# The third approximation to the root of equation, x 5 + 2 = 0 where x 1 = − 1 .

### 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 8E
To determine

## To calculate: The third approximation to the root of equation,x5+2=0 where x1=−1 .

Expert Solution

The third approximation is x3=1.6165 .

### Explanation of Solution

Given information:

The equation is given as:

x5+2=0

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,

x5+2=0 and x1=1

This implies −

f(x)=x5+2f'(x)=5x4

Now, let n=1 ,

x2=x1f(x1)f'(x1)

x2=x1x15+25x14

x2=(1)((1)5+25(1)4)x2=(1)(1+25)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=x2x25+25x24

x3=(0.8)((0.8)5+25(0.8)4)x3=(0.8)(0.32768+22.048)x3=0.8(1.672322.048)

x3=0.8(0.8165625)x3=0.80.8165625x3=1.6165625

The third approximation with four decimal places is x3=1.6165 .

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