Answer
$$x \approx 2.273791732$$
Work Step by Step
$$\eqalign{
& {x^5} - 5{x^3} - 2 = 0;{\text{ }}x > 0 \cr
& {\text{From the graph we can see that the equation has 3 solutions}} \cr
& {\text{Let }}f\left( x \right) = {x^5} - 5{x^3} - 2 \cr
& f'\left( x \right) = 5{x^4} - 15{x^2} \cr
& {\text{Using the Newton's Method }} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}},{\text{ }}n = 1,2,3,...{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^5 - 5x_n^3 - 2}}{{5x_n^4 - 15x_n^2}}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{From the graph we can see that the possible initial }} \cr
& {\text{approximation is }}{x_1} = 2,{\text{ substituting into }}\left( {\bf{2}} \right) \cr
& {x_2} = 2 - \frac{{{{\left( 2 \right)}^5} - 5{{\left( 2 \right)}^3} - 2}}{{5{{\left( 2 \right)}^4} - 15{{\left( 2 \right)}^2}}} \cr
& {x_2} \approx 2.500000000 \cr
& {\text{Continuing the process we obtain}} \cr
& {x_3} \approx 2.327384615 \cr
& {x_4} \approx 2.277677678 \cr
& {x_5} \approx 2.273814035 \cr
& {x_6} \approx 2.273791733 \cr
& {x_7} \approx 2.273791732 \cr
& {x_8} \approx 2.273791732 \cr
& {x_7} = {x_8}{\text{ into 9 decimal places}}{\text{, then the solution for }}x > 0{\text{ of}} \cr
& {\text{the equation is }} \cr
& x \approx 2.273791732 \cr} $$
