Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 19

$$\eqalign{ & x = - 1.257691 \cr & x = 0.653483 \cr} $$

$$\eqalign{ & {2^x} = 2 - {x^2} \cr & {\text{Let }}y = {2^x},{\text{ }}y = 2 - {x^2} \cr & {\text{The graphs of the functions are shown below}} \cr & \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & {2^x} = 2 - {x^2} \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 & {2^x} - 2 + {x^2} = 0 \cr & \cr & {\text{Let }}f\left( x \right) = \cos 2x - {x^3},{\text{ and differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{2^x} - 2 + {x^2}} \right] \cr & f'\left( x \right) = {2^x}\ln 2 + 2x \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{{{2^{{x_n}}} - 2 + x_n^2}}{{{2^{{x_n}}}\ln 2 + 2{x_n}}} \cr & \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x = - 1.5 \cr & {\text{If we choose }}{x_1} = 2{\text{ as the initial approximation}},{\text{then we obtain}} \cr & {x_2} \approx - 1.5 - \frac{{{2^{ - 1.5}} - 2 + {{\left( { - 1.5} \right)}^2}}}{{{2^{ - 1.5}}\left( {\ln 2} \right) + 2\left( { - 1.5} \right)}} \cr & {x_2} \approx - 1.280919 \cr & {x_3} \approx - 1.257951 \cr & {x_4} \approx - 1.257691 \cr & {x_5} \approx - 1.257691 \cr & {x_4}{\text{ and }}{x_5}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation }}{2^x} = 2 - {x^2}{\text{ is}} \cr & x = - 1.257691 \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 0.6 - \frac{{{2^{0.6}} - 2 + {{\left( {0.6} \right)}^2}}}{{{2^{0.6}}\left( {\ln 2} \right) + 2\left( {0.6} \right)}} \cr & {x_2} \approx 0.655221 \cr & {x_3} \approx 0.653484 \cr & {x_4} \approx 0.653483 \cr & {x_5} \approx 0.653483 \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 }}{2^x} = 2 - {x^2}{\text{ is}} \cr & x = 0.653482 \cr & \cr & {\text{The solutions are:}} \cr & x = - 1.257691 \cr & x = 0.653483 \cr} $$