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 19E
To determine

To calculate: The roots of the equation,

  x22xx2=1 .

Expert Solution

Answer to Problem 19E

The roots are

  x2=1.31317864

  x3=0.72292771 .

Explanation of Solution

Given information:

The equation is given as:

  x22xx2=1 .

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,

  x22xx2=1

The initial approximation x1 can be guessed by sketching both the graphs

  y=x22xx2 and y=1 .

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 19E

Both the graph intersect at two points whose x-coordinate can be seen in graph.

Let x1=1.978 where

  f(x)=x22xx21f'(x)=x2[122xx2(12x)]+2x(2xx2)

  =x22x322xx2+2x(2xx2)

Now, at n=1

  x2=x1f(x1)f'(x1)

  x2=x1{x122x1x121x122x1322x1x12+2x1(2x1x12)}

  x2=(1.978)((1.978)22(1.978)(1.978)21(1.978)22(1.978)322(1.978)(1.978)2+2(1.978)(2(1.978)(1.978)2))x2=(1.978)(3.0355497633.912484+15.47778671.040540875+2.058189851)x2=(1.978)(3.03554976311.11470292+2.058189851)x2=(1.978)(3.0355497639.056513069)

  x2=1.978(0.335178643)x2=1.9780.335178643x2=1.313178643

The second approximation is x2=1.31317864 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x2{x222x2x221x222x2322x2x22+2x2(2x2x22)}

  x3=(1.31317864)(0.72953222371.724438141+4.5289906662.005908123+(2.634115701))x3=(1.31317864)(0.72953222371.3981460532.634115701)x3=(1.31317864)(0.72953222371.235969648)

  x3=1.31317864(0.590250921)x3=1.31317864+0.590250921x3=0.722927718

The third approximation with six decimal places is x3=0.72292771 .

Therefore, the roots with eight decimal places are :-

  x2=1.31317864

  x3=0.72292771

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