   Chapter 4.8, Problem 33E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Explain why Newton’s method fails when applied to the equation x 3 = 0 with any initial approximation x1 ≠ 0. Illustrate your explanation with a sketch.

To determine

To find: The Newton’s method fails for the given equation with any approximation x10 with sketch the graph.

Explanation

Given:

The equation is x13=0.

The initial approximation is x1=0.5.

Result Used:

The Newton’s method is xn+1=xnf(xn)f(xn).

Graph:

From the Figure 1, it is observed that the value gets doubled.

Calculation:

Let f(x)=x13, then f(x)=0 by the given condition.

Calculate the derivative of f(x).

Let f(x)=df(x)dx.

f(x)=d(x13)dx=13x23

Substitute the value of f(x) and f(x) in the above mentioned result to obtain xn+1

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