## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 5

#### Answer

$$2.24$$

#### Work Step by Step

\eqalign{ & - 3{x^3} + 5{x^2} + 3x + 2 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {2,3} \right] \cr & {\text{let }}f\left( x \right) = - 3{x^3} + 5{x^2} + 3x + 2,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( { - 3{x^3} + 5{x^2} + 3x + 2} \right) \cr & f'\left( x \right) = - 9{x^2} + 10x + 3 \cr & {\text{interval }}\left[ {2,3} \right]{\text{ then }}a = 2{\text{ and }}b = 3 \cr & {\text{check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 2 \right) = - 3{\left( 2 \right)^3} + 5{\left( 2 \right)^2} + 3\left( 2 \right) + 2 = 4 > 0 \cr & f'\left( 3 \right) = - 3{\left( 3 \right)^3} + 5{\left( 3 \right)^2} + 3\left( 3 \right) + 2 = - 25 < 0 \cr & f\left( 2 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{, then}} \cr & {\text{there is a solution for the equaion in the interval }}\left( {2,3} \right) \cr & {\text{As an initial guess }}{c_1} = 2 \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)}} = 2 - \frac{{ - 3{{\left( 2 \right)}^3} + 5{{\left( 2 \right)}^2} + 3\left( 2 \right) + 2}}{{ - 15{{\left( 2 \right)}^2} + 10\left( 2 \right) + 3}} = 2.3076 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 2.3076 - \frac{{ - 3{{\left( {2.3076} \right)}^3} + 5{{\left( {2.3076} \right)}^2} + 3\left( {2.3076} \right) + 2}}{{ - 15{{\left( {2.3076} \right)}^2} + 10\left( {2.3076} \right) + 3}} = 2.2473 \cr & {\text{In the same way}} \cr & {c_4} = 2.2473 - \frac{{ - 3{{\left( {2.2473} \right)}^3} + 5{{\left( {2.2473} \right)}^2} + 3\left( {2.2473} \right) + 2}}{{ - 15{{\left( {2.2473} \right)}^2} + 10\left( {2.2473} \right) + 3}} = 2.2445 \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 2.24 \cr}

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