Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 18

$$x \approx 1.465571$$

$$\eqalign{ & y = {x^3},{\text{ }}y = {x^2} + 1 \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & {x^3} = {x^2} + 1 \cr & {\text{Subtract }}{x^2} + 1{\text{ from both sides of equation to write the functions}} \cr & {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr & f\left( x \right) = 0 \cr & {x^3} - {x^2} - 1 = 0 \cr & {\text{Let }}f\left( x \right) = {x^3} - {x^2} - 1,{\text{ differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - {x^2} - 1} \right] \cr & f'\left( x \right) = 3{x^2} - 2x \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 - x_n^2 - 1}}{{3x_n^2 - 2{x_n}}} \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_1} \approx 1 - \frac{{{{\left( 1 \right)}^3} - {{\left( 1 \right)}^2} - 1}}{{3{{\left( 1 \right)}^2} - 2\left( 1 \right)}} \approx 2 \cr & n = 1,{\text{ }}{x_1} = 2 \cr & {x_2} \approx 1 - \frac{{{{\left( 2 \right)}^3} - {{\left( 2 \right)}^2} - 1}}{{3{{\left( 2 \right)}^2} - 2\left( 2 \right)}} \approx 1.625 \cr & n = 2,{\text{ }}{x_2} = 1.625 \cr & {x_3} \approx 1.625 - \frac{{{{\left( {1.625} \right)}^3} - {{\left( {1.625} \right)}^2} - 1}}{{3{{\left( {1.625} \right)}^2} - 2\left( {1.625} \right)}} \approx 1.485785 \cr & n = 3,{\text{ }}{x_2} = 1.485785 \cr & {x_4} \approx 1.485785 - \frac{{{{\left( {1.485785} \right)}^3} - {{\left( {1.485785} \right)}^2} - 1}}{{3{{\left( {1.485785} \right)}^2} - 2\left( {1.485785} \right)}} \cr & {x_4} \approx 1.465955 \cr & n = 4,{\text{ }}{x_2} = 1.465955 \cr & {x_5} \approx 1.465955 - \frac{{{{\left( {1.465955} \right)}^3} - {{\left( {1.465955} \right)}^2} - 1}}{{3{{\left( {1.465955} \right)}^2} - 2\left( {1.465955} \right)}} \cr & {x_5} \approx 1.465571 \cr & \cr & {\text{The approximation of the solutions is:}} \cr & x \approx 1.465571 \cr} $$