Problem

If $ f(x) = \left\{ \begin{array}{ll}…

03:31

Problem 33 Hard Difficulty

Explain why Newton's method fails when applied to the equation $ \sqrt[3]{x} = 0 $ with any initial approximation $ x_1 \not= 0 $. Illustrate your explanation with a sketch.

Answer

not converge

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 8

Newton's Method

Discussion

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Watch More Solved Questions in Chapter 4

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Problem 32

Problem 33

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Problem 42

Video Transcript

question 33. We're looking at our function as extra 1/3 but a key root of X and a prime of X and, of course, is 1/3 x to the negative 2/3. So x to the n plus one is an equal x minus x to the 1/3 over 1/3 X to the negative 2/3. So simplify that I would have, um, one there x to the to certain so back to the I'm just simply crying this part right here, x and we're when they're x 2/3 I'm going to invert and multiply, So I'm gonna have X to the 1/3 times three x to the 2/3 over one. So I'm just gonna multiply them when I must play and the exponents. So I have three x. So now I have X minus three X, which gives me and negative to X and negative X will not converge

Sherrie F.

University of Houston

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 8

Newton's Method

Top Calculus 2 / BC Educators

Catherine R.

Missouri State University

Caleb E.

Baylor University

Michael J.

Idaho State University

Joseph L.

Boston College