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The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton-Raphson method. If x = 2 is taken as the initial approximation of the solution, the next approximation using this method will be (Answer up to two decimal places)
Correct answer is '1.33'. Can you explain this answer?
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The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton-Ra...
We have
f(x) = x3- x2 + 4x- 4
f ’(x) =3x2 -2x +4
Taking x0 = 2 in Newton-Raphosn method
=1.33
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The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton-Ra...
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 approximation and iteratively refines the solution until it converges to the desired accuracy. The method involves linearizing the equation and solving for the next approximation using the tangent line.
Given Equation
The given equation is x^3 - x^2 + 4x - 4 = 0.
Step 1: Derivative of the Equation
To apply the Newton-Raphson method, we need to find the derivative of the equation. Taking the derivative of the given equation, we get:
f'(x) = 3x^2 - 2x + 4
Step 2: Initial Approximation
The initial approximation provided is x = 2.
Step 3: Iterative Calculation
Using the Newton-Raphson formula, we can calculate the next approximation:
x1 = x0 - f(x0) / f'(x0)
where x1 is the next approximation, x0 is the initial approximation, f(x) is the given equation, and f'(x) is the derivative of the equation.
Plugging in the values, we have:
x1 = 2 - (2^3 - 2^2 + 4(2) - 4) / (3(2)^2 - 2(2) + 4)
Simplifying the equation, we get:
x1 = 2 - (8 - 4 + 8 - 4) / (12 - 4 + 4)
x1 = 2 - 8 / 12
x1 = 2 - 0.67
x1 = 1.33
Therefore, the next approximation using the Newton-Raphson method is 1.33.
Summary
- The Newton-Raphson method is an iterative numerical method used to find the roots of an equation.
- The given equation is x^3 - x^2 + 4x - 4 = 0.
- The derivative of the equation is f'(x) = 3x^2 - 2x + 4.
- The initial approximation provided is x = 2.
- Using the Newton-Raphson formula, we calculate the next approximation as x1 = 1.33.
- The Newton-Raphson method is an efficient method for finding roots of equations, but it requires an initial approximation close to the actual root for convergence.
The Newton-Raphson method is an iterative numerical method used to find the roots of a given equation. It starts with an initial approximation and iteratively refines the solution until it converges to the desired accuracy. The method involves linearizing the equation and solving for the next approximation using the tangent line.
Given Equation
The given equation is x^3 - x^2 + 4x - 4 = 0.
Step 1: Derivative of the Equation
To apply the Newton-Raphson method, we need to find the derivative of the equation. Taking the derivative of the given equation, we get:
f'(x) = 3x^2 - 2x + 4
Step 2: Initial Approximation
The initial approximation provided is x = 2.
Step 3: Iterative Calculation
Using the Newton-Raphson formula, we can calculate the next approximation:
x1 = x0 - f(x0) / f'(x0)
where x1 is the next approximation, x0 is the initial approximation, f(x) is the given equation, and f'(x) is the derivative of the equation.
Plugging in the values, we have:
x1 = 2 - (2^3 - 2^2 + 4(2) - 4) / (3(2)^2 - 2(2) + 4)
Simplifying the equation, we get:
x1 = 2 - (8 - 4 + 8 - 4) / (12 - 4 + 4)
x1 = 2 - 8 / 12
x1 = 2 - 0.67
x1 = 1.33
Therefore, the next approximation using the Newton-Raphson method is 1.33.
Summary
- The Newton-Raphson method is an iterative numerical method used to find the roots of an equation.
- The given equation is x^3 - x^2 + 4x - 4 = 0.
- The derivative of the equation is f'(x) = 3x^2 - 2x + 4.
- The initial approximation provided is x = 2.
- Using the Newton-Raphson formula, we calculate the next approximation as x1 = 1.33.
- The Newton-Raphson method is an efficient method for finding roots of equations, but it requires an initial approximation close to the actual root for convergence.
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The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton-Raphson method. If x = 2 is taken as the initial approximation of the solution, the next approximation using this method will be (Answer up to two decimal places)Correct answer is '1.33'. Can you explain this answer?
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