## Calculus with Applications (10th Edition)

$$4.558$$
\eqalign{ & \root 3 \of {94.7} \cr & {\text{We can note that }}\root 3 \of {94.7} {\text{ is a solution of the equation }}{x^3} - 94.7 = 0.{\text{ Then}} \cr & {\text{let }}f\left( x \right) = {x^3} - 94.7,\,\,\,\,\,\,f'\left( x \right) = 3{x^2} \cr & \cr & {\text{Since 4 < }}\root 3 \of {94.7} < 5,{\text{ we can use }}{c_1} = 4{\text{ as the first approximation}}{\text{. }} \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} - 94.7}}{{3{{\left( 4 \right)}^2}}} = 4.6395 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 4.6395 - \frac{{{{\left( {4.6395} \right)}^3} - 94.7}}{{3{{\left( {4.6395} \right)}^2}}} = 4.5595 \cr & {\text{In the same way}} \cr & {c_4} = 4.5595 - \frac{{{{\left( {4.5595} \right)}^3} - 94.7}}{{3{{\left( {4.5595} \right)}^2}}} = 4.5580 \cr & {c_5} = 4.5580 - \frac{{{{\left( {4.5580} \right)}^3} - 94.7}}{{3{{\left( {4.5580} \right)}^2}}} = 4.5580 \cr & \cr & {\text{Since }}{c_4} = {c_5} = 94.7,{\text{ to the nearest thousand}}{\text{, }}\root 3 \of {94.7} = 4.558 \cr}