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[[Category:Newton-Raphson method]]
 
[[Category:Newton-Raphson method]]
 
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[[Category: Mathematics questions]]
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[[Category: Calculus questions]]

Revision as of 17:25, 15 April 2014

Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is

(A) 3.575

(B) 3.676

(C) 3.667

(D) 3.607

Solution by Happy Mittal

We know that, according to Newton-Raphson method, $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ In this question $f(x) = x^2-13$, and so $f'(x) = 2x$.
Here current $x_n = 3.5$, $f(x_n) = f(3.5) = -0.75$, and $f'(x_n) = f'(3.5) = 7$. So $$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$ So option (D) is correct.



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Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is

(A) 3.575

(B) 3.676

(C) 3.667

(D) 3.607

Solution by Happy Mittal[edit]

We know that, according to Newton-Raphson method, $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ In this question $f(x) = x^2-13$, and so $f'(x) = 2x$.
Here current $x_n = 3.5$, $f(x_n) = f(3.5) = -0.75$, and $f'(x_n) = f'(3.5) = 7$. So $$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$ So option (D) is correct.



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