Answer
$ - 0.453$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} + 2x + 1 \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = 3{x^2} + 2 \cr
& {\text{Using the Newton's Method}} \cr
& {\text{The iterative formula is}} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f\left( {{x_n}} \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^3 + 2{x_n} + 1}}{{3x_n^2 + 2}} \cr
& {\text{From the graph we can see that the first possible initial }} \cr
& {\text{approximation is }}{x_1} \approx - 0.500 \cr
& {\text{The calculations for the iterations are shown below}} \cr
& {x_1} \approx - 0.500 \cr
& {x_2} \approx - 0.455 \cr
& {\text{Continuing the iterations we obtain}} \cr
& {x_3} \approx - 0.453 \cr
& {x_4} \approx - 0.453 \cr
& {\text{The successive approximations }}{x_4}{\text{ and }}{x_3}{\text{ differ by}} \cr
& \left| {{x_4} - {x_3}} \right| < 0.001 \cr
& {\text{We can estimate the first zero of }}f\left( x \right){\text{ to be }} \cr
& x \approx - 0.453 \cr
& {\text{Using a graphing utility we obtain}} \cr
& x = - 0.453397651 \cr} $$
