Chapter 4 - Section 4.8 - Newton''s Method - 4.8 Exercises - Page 355: 23

$$\eqalign{ & - 1.69312029 \cr & - 0.74466668 \cr & 1.26587094 \cr} $$

$$\eqalign{ & - 2{x^7} - 5{x^4} + 9{x^3} + 5 = 0 \cr & {\text{Let }}f\left( x \right) = - 2{x^7} - 5{x^4} + 9{x^3} + 5 \cr & {\text{The graph of the equation is shown below}} \cr & f\left( x \right) = 0 \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ { - 2{x^7} - 5{x^4} + 9{x^3} + 5} \right] \cr & f'\left( x \right) = - 14{x^6} - 20{x^3} + 27{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{{ - 2x_n^7 - 5x_n^4 + 9x_n^3 + 5}}{{ - 14x_n^6 - 20x_n^3 + 27x_n^2}} \cr & \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}x \approx - 1.5 \cr & {\text{If we choose }}{x_1} \approx - 1.5{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx - 2.02902903 \cr & {x_3} \approx - 1.84548168 \cr & {x_4} \approx - 1.73741221 \cr & {x_5} \approx - 1.69807533 \cr & {x_6} \approx - 1.69319063 \cr & {x_7} \approx - 1.69312031 \cr & {x_8} \approx - 1.69312029 \cr & {x_9} \approx - 1.69312029 \cr & {x_9}{\text{ and }}{x_8}{\text{ are agree to eight decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is }} \approx - 1.69312029 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}x \approx - 0.75 \cr & {\text{If we choose }}{x_1} \approx - 0.75{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx - 0.74470322 \cr & {x_3} \approx - 0.74466668 \cr & {x_4} \approx - 0.74466668 \cr & {x_4}{\text{ and }}{x_3}{\text{ are agree to six decimal places}}{\text{, we conclude that}} \cr & {\text{the first solution of the equation is }} \approx - 0.74466668 \cr & \cr & {\text{From the graph we can see that the third possible initial }} \cr & {\text{approximation is }}x \approx 1.25 \cr & {\text{If we choose }}{x_1} \approx 1.25{\text{ as the initial approximation}},{\text{then }} \cr & {\text{we obtain:}} \cr & {x_2} \approx 1.26659383 \cr & {x_3} \approx 1.26587237 \cr & {x_4} \approx 1.26587094 \cr & {x_5} \approx 1.26587094 \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 1.26587094 \cr} $$