Calculus with Applications (10th Edition)

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

Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 24

Answer

$$2.466$$

Work Step by Step

$$\eqalign{ & {\text{We can note that }}\root 3 \of {15} {\text{ is a solution of the equation }}{x^3} - 15 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^3} - 15,\,\,\,\,\,\,f'\left( x \right) = 3{x^2} \cr & {\text{Since 3 < }}\root 3 \of {15} < 4,{\text{ we can use }}{c_1} = 3{\text{ as the first approximation }} \cr & {\text{Then using Newton's method we have}}: \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 3 - \frac{{{{\left( 3 \right)}^3} - 15}}{{3{{\left( 3 \right)}^2}}} = 2.5555 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 2.5555 - \frac{{{{\left( {2.5555} \right)}^3} - 15}}{{3{{\left( {2.5555} \right)}^2}}} = 2.4692 \cr & {\text{In the same way}} \cr & {c_4} = 2.4692 - \frac{{{{\left( {2.4692} \right)}^3} - 15}}{{3{{\left( {2.4692} \right)}^2}}} = 2.4662 \cr & {c_5} = 2.4662 - \frac{{{{\left( {2.4662} \right)}^3} - 15}}{{3{{\left( {2.4662} \right)}^2}}} = 2.4662 \cr & {\text{Since }}{c_4} = {c_5} = 2.466,{\text{ to the nearest thousand}}{\text{, }}\root 3 \of {15} = 2.466 \cr} $$
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