The approximated value up to four decimal places

Question

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Answer 1

Question

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

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Answer 2

Question

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

Answer 6

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

Answer 7

Chapter 4.8, Problem 1E

To determine

The approximated value up to four decimal places

Answer 8

Expert Solution & Answer

Answer to Problem 1E

23=1.2599

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The given expression is

23

Calculation:

The given problem is equivalent to that of approximating the positive real zero r of f(x)=x3−2

The graph of ‘f’ can be sketched as

Calculus : The Classic Edition (with Make the Grade and Infotrac), Chapter 4.8, Problem 1E

Since f(1)=−1 and f(2)=6 , it follows from the continuity of f that 1 < r < 2. Moreover, since f is increasing, there can be only one zero in the open interval (1, 2).

If xn is the any approximation to r, then by Newton’s method, the next approximation is given by

xn+1=xn−f( x n )f'( x n )=xn−xn3−23xn2(since f'(x) = 3x2 )

Let us choose x1=1 as the first approximation and proceed as follows:

x2=1− ( 1 )3−23 ( 1 )2≈1.3333x3=1.3333− ( 1.3333 )3−23 ( 1.3333 )2≈1.2638x4=1.2638− ( 1.2638 )3−23 ( 1.2638 )2≈1.2599x5=1.2599− ( 1.2599 )3−23 ( 1.2599 )2≈1.2599

Since, two consecutive values of xn are same, therefore, 23=1.2599 . It can be shown to eight decimal places

23=1.25992105

Answer 12

Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Prob. 3E Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E

The approximated value up to four decimal places

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The approximated value up to four decimal places

Answer to Problem 1E

Explanation of Solution

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Chapter 4.8 Solutions