Calculus with Applications (10th Edition)

$$3.873$$
\eqalign{ & {\text{We can note that }}\sqrt {15} {\text{ is a solution of the equation }}{x^2} - 15 = 0.{\text{ }} \cr & {\text{let }}f\left( x \right) = {x^2} - 15,\,\,\,\,\,\,f'\left( x \right) = 2x \cr & {\text{Since 3 < }}\sqrt {15} < 4,{\text{ we can use }}{c_1} = 3{\text{ as the first approximation }} \cr & {\text{Then using Newton's method we have}}: \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 3 - \frac{{{{\left( 3 \right)}^2} - 15}}{{2\left( 3 \right)}} = 3.9196 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 3.9196 - \frac{{{{\left( {3.9196} \right)}^2} - 15}}{{2\left( {3.9196} \right)}} = 3.8732 \cr & {\text{In the same way}} \cr & {c_4} = 3.8732 - \frac{{{{\left( {3.8732} \right)}^2} - 15}}{{2\left( {3.8732} \right)}} = 3.8729 \cr & {c_5} = 3.8729 - \frac{{{{\left( {3.8729} \right)}^2} - 15}}{{2\left( {3.8729} \right)}} = 3.8729 \cr & {\text{Since }}{c_4} = {c_5} = 3.872,{\text{ to the nearest thousand}}{\text{, }}\sqrt {11} {\text{ is }}3.873 \cr}