Answer
$1.259$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2 - {x^3} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = - 3{x^2} \cr
& {\text{Using the Newton's Method}} \cr
& {\text{The iterative formula is}} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f\left( {{x_n}} \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{2 - x_n^3}}{{ - 3x_n^2}} \cr
& {x_{n + 1}} = {x_n} + \frac{{2 - x_n^3}}{{3x_n^2}} \cr
& {\text{From the graph we can see that the possible initial }} \cr
& {\text{approximation is }}{x_1} = 1 \cr
& {\text{The calculations for the iterations are shown below}} \cr
& {x_2} = 1 + \frac{{2 - {{\left( 1 \right)}^3}}}{{3{{\left( 1 \right)}^2}}} \approx 1.33333 \cr
& {\text{Continuing the iterations we obtain}} \cr
& {x_3} \approx 1.26389 \cr
& {x_4} \approx 1.25993 \cr
& {x_5} \approx 1.25992 \cr
& {\text{The successive approximations }}{x_5}{\text{ and }}{x_4}{\text{ differ by}} \cr
& \left| {{x_5} - {x_4}} \right| \approx 0.00001 < 0.001 \cr
& {\text{We can estimate the zero of }}f\left( x \right){\text{ to be }} \cr
& x \approx 1.25992 \cr
& {\text{taking three decimal places}} \cr
& x \approx 1.259 \cr
& {\text{Using a graphing utility we obtain}} \cr
& x \approx 1.25992105 \cr} $$
