Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{4.000000} \\
1&{3.250000} \\
2&{3.163462} \\
3&{3.162278} \\
4&{3.162278} \\
5&{3.162278} \\
6&{3.162278} \\
7&{3.162278} \\
8&{3.162278} \\
9&{3.162278} \\
{10}&{3.162278}
\end{array}}\]
Work Step by Step
\[\begin{gathered}
{\text{Let }}f\left( x \right) = {x^2} - 10,{\text{ and }}{x_0} = 4 \hfill \\
{\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \hfill \\
f'\left( x \right) = 2{x_n} \hfill \\
{\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \hfill \\
{x_{n + 1}} = {x_n} - \frac{{x_n^2 - 10}}{{2{x_n}}} \hfill \\
\hfill \\
{\text{Taking }}{x_0} = 4 \hfill \\
{x_0} = 4 \hfill \\
{x_{0 + 1}} = {x_1} = 4 - \frac{{{{\left( 4 \right)}^2} - 10}}{{2\left( 4 \right)}} = 3.25 \hfill \\
{x_{1 + 1}} = {x_2} = 3.25 - \frac{{{{\left( {3.25} \right)}^2} - 10}}{{2\left( {3.25} \right)}} \approx 3.163462 \hfill \\
{x_{1 + 2}} = {x_3} = 3.163462 - \frac{{{{\left( {3.163462} \right)}^2} - 10}}{{2\left( {3.163462} \right)}} \approx 3.162277 \hfill \\
{x_{1 + 3}} = {x_4} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
{x_{1 + 4}} = {x_5} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
{x_{1 + 5}} = {x_6} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
{x_{1 + 6}} = {x_7} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
{x_{1 + 7}} = {x_8} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
{x_{1 + 8}} = {x_9} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
{x_{1 + 9}} = {x_{10}} = 3.162278 - \frac{{{{\left( {3.162278} \right)}^2} - 10}}{{2\left( {3.162278} \right)}} \approx 3.162278 \hfill \\
\hfill \\
{\text{Thus}}{\text{, we obtain}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{4.000000} \\
1&{3.250000} \\
2&{3.163462} \\
3&{3.162278} \\
4&{3.162278} \\
5&{3.162278} \\
6&{3.162278} \\
7&{3.162278} \\
8&{3.162278} \\
9&{3.162278} \\
{10}&{3.162278}
\end{array}} \hfill \\
\end{gathered} \]
