Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 4

$$2.69$$

$$\eqalign{ & - {x^3} + 4{x^2} - 5x + 4 = 0;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{in the interval }}\left[ {2,3} \right] \cr & {\text{let }}f\left( x \right) = - {x^3} + 4{x^2} - 5x + 4,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( { - {x^3} + 4{x^2} - 5x + 4} \right) \cr & f'\left( x \right) = - 3{x^2} + 8x - 5 \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) = - {\left( 2 \right)^3} + 4{\left( 2 \right)^2} - 5\left( 2 \right) + 4 = 2 < 0 \cr & f'\left( 3 \right) = - {\left( 3 \right)^3} + 4{\left( 3 \right)^2} - 5\left( 3 \right) + 4 = - 2 > 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{{ - {{\left( 2 \right)}^3} + 4{{\left( 2 \right)}^2} - 5\left( 2 \right) + 4}}{{ - 3{{\left( 2 \right)}^2} + 8\left( 2 \right) - 5}} = 4 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 4 - \frac{{ - {{\left( 4 \right)}^3} + 4{{\left( 4 \right)}^2} - 5\left( 4 \right) + 4}}{{ - 3{{\left( 4 \right)}^2} + 8\left( 4 \right) - 5}} = 3.238 \cr & {\text{In the same way}} \cr & {c_4} = 3.238 - \frac{{ - {{\left( {3.238} \right)}^3} + 4{{\left( {3.238} \right)}^2} - 5\left( {3.238} \right) + 4}}{{ - 3{{\left( {3.238} \right)}^2} + 8\left( {3.238} \right) - 5}} = 2.839 \cr & {c_5} = 2.839 - \frac{{ - {{\left( {2.839} \right)}^3} + 4{{\left( {2.839} \right)}^2} - 5\left( {2.839} \right) + 4}}{{ - 3{{\left( {2.839} \right)}^2} + 8\left( {2.839} \right) - 5}} = 2.7095 \cr & {c_5} = 2.7095 - \frac{{ - {{\left( {2.7095} \right)}^3} + 4{{\left( {2.7095} \right)}^2} - 5\left( {2.7095} \right) + 4}}{{ - 3{{\left( {2.7095} \right)}^2} + 8\left( {2.7095} \right) - 5}} = 2.6957 \cr & {c_6} = 2.6957 - \frac{{ - {{\left( {2.6957} \right)}^3} + 4{{\left( {2.6957} \right)}^2} - 5\left( {2.6957} \right) + 4}}{{ - 3{{\left( {2.6957} \right)}^2} + 8\left( {2.6957} \right) - 5}} = 2.6956 \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 2.69 \cr} $$