Chapter 3 - The Derivative In Graphing And Applications - 3.7 Newton's Method - Exercises Set 3.7 - Page 250: 12

$$x \approx 0.876726215$$

$$\eqalign{ & \sin x = {x^2};{\text{ }}x > 0{\text{ }} \cr & {\text{From the graph we can see that the equation has two solutions}} \cr & {\text{Subtract }}x{\text{ from both sides of equation to write the functions}} \cr & {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr & \sin x - {x^2} = 0 \cr & {\text{Let }}f\left( x \right) = \sin x - {x^2},{\text{ differentiating}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x - {x^2}} \right] \cr & f'\left( x \right) = \cos x - 2x \cr & {\text{Using the Newton's Method }} \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}},{\text{ }}n = 1,2,3,...{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{\sin {x_n} - x_n^2}}{{\cos {x_n} - 2{x_n}}}{\text{ }}\left( {\bf{2}} \right) \cr & {\text{From the graph we can see that the possible initial }} \cr & {\text{approximation is }}{x_1} = 0.8,{\text{ substituting into }}\left( {\bf{2}} \right) \cr & {x_{n + 1}} = \left( {0.8} \right) - \frac{{\sin \left( {0.8} \right) - {{\left( {0.8} \right)}^2}}}{{\cos \left( {0.8} \right) - 2\left( {0.8} \right)}} \cr & {x_2} \approx 0.885637845 \cr & {\text{Continuing the process we obtain}} \cr & {x_3} \approx 0.876822914 \cr & {x_4} \approx 0.876726227 \cr & {x_5} \approx 0.876726215 \cr & {x_6} \approx 0.876726215 \cr & {x_5} = {x_6}{\text{ into 9 decimal places}}{\text{, then the solution for }}x > 0{\text{ of}} \cr & {\text{the equation is }} \cr & x \approx 0.876726215 \cr} $$