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When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 is
- a)-0.82
- b)0.49
- c)0,705
- d)1.69
Correct answer is option 'C'. Can you explain this answer?
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When the Newton-Raphson method is applied to solve the equation f(x) =...
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When the Newton-Raphson method is applied to solve the equation f(x) =...
Given equation: f(x) = x^3 - 2x - 1 = 0
Newton-Raphson method: The Newton-Raphson method is an iterative numerical method used to find the roots of a given equation. It starts with an initial guess value and improves the approximation of the root with each iteration.
Algorithm:
1. Start with an initial guess value x0.
2. Calculate the function value f(x0) and its derivative f'(x0) at x0.
3. Update the guess value using the formula:
x1 = x0 - f(x0)/f'(x0)
4. Repeat steps 2 and 3 until the desired level of accuracy is achieved.
Applying the Newton-Raphson method:
Given equation: f(x) = x^3 - 2x - 1 = 0
Step 1: Initial guess value: x0 = 1.2
Step 2: Calculate f(x0) and f'(x0):
f(x0) = (1.2)^3 - 2(1.2) - 1 = 1.728 - 2.4 - 1 = -1.672
f'(x0) = 3(1.2)^2 - 2 = 4.32 - 2 = 2.32
Step 3: Update the guess value using the formula:
x1 = x0 - f(x0)/f'(x0)
= 1.2 - (-1.672)/2.32
= 1.2 + 0.721
= 1.921
Therefore, the solution at the end of the first iteration with the initial guess value x0 = 1.2 is x1 = 1.921.
Conclusion:
The correct answer is option 'C' (0.705) which is not mentioned in the given options. However, based on the calculations, the solution at the end of the first iteration is x1 = 1.921.
Newton-Raphson method: The Newton-Raphson method is an iterative numerical method used to find the roots of a given equation. It starts with an initial guess value and improves the approximation of the root with each iteration.
Algorithm:
1. Start with an initial guess value x0.
2. Calculate the function value f(x0) and its derivative f'(x0) at x0.
3. Update the guess value using the formula:
x1 = x0 - f(x0)/f'(x0)
4. Repeat steps 2 and 3 until the desired level of accuracy is achieved.
Applying the Newton-Raphson method:
Given equation: f(x) = x^3 - 2x - 1 = 0
Step 1: Initial guess value: x0 = 1.2
Step 2: Calculate f(x0) and f'(x0):
f(x0) = (1.2)^3 - 2(1.2) - 1 = 1.728 - 2.4 - 1 = -1.672
f'(x0) = 3(1.2)^2 - 2 = 4.32 - 2 = 2.32
Step 3: Update the guess value using the formula:
x1 = x0 - f(x0)/f'(x0)
= 1.2 - (-1.672)/2.32
= 1.2 + 0.721
= 1.921
Therefore, the solution at the end of the first iteration with the initial guess value x0 = 1.2 is x1 = 1.921.
Conclusion:
The correct answer is option 'C' (0.705) which is not mentioned in the given options. However, based on the calculations, the solution at the end of the first iteration is x1 = 1.921.
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When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer?
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When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer?.
When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer?.
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When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer?, a detailed solution for When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 isa)-0.82b)0.49c)0,705d)1.69Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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