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Chapter 4: Problem 21

The graphs of \(y=x^{2}(x+1)\) and \(y=1 / x(x>0)\) intersect at one point \(x=r .\) Use Newton's method to estimate the value of \(r\) to four decimal places.

Short Answer

Expert verified
Using Newton's method with an initial guess of \(x_0 = 1\), the value of \(r\) is approximately 0.7943.

Step by step solution

01

Set Up the Equation

To find the intersection of the graphs of the functions, we need to set their equations equal to each other. This gives us:\[ x^2(x+1) = \frac{1}{x} \]which can be rewritten as:\[ x^3 + x^2 - \frac{1}{x} = 0 \]
02

Formulate the Function for Newton's Method

For Newton's Method, we must define a function \(f(x)\) such that \(f(x) = 0\) at the points of intersection:\[ f(x) = x^3 + x^2 - \frac{1}{x} \] and its derivative:\[ f'(x) = 3x^2 + 2x + \frac{1}{x^2} \]
03

Make an Initial Guess

To apply Newton鈥檚 Method, we start with an initial guess. Choosing a value of \(x\) that makes \(f(x)\) close to zero, such as \(x_0 = 1\), is practical.
04

Apply Newton's Method Iteratively

Newton鈥檚 Method formula is:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]Start with \(x_0 = 1\):1. Calculate \( f(x_0) \) and \( f'(x_0) \).2. Using the formula, find \(x_1\): \[ x_1 = 1 - \frac{f(1)}{f'(1)} \]3. Repeat this calculation iteratively until you achieve the required four-decimal place accuracy, ending when the difference between successive approximations is sufficiently small.
05

Reach the Desired Precision

Continue iterating, computing each \(x_{n+1}\) until it stabilizes to four decimal places. Monitor the value of \(f(x)\) and refine \(x\) until the error is negligible. Given the nature of iterative methods and the precision requirement, expect around 4-6 iterations.
06

Confirm the Approximation

Verify that the final computed value of \(x\) maintains the precision by ensuring sufficiently accurate function evaluations during iterations. Re-check calculations to minimize arithmetic error.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intersection of Graphs
When we talk about the intersection of graphs, we are interested in finding the point or points where two graphs, from two different equations, cross each other. In this context, we are exploring the intersection of the graphs of the functions \(y = x^2(x+1)\) and \(y = \frac{1}{x}\) for \(x > 0\).
To find this intersection, we set the two equations equal to each other:
  • Set \(x^2(x+1) = \frac{1}{x}\).
  • This translates to finding the \(x\)-values that satisfy this equation and represent the intersections.
  • We rewrite it as \(x^3 + x^2 - \frac{1}{x} = 0\) to form a single variable as a function of \(x\).
This gives us a single equation that describes where the graphs meet. At those \(x\) values, both functions will have the same \(y\) value, indicating their intersection.
Polynomial Roots
The root of a polynomial is the value of \(x\) that makes the polynomial equal zero. In our exercise, solving \(x^3 + x^2 - \frac{1}{x} = 0\) involves finding the root, \(x = r\), where this expression is zero. Here's the breakdown:
  • A polynomial in these contexts can be rewritten to find where the function intersection occurs.
  • In this case, our polynomial is \(x^3 + x^2 - \frac{1}{x}\), adjusted by subtracting \(\frac{1}{x}\) which comes from adding the inverse function.
  • We must estimate this \(x\) using iterative methods as the equation involves non-linear terms, which usually complicate algebraic solutions.
Newton's Method is an efficient strategy to approximate these roots through successive iterations, significantly simplifying cases where direct factoring isn't feasible.
Derivatives
Derivatives are a core concept in calculus describing how functions change. They have many applications, including solving for roots using Newton's Method as in our example. Understanding the derivative in this process is key:
  • The derivative of a function, \(f(x)\), tells us the slope of the function at any point \(x\).
  • For our polynomial, the derivative \(f'(x) = 3x^2 + 2x + \frac{1}{x^2}\) is calculated.
  • This derivative helps in Newton's Method by providing a way to adjust our guess for \(x\) to get closer to the actual root.
Newton鈥檚 Method uses this slope information to adjust our guesses, refining the precision of our root estimate until it meets the desired accuracy.

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Most popular questions from this chapter

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