Find the third approximation x3 to the root of the equation f(x) = x3 – 2x – 7, setting x1 = 2

Question

Find the third approximation x3 to the root of the equation f(x) = x3 &#8211; 2x &#8211; 7, setting x1 = 2

Suresh Chand · Accepted Answer

Solution: Given f(x) = x3 – 2x – 7 = 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{-5}{25} = 2.2) Then x3 = (2.1 – frac{f(2.2)}{f^{‘}(2.2)} = 2 + frac{0.752}{12.52} = 2.06) Therefore, The value of third approximation x3 is 2.06

Find the third approximation x₃ to the root of the equation f(x) = x³ – 2x – 7, setting x₁ = 2.

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Find the third approximation x3 to the root of the equation f(x) = x3 – 2x – 7, setting x1 = 2.

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Find the third approximation x₃ to the root of the equation f(x) = x³ – 2x – 7, setting x₁ = 2.