Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 87

$x \approx - 0.795\text{ and }x \approx - 2.182$

$$\eqalign{ & f\left( x \right) = {x^4} + {x^3} - 3{x^2} + 2 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = 4{x^3} + 3{x^2} - 6x \cr & {\text{Using the Newton's Method}} \cr & {\text{The iterative formula is}} \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f\left( {{x_n}} \right){\text{ into }}\left( {\bf{1}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{x_n^4 + x_n^3 - 3x_n^2 + 2}}{{4x_n^3 + 3x_n^2 - 6{x_n}}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} \approx - 1.000 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx - 1.000 \cr & {x_2} \approx - 0.800 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx - 0.795 \cr & {x_4} \approx - 0.795 \cr & {\text{The successive approximations }}{x_4}{\text{ and }}{x_3}{\text{ differ by}} \cr & \left| {{x_4} - {x_3}} \right| < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx - 0.795 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx - 0.795197727 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}{x_1} \approx - 2.000 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx - 2.000 \cr & {x_2} \approx - 2.250 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx - 2.188 \cr & {x_4} \approx - 2.182 \cr & {x_5} \approx - 2.182 \cr & {\text{The successive approximations }}{x_4}{\text{ and }}{x_5}{\text{ differ by}} \cr & \left| {{x_5} - {x_4}} \right| < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx - 2.182 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx - 2.18227091 \cr & \cr & {\text{The solutions are}} \cr & x \approx - 0.795{\text{ and }}x \approx - 2.182 \cr} $$