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

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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

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