# The roots of the equation, x 4 = 1 + x .

### 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 13E
To determine

## To calculate: The roots of the equation,  x4=1+x .

Expert Solution

The roots are

x20.8

x30.717350

x40.724555 .

### Explanation of Solution

Given information:

The equation is given as:

x4=1+x .

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,

x4=1+x

Now,

x41x=0f(x)=x4x1f'(x)=4x31

Let the initial approximation x1=1

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1x14x114x131

x2=(1)((1)4(1)14(1)31)x2=(1)(1+1141)x2=(1)(15)x2=1+15x2=45

x2=0.8

The second approximation is x2=0.8 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2x24x214x231

x3=(0.8)((0.8)4(0.8)14(0.8)31)x3=(0.8)(0.4096+0.811.5361)x3=0.8(0.20962.536)

x3=0.8+0.082649842x3=0.717350157

The third approximation with six decimal places is x3=0.717350 .

Let n=3 ,

x4=x3f(x3)f'(x3)

x4=x3x34x314x331

x4=(0.717350)((0.717350)4(0.717350)14(0.717350)31)x4=(0.717350)(0.26480392+0.71735011.476567481)x4=0.717350(0.017846082.47656748)

x4=0.7173500.007205973649x4=0.724555973

The fourth approximation with six decimal places is x4=0.724555 .

Therefore, the roots with six decimal places are :-

x20.8

x30.717350

x40.724555

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