Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 317: 37

$${x_1} = 0.620723,{\text{ }}{x_2} \approx 3.036454$$

$$\eqalign{ & f\left( x \right) = \ln x - {x^2} + 3x - 1 \cr & {\text{Let }}f\left( x \right) = 0 \cr & \ln x - {x^2} + 3x - 1 = 0 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x - {x^2} + 3x - 1} \right] \cr & f'\left( x \right) = \frac{1}{x} - 2x + 3 \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 311}}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr & {\text{form}} \cr & {x_{n + 1}} = {x_n} - \frac{{\ln \left( {{x_n}} \right) - x_n^2 + 3{x_n} - 1}}{{\frac{1}{{{x_n}}} - 2{x_n} + 3}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = 0.5 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 0.5 \cr & {x_{n + 1}} = \left( {0.5} \right) - \frac{{\ln \left( {0.5} \right) - {{\left( {0.5} \right)}^2} + 3\left( {0.5} \right) - 1}}{{\frac{1}{{\left( {0.5} \right)}} - 2\left( {0.5} \right) + 3}} \approx 0.610787 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 0.620655 \cr & {x_3} \approx 0.620723 \cr & {x_4} \approx 0.620723 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 3 \cr & {\text{Therefore}}{\text{,}} \cr & n = 0,{\text{ }}{x_0} = 3 \cr & {x_{n + 1}} = \left( 3 \right) - \frac{{\ln \left( 3 \right) - {{\left( 3 \right)}^2} + 3\left( 3 \right) - 1}}{{\frac{1}{{\left( 3 \right)}} - 2\left( 3 \right) + 3}} \approx 3.036980 \cr & {\text{Computing the same calculation we obtain:}}{\text{.}} \cr & {x_2} \approx 3.036454 \cr & {x_3} \approx 3.036454 \cr & \cr & {\text{The approximation of the roots are:}} \cr & {x_1} = 0.620723,{\text{ }}{x_2} \approx 3.036454 \cr} $$