Chapter 3 - Applications of Differentiation - 3.8 Exercises - Page 229: 11

$$\eqalign{ & 0.900 \cr & 1.100 \cr & 1.900 \cr} $$

$$\eqalign{ & f\left( x \right) = {x^3} - 3.9{x^2} + 4.79x - 1.881 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = 3{x^2} - 7.8x + 4.79 \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 - 3.9x_n^2 + 4.79{x_n} - 1.881}}{{3x_n^2 - 7.8{x_n} + 4.79}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} = 0.8 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} = 0.8 \cr & {x_2} = 0.8702 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 0.8959 \cr & {x_4} \approx 0.8999 \cr & {x_5} \approx 0.9000 \cr & {x_6} \approx 0.9000 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_6} - {x_5}} \right| \approx 0 < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 0.9000 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 0.9 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}{x_1} \approx 1.05 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 1.05 \cr & {x_2} \approx 1.1189 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 1.1011 \cr & {x_4} \approx 1.1000 \cr & {x_5} \approx 1.1000 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_4} - {x_5}} \right| \approx 0 < 0.001 \cr & {\text{We can estimate the second zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 1.1000 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 1.10 \cr & \cr & {\text{From the graph we can see that the third possible initial }} \cr & {\text{approximation is }}{x_1} \approx 1.8 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 1.8 \cr & {x_2} \approx 1.9340 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 1.9023 \cr & {x_4} \approx 1.9000 \cr & {x_5} \approx 1.9000 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_4} - {x_5}} \right| \approx 0 < 0.001 \cr & {\text{We can estimate the third zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 1.9000 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 1.9 \cr & \cr & {\text{The solutions are:}} \cr & x = 0.90,{\text{ }}x = 1.100{\text{ and }}x = 1.900 \cr} $$