Chapter 3 - Applications of Differentiation - 3.8 Exercises - Page 229: 14

$0.865$

$$\eqalign{ & f\left( x \right) = {x^3} - \cos x \cr & {\text{Differentiating}} \cr & f\left( x \right) = 3{x^3} - \left( { - \sin x} \right) \cr & f'\left( x \right) = 3{x^2} + \sin x \cr & {\text{Using the Newton's Method}} \cr & {\text{The iterative formula is}} \cr & {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f\left( {{x_n}} \right){\text{ into }}\left( {\bf{1}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{x_n^3 - \cos {x_n}}}{{3x_n^2 + \sin {x_n}}} \cr & {\text{From the graph we can see that the first possible initial }} \cr & {\text{approximation is }}{x_1} \approx 0.8 \cr & {\text{The calculations for the iterations are shown below}} \cr & {x_1} \approx 0.8 \cr & {x_2} = 0.8 - \frac{{{{\left( {0.8} \right)}^3} - \cos \left( {0.8} \right)}}{{3{{\left( {0.8} \right)}^2} + \sin \left( {0.8} \right)}} \cr & {x_2} \approx 0.8700 \cr & {\text{Continuing the iterations we obtain}} \cr & {x_3} \approx 0.8655 \cr & {x_4} \approx 0.8655 \cr & {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr & \left| {{x_4} - {x_3}} \right| \approx 0 < 0.001 \cr & {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr & x \approx 0.8655 \cr & {\text{Taking three decimal plases}} \cr & x \approx 0.865 \cr & {\text{Using a graphing utility we obtain}} \cr & x \approx 0.86547 \cr} $$