BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4.7, Problem 6E
To determine

To calculate: The third approximation to the root of equation,

13x3+12x2+3=0 where x1=3 .

Expert Solution

Answer to Problem 6E

The third approximation is x3=2.7186 .

Explanation of Solution

Given information:

The equation is given as:

  13x3+12x2+3=0

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

  xn+1=xnf(xn)f'(xn) and so on.

Calculation:

Consider the equation,

13x3+12x2+3=0 and x1=3

This implies −

  f(x)=13x3+12x2+3f'(x)=x2+x

Now, let n=1 ,

  x2=x1f(x1)f'(x1)

  x2=x113x13+12x12+3x12+x1

  x2=(3)(13(3)3+12(3)2+3(3)23)x2=(3)(9+92+393)x2=(3)(18+9+626)x2=(3)(326)x2=(3)(32×16)

  x2=(3)(14)x2=3+14x2=12+14x2=114x2=2.75

The second approximation is x2=2.75 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x213x23+12x22+3x22+x2

  x3=(2.75)(13(2.75)3+12(2.75)2+3(2.75)2+(2.75))x3=(2.75)(6.932291+3.78125+37.56252.75)x3=(2.75)(6.932291+6.781254.8125)

  x3=2.75(0.1510414.8125)x3=2.75+0.031385x3=2.718615

The third approximation with four decimal places is x3=2.7186 .

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