The Newton-Raphson method is used to solve the equation $f (x) = x^{3} - 5 x^{2} + 6 x - 8 = 0$ . Taking the initial guess as x = 5, The solution obtained at the end of the first iteration is
4.2903

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Solution

The correct option is A 4.2903
$f (x) = x^{3} - 5 x^{2} + 6 x - 8$
$f^{'} (x) = 3 x^{3} - 10 x + 6$
At $x_{0} = 5, f (x_{0}) = f (5) = 22$ &
$f^{'} (x_{0}) = f^{'} (5) = 31$
By Newton-Raphson Method
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})} = 5 - \frac{22}{31}$
= 4.2903

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The Newton-Raphson method is used to solve the equation f(x)=x3−5x2+6x−8=0. Taking the initial guess as x = 5, The solution obtained at the end of the first iteration is4.2903

The correct option is A 4.2903 f(x)=x3−5x2+6x−8 f′(x)=3x3−10x+6 At x0=5,f(x0)=f(5)=22 & f′(x0)=f′(5)=31 By Newton-Raphson Method x1=x0−f(x0)f′(x0)=5−2231 = 4.2903

The Newton-Raphson method is used to solve the equation $f (x) = x^{3} - 5 x^{2} + 6 x - 8 = 0$ . Taking the initial guess as x = 5, The solution obtained at the end of the first iteration is
4.2903

The correct option is A 4.2903
$f (x) = x^{3} - 5 x^{2} + 6 x - 8$
$f^{'} (x) = 3 x^{3} - 10 x + 6$
At $x_{0} = 5, f (x_{0}) = f (5) = 22$ &
$f^{'} (x_{0}) = f^{'} (5) = 31$
By Newton-Raphson Method
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})} = 5 - \frac{22}{31}$
= 4.2903