Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=11$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 11$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + 2 + e^{- (11.0000000000)}}{- \frac{3 (11.0000000000)^{2}}{1000} - e^{- (11.0000000000)}} = 12.8429364229$ $LaTeX: x_{2} = (12.8429364229) - \frac{- \frac{(12.8429364229)^{3}}{1000} + 2 + e^{- (12.8429364229)}}{- \frac{3 (12.8429364229)^{2}}{1000} - e^{- (12.8429364229)}} = 12.6038131547$ $LaTeX: x_{3} = (12.6038131547) - \frac{- \frac{(12.6038131547)^{3}}{1000} + 2 + e^{- (12.6038131547)}}{- \frac{3 (12.6038131547)^{2}}{1000} - e^{- (12.6038131547)}} = 12.5992192606$ $LaTeX: x_{4} = (12.5992192606) - \frac{- \frac{(12.5992192606)^{3}}{1000} + 2 + e^{- (12.5992192606)}}{- \frac{3 (12.5992192606)^{2}}{1000} - e^{- (12.5992192606)}} = 12.5992175853$ $LaTeX: x_{5} = (12.5992175853) - \frac{- \frac{(12.5992175853)^{3}}{1000} + 2 + e^{- (12.5992175853)}}{- \frac{3 (12.5992175853)^{2}}{1000} - e^{- (12.5992175853)}} = 12.5992175853$