Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 22

$$\eqalign{ & x = - 2.490864 \cr & x = 0.656620 \cr & x = 1.834243 \cr} $$

$$\eqalign{ & {x^3} = 5x - 3 \cr & {\text{Let }}y = {x^3},{\text{ }}y = 5x - 3 \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & \arctan {x^3} = 5x - 3 \cr & {\text{Subtract }}x - 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} - 5x + 3 = 0 \cr & \cr & {\text{Let }}f\left( x \right) = {x^3} - 5x + 3,{\text{ and differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 5x + 3} \right] \cr & f'\left( x \right) = 3{x^2} - 5 \cr & {\text{Use the procedure Newton's Method for Approximations}} \cr & {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 352}}} \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 - 5{x_n} + 3}}{{3x_n^2 - 5}} \cr & \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = - 2.5 \cr & {\text{If we choose }}{x_1} = - 2.5{\text{ as the initial approximation}},{\text{we obtain}} \cr & {x_2} \approx - 2.490909 \cr & {x_3} \approx - 2.490864 \cr & {x_4} \approx - 2.490864 \cr & {x_3}{\text{ and }}{x_4}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is}} \cr & x = - 2.490864 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x = 0.6 \cr & {\text{Choosing }}{x_1} = 0.6{\text{ as the initial approximation}},{\text{then we obtain}} \cr & {x_2} \approx 0.655102 \cr & {x_3} \approx 0.656619 \cr & {x_4} \approx 0.656620 \cr & {x_5} \approx 0.656620 \cr & {x_4}{\text{ and }}{x_5}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the second solution of the equation is}} \cr & x = 0.656620 \cr & \cr & {\text{From the graph we can see that the third possible initial }} \cr & {\text{approximation is }}x = 1.8 \cr & {\text{Choosing }}{x_1} = 1.8{\text{ as the initial approximation}},{\text{then we obtain}} \cr & {x_2} \approx 1.835593 \cr & {x_3} \approx 1.834245 \cr & {x_4} \approx 1.834243 \cr & {x_5} \approx 1.834243 \cr & {x_4}{\text{ and }}{x_5}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the third solution of the equation is}} \cr & x = 0.656620 \cr & \cr & {\text{The solutions are:}} \cr & x = - 2.490864 \cr & x = 0.656620 \cr & x = 1.834243 \cr} $$