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 22E
To determine

To calculate: The roots of the equation,

  earctanx=x3+1 .

Expert Solution

Answer to Problem 22E

The roots are

  x2=1.24593328

  x3=2.49215474 .

Explanation of Solution

Given information:

The equation is given as:

  earctanx=x3+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,

  earctanx=x3+1

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

  y=earctanx and y=x3+1 .

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 22E

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

Let x1=0.918 where

  f(x)=earctanxx3+1f'(x)=earctanx1+x232(x2x3+1)

Now, at n=1

  x2=x1f(x1)f'(x1)

  x2=x1{earctanx1x13+1earctanx11+x1232(x12x13+1)}

  x2=(0.918)(earctan(0.918)(0.918)3+1earctan(0.918)1+(0.918)232((0.918)2(0.918)3+1))x2=(0.918)(0.4758410940.8427239990.2582270021.377000002)x2=(0.918)(0.3668829051.118773)

  x2=0.918(0.327933284)x2=0.9180.327933284x2=1.245933285

The second approximation is x2=1.24593328 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x2{earctanx2x23+1earctanx21+x2232(x22x23+1)}

  x3=(1.24593328)(earctan(1.24593328)(1.24593328)3+1earctan(1.24593328)1+(1.24593328)232((1.24593328)2(1.24593328)3+1))x3=(1.24593328)(0.4088261731.7129285450.3129370961.359382219)x3=(1.24593328)(1.3041023721.046445123)

  x3=1.24593328(1.246221463)x3=1.245933281.246221463x3=2.492154743

The third approximation with six decimal places is x3=2.49215474 .

Therefore, the roots with eight decimal places are :-

  x2=1.24593328

  x3=2.49215474

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