Chapter 3 - Applications of Differentiation - 3.8 Exercises - Page 229: 12

$$\eqalign{ & - 1.3803 \cr & 0.8192 \cr} $$

$$\eqalign{ & f\left( x \right) = {x^4} + {x^3} - 1 \cr & {\text{Differentiating}} \cr & f'\left( x \right) = 4{x^3} + 3{x^2} \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 - 1}}{{4x_n^3 + 3x_n^2}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} = - 1.5 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} = - 1.5 \cr & {x_2} = - 1.398 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx - 1.3807 \cr & {x_4} \approx - 1.3803 \cr & {x_5} \approx - 1.3803 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_5} - {x_4}} \right| \approx 0 < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx - 1.3803 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx - 1.380277 \cr & \cr & {\text{From the graph we can see that the second possible initial }} \cr & {\text{approximation is }}{x_1} \approx 0.7 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 0.7 \cr & {x_2} \approx 0.8467 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 0.8203 \cr & {x_4} \approx 0.8192 \cr & {x_5} \approx 0.8192 \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 0.8192 \cr & {\text{Using a graphing utility we obtain}} \cr & x = 0.8191725134 \cr & {\text{The solutions are:}} \cr & x = - 1.3803{\text{ and }}x = 0.8192 \cr} $$