BuyFind

Single Variable Calculus: Concepts...

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

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

Answer to Problem 10E

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.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 10E

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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