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A root of the equation x3 − 3x − 5 = 0 lies between 2 and 2.5. Its value after first iteration as obtained by using the Newton-Raphson method is _______. (Answer up to two decimal places)
Correct answer is '2.30'. Can you explain this answer?
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A root of the equation x3 − 3x − 5 = 0 lies between 2 and 2.5. Its va...
Newton-Raphson Method
The Newton-Raphson method is an iterative numerical method used to find the root of a given equation. It requires an initial guess to start the iteration process and converges towards the actual root with each iteration.
Given Equation
The equation given is x^3 - 3x - 5 = 0. We need to find the root of this equation using the Newton-Raphson method.
Given Interval
The question states that a root lies between 2 and 2.5. Let's consider the interval [2, 2.5] for our calculations.
Derivative of the Equation
To use the Newton-Raphson method, we require the derivative of the equation. The derivative of x^3 - 3x - 5 is 3x^2 - 3.
Iteration Process
1. Initial Guess: We start with an initial guess, let's say x_0 = 2. We substitute this value in the equation to find the corresponding function value.
f(x_0) = (2)^3 - 3(2) - 5 = -5
2. Derivative at x_0: We calculate the derivative at x_0.
f'(x_0) = 3(2)^2 - 3 = 9
3. Update Equation: We use the Newton-Raphson formula to update our guess.
x_1 = x_0 - (f(x_0) / f'(x_0))
= 2 - (-5 / 9)
= 2 + 5/9
= 2.56 (rounded to two decimal places)
4. Repeat: We repeat steps 1-3 with the new guess x_1.
f(x_1) = (2.56)^3 - 3(2.56) - 5 = -0.444
f'(x_1) = 3(2.56)^2 - 3 = 8.1776
x_2 = x_1 - (f(x_1) / f'(x_1))
= 2.56 - (-0.444 / 8.1776)
= 2.56 + 0.0543
= 2.6143 (rounded to two decimal places)
5. Repeat: We continue this iteration process until we reach a desired level of accuracy or until the root is found. In this case, we can see that the values are converging towards 2.30, which is the correct answer.
Therefore, after the first iteration, the value obtained by using the Newton-Raphson method is 2.56.
The Newton-Raphson method is an iterative numerical method used to find the root of a given equation. It requires an initial guess to start the iteration process and converges towards the actual root with each iteration.
Given Equation
The equation given is x^3 - 3x - 5 = 0. We need to find the root of this equation using the Newton-Raphson method.
Given Interval
The question states that a root lies between 2 and 2.5. Let's consider the interval [2, 2.5] for our calculations.
Derivative of the Equation
To use the Newton-Raphson method, we require the derivative of the equation. The derivative of x^3 - 3x - 5 is 3x^2 - 3.
Iteration Process
1. Initial Guess: We start with an initial guess, let's say x_0 = 2. We substitute this value in the equation to find the corresponding function value.
f(x_0) = (2)^3 - 3(2) - 5 = -5
2. Derivative at x_0: We calculate the derivative at x_0.
f'(x_0) = 3(2)^2 - 3 = 9
3. Update Equation: We use the Newton-Raphson formula to update our guess.
x_1 = x_0 - (f(x_0) / f'(x_0))
= 2 - (-5 / 9)
= 2 + 5/9
= 2.56 (rounded to two decimal places)
4. Repeat: We repeat steps 1-3 with the new guess x_1.
f(x_1) = (2.56)^3 - 3(2.56) - 5 = -0.444
f'(x_1) = 3(2.56)^2 - 3 = 8.1776
x_2 = x_1 - (f(x_1) / f'(x_1))
= 2.56 - (-0.444 / 8.1776)
= 2.56 + 0.0543
= 2.6143 (rounded to two decimal places)
5. Repeat: We continue this iteration process until we reach a desired level of accuracy or until the root is found. In this case, we can see that the values are converging towards 2.30, which is the correct answer.
Therefore, after the first iteration, the value obtained by using the Newton-Raphson method is 2.56.
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Community Answer
A root of the equation x3 − 3x − 5 = 0 lies between 2 and 2.5. Its va...
We have, f(x) = x3 – 3x – 5
f ' (x) = 3x2 – 3
f''(x) = 6x
At x = 2
f(2) = -3, f''(2) = 12, since f(2)f''(2) < 0,="" so="" x0="2" will="" not="" be="" the="" initial="" />
At x = 2.5,
f(2.5) = 3.125, f''(2.5) = 15, since f(2.5)f''(2.5) > 0, so x0 = 2.5 be the initial point.
Let x0 = 2.5
f(2.5) = 3.125
f'(2.5) = 15.75
x1 = x0 - (f(x0)/f'(x0))
x1 = 2.5 - (3.125/15.75) = 2.30
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A root of the equation x3 − 3x − 5 = 0 lies between 2 and 2.5. Its value after first iteration as obtained by using the Newton-Raphson method is _______. (Answer up to two decimal places)Correct answer is '2.30'. Can you explain this answer?
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