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 9E
To determine

To Find: The second approximation x2.

Expert Solution

Answer to Problem 9E

The second approximation is x21.25.

Explanation of Solution

Formula used:

The Newton’s formula is,xn+1=xnf(xn)f(xn) where f(x) is the given function.

Given:

The initial approximation is x1=1.

The function is f(x)=x3+x+3.

Calculation:

Calculate the derivative of f(x).

f(x)=ddx(x3+x+3)=2ddx(x3)+ddx(x)+ddx(3)=3x2+1+0=3x2+1

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

f(1)=3(1)2+1=3(1)+1=3+1=4

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

f(1)=(1)3+(1)+3=11+3=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)=4,

x2=114x2=10.25x21.25

Therefore, the second approximation to the root of the given equation is x2=1.25.

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 9E

From Figure 1, it is observed that the Newton’s method achieve the tangent line at (1,1) and tangent line intersects the x-axis at the point (1.25,0).

Moreover, from the graph it is noticed that the tangent line gives the second approximation at x2=1.25.

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