Chapter 4 - Applications of the Derivative - Review Exercises - Page 331: 35

$${x_1} = 1,{\text{ }}{x_2} \approx - 0.434259,{\text{ }}{x_3} = 0.767592$$

$$\eqalign{ & f\left( x \right) = 3{x^3} - 4{x^2} + 1 \cr & {\text{Let }}f\left( x \right) = 0 \cr & 3{x^3} - 4{x^2} + 1 = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^3} - 4{x^2} + 1} \right] \cr & f'\left( x \right) = 9{x^2} - 8x \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 311}}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr & {\text{form}} \cr & {x_{n + 1}} = {x_n} - \frac{{3x_n^3 - 4x_n^2 + 1}}{{9x_n^2 - 8{x_n}}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = - 0.4 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = - 0.4 \cr & {x_{n + 1}} = \left( { - 0.4} \right) - \frac{{3{{\left( { - 0.4} \right)}^3} - 4{{\left( { - 0.4} \right)}^2} + 1}}{{9{{\left( { - 0.4} \right)}^2} - 8\left( { - 0.4} \right)}} \approx - 0.436207 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx - 0.434264 \cr & {x_3} \approx - 0.434259 \cr & {x_4} \approx - 0.434259 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 0.8 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 0.8 \cr & {x_{n + 1}} = \left( {0.8} \right) - \frac{{3{{\left( {0.8} \right)}^3} - 4{{\left( {0.8} \right)}^2} + 1}}{{9{{\left( {0.8} \right)}^2} - 8\left( {0.8} \right)}} \approx 0.762500 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 0.767506 \cr & {x_3} \approx 0.767592 \cr & {x_4} \approx 0.767592 \cr & \cr & {\text{From the graph we can see that the third possible initial }} \cr & {\text{approximation is }}x = 1 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 1 \cr & {x_{n + 1}} = \left( 1 \right) - \frac{{3{{\left( 1 \right)}^3} - 4{{\left( 1 \right)}^2} + 1}}{{9{{\left( 1 \right)}^2} - 8\left( 1 \right)}} \approx 1 \cr & x = 1 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} = 1,{\text{ }}{x_2} \approx - 0.434259,{\text{ }}{x_3} = 0.767592 \cr} $$