Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-14, use Newton's Method to approximate the indicated root of the given equation accurate to five decimal places. Begin by sketching a graph. The root of \(x \ln x=2\)

Step-by-Step Solution

Answer

The approximate root of the equation \(x \ln x = 2\) is around 2.857.

Step 1: Understanding the Problem

We need to find the root of the equation \(x \ln x = 2\). This means we need to find the value of \(x\) that satisfies this equation.

Step 2: Express as a Function

Rewrite the equation \(x \ln x = 2\) in terms of a function. We let \(f(x) = x \ln x - 2\). This transforms the problem into finding the root of \(f(x) = 0\).

Step 3: Identify Initial Guess

To use Newton's Method, we start with an initial guess. By sketching or using a graphing tool, assume the starting point to be \(x_0 = 3\) since \(\ln(3)\) is slightly less than 2, making \(3 \ln(3)\) close to 2.

Step 4: Differentiate the Function

Find the derivative of \(f(x)\), which is needed for Newton's Method. The derivative is \(f'(x) = \ln x + 1\).

Step 5: Apply Newton's Method

Use the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) to iterate and find a more accurate root. Start with \(x_0 = 3\), then compute:- \(x_1 = 3 - \frac{3 \ln 3 - 2}{\ln 3 + 1}\)- Continue iterations until the result converges to five decimal places.

Step 6: Iteration Calculations

Perform the iterations:- \(x_1 \approx 3 - \frac{3(1.0986) - 2}{1.0986 + 1} = 3 - \frac{0.2958}{2.0986} \approx 2.8595\)- Using \(x_1 = 2.8595\), compute \(x_2\) similarly.- Continue this process until \(x_n\) converges to a stable value up to five decimal places.

Step 7: Verify the Solution

Check the solution by substituting back into the original equation \(x \ln x = 2\) to ensure it’s close to zero when \(x =\) the approximate root. Verify that after several iterations the solution stabilizes and the function \(f(x)\) is close to zero.

Key Concepts

Root Approximation

In mathematics, root approximation is a crucial concept used to determine the zeros of functions, or find solution points where the function value equals zero. When you say you are approximating a root, you are attempting to find values that make the equation true. This method is useful because not all equations have roots that can be solved exactly or easily using algebraic manipulation.

There are various techniques to approximate roots, and Newton's Method is a popular choice. It's particularly useful for its speed and efficiency when dealing with complex functions.

This method involves beginning with an 'initial guess' and refining that guess over iterations until you reach a value sufficiently close to the actual root.

• Choose a starting value (initial guess) that is reasonably close to the expected root.
• Apply the Newton's formula repeatedly.
• Continue until the approximation stabilizes to the desired level of accuracy.

It's important to remember that while Newton's Method is quite powerful, it may not work well on certain types of functions, such as those with horizontal tangents or discontinuities.

Calculus Problems

Calculus is a branch of mathematics that deals with the rates of change and the accumulation of quantities. It plays a significant role in solving problems related to rates, areas under curves, and much more.

When faced with calculus problems, understanding derivatives and their applications is vital. In Newton's Method, derivatives are central to refining guesses toward a function's root.

• The derivative of a function, denoted as \( f'(x) \), gives us the slope of the tangent line at any given point on the function.
• This slope information is used to improve our initial guess in Newton's Method to move closer to the true root.

Being adept with derivatives allows for solving a diverse array of calculus problems, from simple ones to more advanced topics like optimization and differential equations. Solving calculus problems effectively often involves a mix of algebraic rearrangements, understanding functional behavior, and successful application of derivatives or integrals.

Numerical Methods

Numerical methods provide approximate solutions to mathematical problems that may not be solvable using standard analytical approaches. They are essential in fields where exact solutions are difficult or impossible to find.

One strength of numerical methods, like Newton's Method, is their ability to handle real-world scenarios with complicated equations. By leveraging computational power, numerical methods allow us to find useful approximations without requiring "closed-form" solutions.

• Useful across various science and engineering fields.
• Capable of iterative refinement of solutions, leading to high precision.
• Flexible to adapt to the complexity levels needed.

In the case of approximating roots, numerical methods are invaluable because they can systematically refine guesses to get to a close approximation, often within the bounds of machine precision. By applying these iterative processes, students and professionals can explore and understand behavior in systems that otherwise would be too complex for traditional methodologies.