## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 23

#### Answer

$$2.080$$

#### Work Step by Step

\eqalign{ & {\text{We can note that }}\root 3 \of 9 {\text{ is a solution of the equation }}{x^3} - 9 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^3} - 9,\,\,\,\,\,\,f'\left( x \right) = 3{x^2} \cr & {\text{Since 2 < }}\root 3 \of 9 < 3,{\text{ we can use }}{c_1} = 2{\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)}} = 2 - \frac{{{{\left( 2 \right)}^3} - 9}}{{3{{\left( 2 \right)}^2}}} = 2.0833 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 2.0833 - \frac{{{{\left( {2.0833} \right)}^3} - 9}}{{3{{\left( {2.0833} \right)}^2}}} = 2.0800 \cr & {\text{In the same way}} \cr & {c_4} = 2.0800 - \frac{{{{\left( {2.0800} \right)}^3} - 9}}{{3{{\left( {2.0800} \right)}^2}}} = 2.0800 \cr & {\text{Since }}{c_3} = {c_4} = 2.080,{\text{ to the nearest thousand}}{\text{, }}\root 3 \of 9 = 2.080 \cr}

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