Stating with x1 = 2, find the third approximation x3 to the root of the equation x3 – 2x – 5 = 0

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Stating with x1 = 2, find the third approximation x3 to the root of the equation x3 &#8211; 2x &#8211; 5 = 0

Suresh Chand · Accepted Answer

Solution: Given f(x) = x3 - 2x - 5 = 0 f'(x) = 3x2 - 2 By Newton's method we have (x_{n+1} = x_n - frac{f(x_n)}{f^{'}(x_n)}) When x1 = 2 x2 = (2 - frac{f(2)}{f^{'}(2)} = 2 + frac{1}{10} = 2.1) Then x3 = (2.1 - frac{f(2.1)}{f^{'}(2.1)} = 2 + frac{0.061}{11.23} = 2.096) Therefore, The value of third approximation x3 is 2.096

Stating with x₁ = 2, find the third approximation x₃ to the root of the equation x³ – 2x – 5 = 0

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Stating with x1 = 2, find the third approximation x3 to the root of the equation x3 – 2x – 5 = 0

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Stating with x₁ = 2, find the third approximation x₃ to the root of the equation x³ – 2x – 5 = 0