Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 20

$${x_1} \approx 0.135673,{\text{ }}{x_2} \approx 1.314978$$

$$\eqalign{ & y = \ln x,{\text{ }}y = {x^3} - 2 \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & \ln x = {x^3} - 2 \cr & {\text{Subtract }}{x^3} - 2{\text{ from both sides of equation to write the }} \cr & {\text{functions in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr & f\left( x \right) = 0 \cr & \ln x - {x^3} + 2 = 0 \cr & {\text{Let }}f\left( x \right) = \ln x - {x^3} + 2,{\text{ differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x - {x^3} + 2} \right] \cr & f'\left( x \right) = \frac{1}{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{{\ln {x_n} - x_n^3 + 2}}{{\frac{1}{{{x_n}}} - 3x_n^2}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = 0.2 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 0.2 \cr & {x_1} \approx \left( {0.2} \right) - \frac{{\ln \left( {0.2} \right) - {{\left( {0.2} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {0.2} \right)}^2}}} \approx 0.121606 \cr & n = 1,{\text{ }}{x_1} = 0.121606 \cr & {x_2} \approx \left( {0.121606} \right) - \frac{{\ln \left( {0.121606} \right) - {{\left( {0.121606} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {0.121606} \right)}^2}}} \cr & {x_2} \approx 0.134904 \cr & n = 2,{\text{ }}{x_2} = 0.134904 \cr & {x_3} \approx \left( {0.134904} \right) - \frac{{\ln \left( {0.134904} \right) - {{\left( {0.134904} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {0.134904} \right)}^2}}} \cr & {x_3} \approx 0.135671 \cr & n = 3,{\text{ }}{x_1} = 0.135671 \cr & {x_4} \approx \left( {0.135671} \right) - \frac{{\ln \left( {0.135671} \right) - {{\left( {0.135671} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {0.135671} \right)}^2}}} \cr & {x_4} \approx 0.135673 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 1.2 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 1.2 \cr & {x_1} \approx \left( {1.2} \right) - \frac{{\ln \left( {1.2} \right) - {{\left( {1.2} \right)}^3} + 2}}{{\frac{1}{{1.2}} - 3{{\left( {1.2} \right)}^2}}} \approx 1.330302 \cr & n = 1,{\text{ }}{x_1} = 1.330302 \cr & {x_2} \approx \left( {1.330302} \right) - \frac{{\ln \left( {1.330302} \right) - {{\left( {1.330302} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {1.330302} \right)}^2}}} \cr & {x_2} \approx 1.315198 \cr & n = 2,{\text{ }}{x_2} = 1.315198 \cr & {x_3} \approx \left( {1.315198} \right) - \frac{{\ln \left( {1.315198} \right) - {{\left( {1.315198} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {1.315198} \right)}^2}}} \cr & {x_3} \approx 1.314978 \cr & n = 3,{\text{ }}{x_1} = 1.314978 \cr & {x_4} \approx \left( {1.314978} \right) - \frac{{\ln \left( {1.314978} \right) - {{\left( {1.314978} \right)}^3} + 2}}{{\frac{1}{{0.2}} - 3{{\left( {1.314978} \right)}^2}}} \cr & {x_4} \approx 1.314978 \cr & \cr & {\text{The approximation of the solutions are:}} \cr & {x_1} \approx 0.135673,{\text{ }}{x_2} \approx 1.314978 \cr} $$