Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 85

$x = - 1.532,{\text{ }}x = - 0.347{\text{ and }}x = 1.879$

$$\eqalign{ & f\left( x \right) = {x^3} - 3x - 1 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = 3{x^2} - 3 \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^3 - 3{x_n} - 1}}{{3x_n^2 - 3}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} \approx - 1.5 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 0.8 \cr & {x_2} \approx - 1.533 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx - 1.532 \cr & {x_4} \approx - 1.532 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\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 - 1.532 \cr & {\text{Using a graphing utility we obtain}} \cr & x = - 1.532088886 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}{x_1} \approx - 0.5 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx - 0.5 \cr & {x_2} \approx 1.1189 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx - 0.333 \cr & {x_4} \approx - 0.347 \cr & {x_5} \approx - 0.347 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_4} - {x_3}} \right| < 0.001 \cr & {\text{We can estimate the second zero of }}f\left( x \right){\text{ to be }} \cr & x \approx - 0.347 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx - 0.347296355 \cr & \cr & {\text{From the graph we can see that the third possible initial }} \cr & {\text{approximation is }}{x_1} \approx 2 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 2 \cr & {x_2} \approx 1.889 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 1.879 \cr & {x_4} \approx 1.879 \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 third zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 1.879 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 1.879385242 \cr & \cr & {\text{The solutions are:}} \cr & x = - 1.532,{\text{ }}x = - 0.347{\text{ and }}x = 1.879 \cr} $$