# The second approximation 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 10E
To determine

## To find: The second approximation x2.

Expert Solution

The second approximation is x21.3¯.

### Explanation of Solution

Formula used:

The newton’s formula is,xn+1=xnf(xn)f(xn)

Given:

The initial approximation is x1=1.

The function is f(x)=x4x1.

Calculation:

Calculate the derivative of f(x).

f(x)=ddx(x4x1)=2ddx(x4)ddx(x)ddx(1)=4x310=4x31

The value of f(x) at x1=1 is,

f(1)=4(1)31=4×11=3

Calculate the value of f(x) at x1=1

f(1)=(1)411=12=1

Calculate x2 by using Newton’s method.

xn+1=xnf(xn)f(xn).

Set n=1, and obtain the value of x2.

x2=x1f(x1)f(x1)

Substitute x1=1,f(x1)=1 and f(x1)=3,

x2=1(1)3=1+13=431.3¯

Therefore, the second approximation to the root of the given equation is x21.3¯.

Graph the given function f(x) and its tangent line at the point (1,1) as shown below in Figure 1.

From Figure 1, it is observed that the Newton’s method achieve the tangent line at (1,1) and it intersects x-axis at (1.3¯,0).

Moreover, from the graph, it is noticed that the tangent line gives the second approximation at x2=1.3¯.

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