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: 69

Answer

$$2.65$$

Work Step by Step

$$\eqalign{ & {x^4} + 3{x^3} - 4{x^2} - 21x - 21 = 0;\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {2,3} \right] \cr & {\text{let }}f\left( x \right) = {x^4} + 3{x^3} - 4{x^2} - 21x - 21.{\text{ So that}}{\text{,}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{x^4} + 3{x^3} - 4{x^2} - 21x - 21} \right) \to f'\left( x \right) = 4{x^3} + 9{x^2} - 8x - 21 \cr & \cr & {\text{Interval }}\left[ {2,3} \right]{\text{ then }}a = 2{\text{ and }}b = 3.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 2 \right) = {\left( 2 \right)^4} + 3{\left( 2 \right)^3} - 4{\left( 2 \right)^2} - 21\left( 2 \right) - 21 = - 39 < 0 \cr & f\left( 3 \right) = {\left( 3 \right)^4} + 3{\left( 3 \right)^3} - 4{\left( 3 \right)^2} - 21\left( 3 \right) - 21 = 42 < 0 \cr & f\left( 2 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr & {\text{There is a solution for the equation in the interval }}\left( {2,3} \right) \cr & \cr & {\text{As an initial guess }}{c_1} = 2.{\text{ A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr & {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 2 - \frac{{{{\left( 2 \right)}^4} + 3{{\left( 2 \right)}^3} - 4{{\left( 2 \right)}^2} - 21\left( 2 \right) - 21}}{{4{{\left( 2 \right)}^3} + 9{{\left( 2 \right)}^2} - 8\left( 2 \right) - 21}} \approx 3.258 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 3.258 - \frac{{{{\left( {3.258} \right)}^4} + 3{{\left( {3.258} \right)}^3} - 4{{\left( {3.258} \right)}^2} - 21\left( {3.258} \right) - 21}}{{4{{\left( {3.258} \right)}^3} + 9{{\left( {3.258} \right)}^2} - 8\left( {3.258} \right) - 21}} \approx 2.805 \cr & {\text{In the same way}} \cr & {c_4} = 2.805 - \frac{{{{\left( {2.805} \right)}^4} + 3{{\left( {2.805} \right)}^3} - 4{{\left( {2.805} \right)}^2} - 21\left( {2.805} \right) - 21}}{{4{{\left( {2.805} \right)}^3} + 9{{\left( {2.805} \right)}^2} - 8\left( {2.805} \right) - 21}} \approx 2.6602 \cr & {c_5} = 2.660 - \frac{{{{\left( {2.6602} \right)}^4} + 3{{\left( {2.6602} \right)}^3} - 4{{\left( {2.6602} \right)}^2} - 21\left( {2.6602} \right) - 21}}{{4{{\left( {2.6602} \right)}^3} + 9{{\left( {2.6602} \right)}^2} - 8\left( {2.6602} \right) - 21}} \approx 2.6458 \cr & {c_6} = 2.6458 - \frac{{{{\left( {2.6458} \right)}^4} + 3{{\left( {2.6458} \right)}^3} - 4{{\left( {2.6458} \right)}^2} - 21\left( {2.6458} \right) - 21}}{{4{{\left( {2.6458} \right)}^3} + 9{{\left( {2.6458} \right)}^2} - 8\left( {2.6458} \right) - 21}} \approx 2.646 \cr & \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}}{\text{. A solution for the given interval is }}x \approx 2.65 \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.