If $f (x) = x^{3} - 9 x + 1$ and $x_{0} = 3$ , then using Newton Raphson method, first iteration is

\frac{53}{18}

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\frac{53}{9}

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\frac{35}{18}

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\frac{35}{9}

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Solution

The correct option is A $\frac{53}{18}$
Given, $f (x) = x^{3} - 9 x + 1$
$f^{'} (x) = 3 x^{2} - 9$
$f (3) = 3^{3} - 9 (3) + 1$
$= 27 - 27 + 1 = 1$
and $f^{'} (3) = 3 {(3)}^{2} - 9$
$= 27 - 9 = 18$
By Newton Raphson method
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$∴ x_{1} = 3 - \frac{f (3)}{f^{'} (3)}$
$= 3 - \frac{1}{18} = \frac{54 - 1}{18} = \frac{53}{18}$

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