Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 27

$$\eqalign{ & - 0.87828292 \cr & 0.79177077 \cr} $$

$$\eqalign{ & \sqrt {4 - {x^3}} = {e^{{x^2}}} \cr & {\text{Subtract }}{e^{{x^2}}}{\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 & \sqrt {4 - {x^3}} - {e^{{x^2}}} = 0 \cr & {\text{Let }}f\left( x \right) = \sqrt {4 - {x^3}} - {e^{{x^2}}} \cr & f\left( x \right) = 0 \cr & {\text{The graph of the equation is shown below}} \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt {4 - {x^3}} - {e^{{x^2}}}} \right] \cr & f'\left( x \right) = \frac{{ - 3{x^2}}}{{2\sqrt {4 - {x^3}} }} - 2x{e^{{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 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{{\sqrt {4 - x_n^3} - {e^{x_n^2}}}}{{\frac{{ - 3x_n^2}}{{2\sqrt {4 - x_n^3} }} - 2{x_n}{e^{x_n^2}}}} \cr & \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x \approx - 0.8 \cr & {\text{If we choose }}{x_1} \approx - 0.8{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx - 0.88815983 \cr & {x_3} \approx - 0.87842996 \cr & {x_4} \approx - 0.87828296 \cr & {x_5} \approx - 0.87828292 \cr & {x_6} \approx - 0.87828292 \cr & {x_6}{\text{ and }}{x_5}{\text{ are agree to eight decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is }} \approx - 0.87828292 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x \approx 0.8 \cr & {\text{If we choose }}{x_1} \approx 0.8{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx 0.79186616 \cr & {x_3} \approx 0.79177078 \cr & {x_4} \approx 0.79177077 \cr & {x_5} \approx 0.79177077 \cr & {x_5}{\text{ and }}{x_4}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is }} \approx 0.79177077 \cr} $$