Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 16

$${x_1} \approx 1.857183,{\text{ }}{x_2} \approx 4.536403$$

$$\eqalign{ & y = {e^x},{\text{ }}y = {x^3} \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & {e^x} = {x^3} \cr & {\text{Subtract }}{x^3}{\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 & {e^x} - {x^3} = 0 \cr & {\text{Let }}f\left( x \right) = {e^x} - {x^3},{\text{ differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x} - {x^3}} \right] \cr & f'\left( x \right) = {e^x} - 3{x^2} \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{{{e^{{x_n}}} - x_n^3}}{{{e^{{x_n}}} - 3x_n^2}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = 2 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 2 \cr & {x_1} \approx 2 - \frac{{{e^2} - {{\left( 2 \right)}^3}}}{{{e^2} - 3{{\left( 2 \right)}^2}}} \approx 1.867501 \cr & n = 1,{\text{ }}{x_1} = 1.867501 \cr & {x_2} \approx 1.867501 - \frac{{{e^{1.867501}} - {{\left( {1.867501} \right)}^3}}}{{{e^{1.867501}} - 3{{\left( {1.867501} \right)}^2}}} \approx 1.857247 \cr & n = 2,{\text{ }}{x_2} = 1.857247 \cr & {x_3} \approx 1.857247 - \frac{{{e^{1.857247}} - {{\left( {1.857247} \right)}^3}}}{{{e^{1.857247}} - 3{{\left( {1.857247} \right)}^2}}} \approx 1.857183 \cr & n = 3,{\text{ }}{x_3} = 1.857183 \cr & {x_4} \approx 1.857183 - \frac{{{e^{1.857247}} - {{\left( {1.857183} \right)}^3}}}{{{e^{1.857247}} - 3{{\left( {1.857183} \right)}^2}}} \approx 1.857183 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 4 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 4 \cr & {x_1} \approx 4 - \frac{{{e^2} - {{\left( 4 \right)}^3}}}{{{e^2} - 3{{\left( 4 \right)}^2}}} \approx 5.244922 \cr & n = 1,{\text{ }}{x_1} = 5.244922 \cr & {x_2} \approx 5.244922 - \frac{{{e^{5.244922}} - {{\left( {5.244922} \right)}^3}}}{{{e^{5.244922}} - 3{{\left( {5.244922} \right)}^2}}} \approx 4.821685 \cr & n = 2,{\text{ }}{x_2} = 4.821685 \cr & {x_3} \approx 4.821685 - \frac{{{e^{4.821685}} - {{\left( {4.821685} \right)}^3}}}{{{e^{4.821685}} - 3{{\left( {4.821685} \right)}^2}}} \approx 4.599805 \cr & n = 3,{\text{ }}{x_3} = 4.599805 \cr & {x_4} \approx 4.599805 - \frac{{{e^{4.599805}} - {{\left( {4.599805} \right)}^3}}}{{{e^{4.599805}} - 3{{\left( {4.599805} \right)}^2}}} \approx 4.540308 \cr & n = 4,{\text{ }}{x_4} = 4.540308 \cr & {x_5} \approx 4.540308 - \frac{{{e^{4.540308}} - {{\left( {4.540308} \right)}^3}}}{{{e^{4.540308}} - 3{{\left( {4.540308} \right)}^2}}} \approx 4.536419 \cr & n = 5,{\text{ }}{x_4} = 4.536419 \cr & {x_5} \approx 4.536419 - \frac{{{e^{4.536419}} - {{\left( {4.536419} \right)}^3}}}{{{e^{4.536419}} - 3{{\left( {4.536419} \right)}^2}}} \approx 4.536403 \cr & \cr & {\text{The approximation of the solutions are:}} \cr & {x_1} \approx 1.857183,{\text{ }}{x_2} \approx 4.536403 \cr} $$