# The approximation of the given number, 20 5 .

### 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 11E
To determine

## To calculate: The approximation of the given number,  205 .

Expert Solution

The approximation is x3=1.6165 .

### Explanation of Solution

Given information:

The equation is given as:

205 .

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 number,

205

The number can be expressed as : −

x520=0f(x)=x520f'(x)=5x4

Let the initial approximation x1=1

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1x15205x14

x2=(1)((1)5205(1)4)x2=(1)(1205)x2=(1)(195)x2=1+195x2=245

x2=4.8

The second approximation is x2=4.8 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2x25205x24

x3=(4.8)((4.8)5205(4.8)4)x3=(4.8)(2548.03968202654.208)x3=4.8(2528.039682654.208)

x3=4.80.952464795x3=3.847535204

The third approximation with eight decimal places is x3=3.84753520 .

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