Chapter 4 - Applications of the Derivative - Review Exercises - Page 331: 37

$${x_1} = 0,{\text{ }}{x_2} \approx - 0.948683,{\text{ }}{x_3} \approx 0.948683$$

$$\eqalign{ & f\left( x \right) = 2{x^5} - 6{x^3} - 4x + 2 \cr & {\text{Calculate the second derivative of the second function}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^5} - 6{x^3} - 4x + 2} \right] \cr & f'\left( x \right) = 10{x^4} - 18{x^2} - 4 \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {10{x^4} - 18{x^2} - 4} \right] \cr & f''\left( x \right) = 40{x^3} - 36x \cr & {\text{Let }}g\left( x \right) = 40{x^3} - 36x \cr & {\text{Let }}g\left( x \right) = 0 \cr & 40{x^3} - 36x = 0 \cr & \cr & {\text{Differentiating}} \cr & g'\left( x \right) = \frac{d}{{dx}}\left[ {40{x^3} - 36x} \right] \cr & g'\left( x \right) = 120{x^2} - 36 \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{{g\left( {{x_n}} \right)}}{{g'\left( {{x_n}} \right)}} \cr & {\text{Substituting }}g\left( {{x_n}} \right){\text{ and }}g'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr & {\text{form}} \cr & {x_{n + 1}} = {x_n} - \frac{{40x_n^3 - 36{x_n}}}{{120x_n^2 - 36}}. \cr & {\text{We have the first root }}x = 0 \cr & {\text{From the graph we can see that the first 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{{40{{\left( { - 1} \right)}^3} - 36\left( { - 1} \right)}}{{120{{\left( { - 1} \right)}^2} - 36}} \approx - 0.952380 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx - 0.948705 \cr & {x_3} \approx - 0.948683 \cr & {x_4} \approx - 0.948683 \cr & \cr & {\text{From the graph we can see that the second 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{{40{{\left( 1 \right)}^3} - 36\left( 1 \right)}}{{120{{\left( 1 \right)}^2} - 36}} \approx 0.952381 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 0.948705 \cr & {x_3} \approx 0.948683 \cr & {x_4} \approx 0.948683 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} = 0,{\text{ }}{x_2} \approx - 0.948683,{\text{ }}{x_3} \approx 0.948683 \cr} $$