Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 71

Answer

$$6.132$$

Work Step by Step

$$\eqalign{ & {\text{We can note that }}\sqrt {37.6} {\text{ is a solution of the equation }}{x^2} - 37.6 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^2} - 37.6,\,\,\,\,\,\,f'\left( x \right) = 2x \cr & {\text{Since 6 < }}\sqrt {37.6} < 7,{\text{ we can use }}{c_1} = 6{\text{ as the first approximation}}{\text{. }} \cr & {\text{Then using Newton's Method we have}}: \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 6 - \frac{{{{\left( 6 \right)}^2} - 37.6}}{{2\left( 6 \right)}} = 6.1333 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 6.1333 - \frac{{{{\left( {6.1333} \right)}^2} - 37.6}}{{2\left( {6.1333} \right)}} = 6.1318 \cr & {\text{In the same way}} \cr & {c_4} = 6.1318 - \frac{{{{\left( {6.1318} \right)}^2} - 37.6}}{{2\left( {6.1318} \right)}} = 6.1318 \cr & {\text{Since }}{c_3} = {c_4} = 6.1318,{\text{ to the nearest thousand}}{\text{, }}\sqrt {37.6} = 6.132 \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.