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: 74

Answer

$$3.183$$

Work Step by Step

$$\eqalign{ & \root 4 \of {102.6} \cr & {\text{We can note that }}\root 4 \of {102.6} {\text{ is a solution of the equation }}{x^4} - 102.4 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^4} - 102.6,\,\,\,\,\,\,f'\left( x \right) = 4{x^3} \cr & \cr & {\text{Since 3 < }}\root 4 \of {102.6} < 4,{\text{ we can use }}{c_1} = 3{\text{ as the first approximation}}{\text{. }} \cr & {\text{Then using Newton's Method we have}}: \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 3 - \frac{{{{\left( 3 \right)}^4} - 102.6}}{{4{{\left( 3 \right)}^3}}} = 3.2 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 3.2 - \frac{{{{\left( {3.2} \right)}^4} - 102.6}}{{4{{\left( {3.2} \right)}^3}}} = 3.1827 \cr & {\text{In the same way}} \cr & {c_4} = 3.1827 - \frac{{{{\left( {3.1827} \right)}^4} - 102.6}}{{4{{\left( {3.1827} \right)}^3}}} = 3.1826 \cr & {c_5} = 3.1826 - \frac{{{{\left( {3.1826} \right)}^4} - 102.6}}{{4{{\left( {3.1826} \right)}^3}}} = 3.1826 \cr & \cr & {\text{Since }}{c_4} = {c_5} = 3.1826,{\text{ to the nearest thousand}}{\text{, }}\root 3 \of {102.6} {\text{ }} = {\text{ }}3.183 \cr} $$
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