## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 3

#### Answer

$$3.06$$

#### Work Step by Step

\eqalign{ & 2{x^3} - 6{x^2} - x + 2 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {3,4} \right] \cr & {\text{let }}f\left( x \right) = 2{x^3} - 6{x^2} - x + 2,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {2{x^3} - 6{x^2} - x + 2} \right) \cr & f'\left( x \right) = 6{x^2} - 12x - 1 \cr & {\text{interval }}\left[ {3,4} \right]{\text{ then }}a = 3{\text{ and }}b = 4 \cr & {\text{check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 3 \right) = 2{\left( 3 \right)^3} - 6{\left( 3 \right)^2} - \left( 3 \right) + 2 = - 1 < 0 \cr & f'\left( 4 \right) = 2{\left( 4 \right)^3} - 6{\left( 4 \right)^2} - \left( 4 \right) + 2 = 30 > 0 \cr & f\left( 3 \right){\text{ and }}f\left( 4 \right){\text{ have opposite signs}}{\text{, then}} \cr & {\text{there is a solution for the equaion in the interval }}\left( {3,4} \right) \cr & {\text{As an initial guess }}{c_1} = 3 \cr & {\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 & {\text{then }} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 3 - \frac{{2{{\left( 3 \right)}^3} - 6{{\left( 3 \right)}^2} - \left( 3 \right) + 2}}{{6{{\left( 3 \right)}^2} - 12\left( 3 \right) - 1}} = 3.0588 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 3.0588 - \frac{{2{{\left( {3.0588} \right)}^3} - 6{{\left( {3.0588} \right)}^2} - \left( {3.0588} \right) + 2}}{{6{{\left( {3.0588} \right)}^2} - 12\left( 3 \right) - 1}} = 3.0565 \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}} \cr & {\text{A solution for the given interval is }}x \approx 3.06 \cr}

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