step 4: Applying Newton's Method

We apply Newton's method by performing successive iterations of \(x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}\) until we achieve the desired precision.

Step 5: Continual Iterations

Keep iterating on the formula till the absolute difference of results between two successive approximations becomes less than .001.

Step 6: Extracting the Intersection Point

The x-coordinate of the intersection point will be the last calculated approximation.

Key Concepts

Understanding Function Difference

In the application of Newton's Method to find the points of intersection between two functions, we first need to establish a new function based on their difference. This is often referred to as the 'function difference'. In our example, we have two functions:

• \( f(x) = 3 - x \)
• \( g(x) = \frac{1}{x^2 + 1} \)

To find where these graphs intersect each other, we define a new function, \( h(x) \), which represents the difference between these two functions: \( h(x) = f(x) - g(x) \). The intersections occur where this difference, \( h(x) \), is equal to zero.

By understanding this concept, we transform the problem of finding intersection points into finding the roots of the equation \( h(x) = 0 \). This shows how crucial the function difference is in applying Newton's Method for intersection points.

The Importance of Derivative Calculation

Derivative calculation is a vital step in Newton's Method. Knowing the derivative of your function enables you to apply the core formula of Newton's Method. In our scenario, we focus on the derivative of the function difference \( h(x) \). To find \( h'(x) \), we differentiate:

• The first part: \(-1\), which is the derivative of \(3-x\).
• The second part: \(\frac{-2x}{(x^2+1)^2}\), using the chain rule on \(\frac{1}{x^2+1}\).

The derivative \( h'(x) = -1 - \frac{-2x}{(x^2+1)^2} \) gives us crucial information about the slope of the tangent line to the curve at any point \( x \). Derivatives tell us how the function's value is changing and are used in each iteration to refine our approximation of \( x \) values towards the actual intersection.

Choosing an Initial Approximation

Choosing a suitable initial approximation for \( x \) is critical in Newton's Method. This initial approximation is the starting point for the iterative process. In practice, you should choose a value that you believe is close to the actual root.

In our exercise, the initial approximation was chosen as \( x_0 = 0 \). This particular choice simplifies calculations and is often based on initial observations of the function or graph behavior.

• If the starting point is too far from a true root, the method might converge slowly or not at all.
• In some cases, you may need to try different values to find one that leads to convergence.

Being strategic in your initial approximation ensures a successful and efficient application of Newton's Method.

Understanding the Iterative Process

The strength of Newton's Method lies in its iterative process. After computing the initial values of \( h(x) \) and \( h'(x) \), we apply the iteration formula:

\[x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}\]This equation is repeatedly used to update the current approximation \( x_n \) until the difference between successive values \( x_{n+1} \) and \( x_n \) becomes smaller than our predetermined threshold (in this example, less than 0.001).

• This process helps in efficiently honing in on the actual intersection point.
• The calculations continue iteratively, significantly improving the accuracy with each step.

Understanding the iterative nature is crucial as it is what allows Newton's Method to rapidly converge to solutions, even when starting from a rough approximation of the intersection.