#### Answer

$$4.946$$

#### Work Step by Step

$$\eqalign{
& \root 3 \of {121} \cr
& {\text{We can note that }}\root 3 \of {121} {\text{ is a solution of the equation }}{x^3} - 121 = 0. \cr
& {\text{Let }}f\left( x \right) = {x^3} - 121,\,\,\,\,\,\,f'\left( x \right) = 3{x^2} \cr
& {\text{Since 4 < }}\root 3 \of {121} < 5,{\text{ we can use }}{c_1} = 4{\text{ as the first approximation }} \cr
& {\text{Then using Newton's method we have}}: \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 4 - \frac{{{{\left( 4 \right)}^3} - 121}}{{3{{\left( 4 \right)}^2}}} = 5.1875 \cr
& {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 5.1875 - \frac{{{{\left( {5.1875} \right)}^3} - 121}}{{3{{\left( {5.1875} \right)}^2}}} = 4.9571 \cr
& {\text{In the same way}} \cr
& {c_4} = 4.9571 - \frac{{{{\left( {4.9571} \right)}^3} - 121}}{{3{{\left( {4.9571} \right)}^2}}} = 4.9461 \cr
& {c_5} = 4.9461 - \frac{{{{\left( {4.9461} \right)}^3} - 121}}{{3{{\left( {4.9461} \right)}^2}}} = 4.9460 \cr
& {\text{Since }}{c_4} = {c_5} = 4.946,{\text{ to the nearest thousand}}{\text{, }}\root 3 \of {121} = 4.946 \cr} $$