# The roots of the equation, 1 x = 1 + x 2 .

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 4.7, Problem 16E
To determine

## To calculate: The roots of the equation,  1x=1+x2 .

Expert Solution

The roots are

x2=0.666666

x3=0.682170

x4=0.682328

### Explanation of Solution

Given information:

The equation is given as:

1x=1+x2 .

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,

1x=1+x2

Now,

1x=1+x2f(x)=x21x+1f'(x)=2x+1x2

Let the initial approximation x1=1

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1x121x1+12x1+1x12

x2=(1)((1)211+12(1)+11)x2=(1)(11+12+1)x2=(1)(13)

x2=113x2=23x2=0.666666666

The second approximation is x2=0.666666 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2x221x2+12x2+1x22

x3=(0.666666)((0.666666)210.666666+12(0.666666)+1(0.666666)2)x3=(0.666666)(0.4444435551.5000015+11.333332+2.250004503)x3=0.666666(0.055557953.583336503)

x3=0.666666+0.01550453x3=0.68217053

The third approximation with six decimal places is x3=0.682170 .

Let n=3 ,

x4=x3f(x3)f'(x3)

x4=x3x321x3+12x3+1x32

x4=(0.682170)((0.682170)210.682170+12(0.682170)+1(0.682170)2)x4=(0.682170)(0.4653559081.465910257+11.36434+2.148892886)x4=0.682170(0.00055543493.513232886)

x4=0.682170+0.0001580979451x4=0.682328097

The fourth approximation with six decimal places is x4=0.682328 .

Therefore, the roots with six decimal places are :-

x2=0.666666

x3=0.682170

x4=0.682328

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