   Chapter 4.8, Problem 34E ### 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

# If f ( x ) = { x if   x   ≥   0 − − x if   x < 0 then the root of the equation f(x) = 0 is x = 0. Explain why Newton’s method fails to find the root no matter which initial approximation x1 ≠ 0 is used. Illustrate your explanation with a sketch.

To determine

To Explain: The Newton’s method fails for the given equation with any approximation x10 by using the graph.

Explanation

Result Used:

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

Given:

The equation is f(x)={x       ifx0x  if x<0

Graph:

Calculation:

Calculate the derivative of f(x) when x>0.

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

f(x)=d(x12)dx=12x12

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

xn+1=xnxn1212xn12=xn2xn=xn

Calculate the derivative of f(x) when x<0

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