## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 18

#### Answer

$$1.732$$

#### Work Step by Step

\eqalign{ & {\text{We can note that }}\sqrt 3 {\text{ is a solution of the equation }}{x^2} - 3 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^2} - 3,\,\,\,\,\,\,f'\left( x \right) = 2x \cr & {\text{Since 1 < }}\sqrt 3 < 2,{\text{ we can use }}{c_1} = 1{\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)}} = 1 - \frac{{{{\left( 1 \right)}^2} - 3}}{{2\left( 1 \right)}} = 2 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 2 - \frac{{{{\left( 2 \right)}^2} - 3}}{{2\left( 2 \right)}} = 1.75 \cr & {\text{In the same way}} \cr & {c_4} = 1.75 - \frac{{{{\left( {1.75} \right)}^2} - 3}}{{2\left( {1.75} \right)}} = 1.7321 \cr & {c_5} = 1.7321 - \frac{{{{\left( {1.7321} \right)}^2} - 3}}{{2\left( {1.7321} \right)}} = 1.7320 \cr & {\text{Since }}{c_4} = {c_5} = 1.732,{\text{ to the nearest thousand}}{\text{, }}\sqrt 3 = 1.732 \cr}

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