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The equation $\mathrm{f}(\mathrm{x})$ is given as $\mathrm{x}^{2}-4=0$. Considering the initial approximation at $\mathrm{x}=6$ then the value of $\mathrm{x}_{1}$ is given as
a) $10 / 3$
b) $4 / 3$
c) $7 / 3$
d) $13 / 3$

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(a)

Explanation:

Iterative formula for Newton Raphson method is given by $\mathrm{x}(1)=\mathrm{x}(0)+\dfrac{f(x(0))}{f^{\prime} x(x(0))}$
Hence $x_{0}=6$ (initial guess), $f\left(x_{0}\right)=32$ and $f^{\prime}\left(x_{0}\right)=12$. Substituting the values in the equation we get $x_{1}=10 / 3$.

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