Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 317: 38

$${x_1} = - 0.099950,{\text{ }}{x_2} \approx 0.100050,{\text{ }}{x_3} \approx 99.999900$$

$$\eqalign{ & f\left( x \right) = {x^2}\left( {x - 100} \right) + 1 \cr & f\left( x \right) = {x^3} - 100{x^2} + 1 \cr & {\text{Let }}f\left( x \right) = 0 \cr & {x^3} - 100{x^2} + 1 = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 100{x^2} + 1} \right] \cr & f'\left( x \right) = 3{x^2} - 200x \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{{x_n^3 - 100x_n^2 + 1}}{{3x_n^2 - 200{x_n}}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = - 0.1 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = - 0.1 \cr & {x_{n + 1}} = \left( { - 0.1} \right) - \frac{{{{\left( { - 0.1} \right)}^3} - 100{{\left( { - 0.1} \right)}^2} + 1}}{{3{{\left( { - 0.1} \right)}^2} - 200\left( { - 0.1} \right)}} \approx - 0.099950 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx - 0.099950 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 0.1 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 0.1 \cr & {x_{n + 1}} = \left( {0.1} \right) - \frac{{{{\left( {0.1} \right)}^3} - 100{{\left( {0.1} \right)}^2} + 1}}{{3{{\left( {0.1} \right)}^2} - 200\left( {0.1} \right)}} \approx 0.100050 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 0.100050 \cr & \cr & {\text{From the graph we can see that the third possible initial }} \cr & {\text{approximation is }}x = 100 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 100 \cr & {x_{n + 1}} = \left( {100} \right) - \frac{{{{\left( {100} \right)}^3} - 100{{\left( {100} \right)}^2} + 1}}{{3{{\left( {100} \right)}^2} - 200\left( {100} \right)}} \approx 99.999900 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 99.999900 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} = - 0.099950,{\text{ }}{x_2} \approx 0.100050,{\text{ }}{x_3} \approx 99.999900 \cr} $$