/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Determine the required values by... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the required values by using Newton's method. Explain why Newton's method does not work for finding the root of \(x^{3}-3 x=5\) if \(x_{1}\) is chosen as 1

Short Answer

Expert verified
Newton's method does not work because the derivative \(f'(1)\) is zero, which results in division by zero.

Step by step solution

01

Understanding Newton's Method

Newton's method is a way to find successively better approximations to the roots (or zeroes) of a real-valued function. It uses the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] where \(x_n\) is the current approximation, \(f(x_n)\) is the function value at \(x_n\), and \(f'(x_n)\) is the derivative of the function at \(x_n\).
02

Identifying the Function and Its Derivative

The given function is \(f(x) = x^3 - 3x - 5\). We first need to find its derivative, \(f'(x)\). The derivative is: \[ f'(x) = 3x^2 - 3 \]
03

Substitute Initial Guess into the Formula

We substitute \(x_1 = 1\) into the Newton's method formula. First, calculate \(f(1)\) and \(f'(1)\):\[ f(1) = 1^3 - 3(1) - 5 = -7 \] \[ f'(1) = 3(1)^2 - 3 = 0 \]
04

Evaluate the Denominator

In Newton's method formula, the denominator \(f'(x_n)\) cannot be zero because it leads to division by zero, indicating the method cannot proceed. With \(f'(1) = 0\), the method encounters a problem.
05

Conclusion on Newton's Method Applicability

Due to division by zero when using \(x_1 = 1\), Newton's method cannot advance to find an appropriate next approximation. Hence, it fails to find the root with this initial guess.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Approximations of Roots
Newton's Method is a widely used technique for approximating the roots of a real-valued function. It is especially effective because it is iterative, meaning that it repeatedly refines estimates to bring the solution closer to the actual root. Each estimation, or approximation, is based on the tangent line to the function at the current point. The intersection of this tangent line with the x-axis gives the next approximation of the root.

During each iteration, the approach is as follows: if you have a guess \( x_n \), the new and improved guess \( x_{n+1} \) is calculated by dropping perpendicular lines down the function, stepping sideways along the tangent there, and where it hits the x-axis, that is your next guess. Thus:
  • The iteration formula is: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)
  • \( f(x_n) \) is the value of the function at the guessed point.
  • \( f'(x_n) \) is the derivative, telling how steep the slope of the function is at that point.
The element of approximating the roots using Newton's Method can effectively navigate through functions with various characteristics. It's important to note, however, that the starting point or the initial guess should be chosen carefully. If not, the method might not work, as was the case in this exercise.
Derivative in Calculus
The derivative plays a critical role in calculus, particularly in Newton's Method, as it is used to analyze the behavior of a function. In simpler terms, the derivative of a function at a point gives the slope of the tangent line to the function at that point. This slope is instrumental when using Newton's Method because it determines how the function rises or falls and, thus, how the next approximation is calculated.

For the given function \( f(x) = x^3 - 3x - 5 \), the derivative is given by:
  • \( f'(x) = 3x^2 - 3 \)
This derivative is critical as it directly influences the denominator in Newton's Method formula. It tells us how fast the function value is changing at any given point. Which means during the application of Newton's Method:
  • If the derivative (slope) is zero, the tangent line becomes horizontal, leading to a situation that halts further approximations.
  • In cases when the derivative gives a non-zero value, the next \( x_{n+1} \) can be calculated straightforwardly.
Understanding derivatives allows us to predict and influence the approximations achieved by Newton's Method.
Division by Zero
Division by zero is a fundamental mathematical problem that disrupts many numerical computations, including those using Newton's Method. When anyone tries to divide a number by zero, the result is undefined, creating significant complications especially in iterative methods where continuous calculations are crucial.

In the case of using Newton's Method, if during an iteration the derivative \( f'(x_n) \) equals zero, the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) encounters a division by zero dilemma.
  • In mathematical operations, having the denominator as zero means the division cannot be executed.
  • Newton's method relies heavily on having a valid, non-zero slope to continue approaching the root.
In this particular exercise, with the initial guess \( x_1 = 1 \), the derivative \( f'(1) = 0 \), leading the method to halt immediately.

This emphasizes the importance of carefully considering initial approximations and inspecting where the derivative could become zero to ensure Newton's Method can be used effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the linearization \(L(x)\) of the given functions for the given values of a. Display \(f(x)\) and \(L(x)\) on the same calculator screen. $$f(x)=6 \sqrt[3]{x}, a=8$$

In Exercises 33 and \(34,\) view the graphs of \(y, y^{\prime}, y^{\prime \prime}\) together on a calculator. State how the graphs of \(y^{\prime}\) and \(y^{\prime \prime}\) are related to the graph of \(y\). $$y=x^{3}-12 x$$

$$\text { Solve the problems in related rates.}$$ The intensity 1 of heat varies directly as the strength of the source and inversely as the square of the distance from the source. If an object approaches a heated object of strength 8.00 units at the rate of \(50.0 \mathrm{cm} / \mathrm{s}\), how fast is the intensity changing when it is \(100 \mathrm{cm}\) from the source?

An airline requires that a carry-on bag has dimensions (length + width + height) that do not exceed 45 in. If a carry-on has a length 2.4 times the width, find the dimensions (to the nearest inch) of this type of carry-on that has the greatest volume.

In Exercises \(43-54,\) sketch the indicated curves by the methods of this section. You may check the graphs by using a calculator. An analysis of data showed that the mean density \(d\) (in \(\mathrm{mg} / \mathrm{cm}^{3}\) ) of a calcium compound in the bones of women was given by \(d=0.00181 x^{3}-0.289 x^{2}+12.2 x+30.4,\) where \(x\) represents the ages of women \((20 < x < 80\) years). (A woman probably has osteoporosis if \(d < 115 \mathrm{mg} / \mathrm{cm}^{3} .\) ) Sketch the graph.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.