Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{811 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{811 x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{2433 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{811 (1.0000000000)^{3}}{1000} + 2 + e^{- (1.0000000000)}}{- \frac{2433 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 1.5558537859$ $LaTeX: x_{2} = (1.5558537859) - \frac{- \frac{811 (1.5558537859)^{3}}{1000} + 2 + e^{- (1.5558537859)}}{- \frac{2433 (1.5558537859)^{2}}{1000} - e^{- (1.5558537859)}} = 1.4176034298$ $LaTeX: x_{3} = (1.4176034298) - \frac{- \frac{811 (1.4176034298)^{3}}{1000} + 2 + e^{- (1.4176034298)}}{- \frac{2433 (1.4176034298)^{2}}{1000} - e^{- (1.4176034298)}} = 1.4043338044$ $LaTeX: x_{4} = (1.4043338044) - \frac{- \frac{811 (1.4043338044)^{3}}{1000} + 2 + e^{- (1.4043338044)}}{- \frac{2433 (1.4043338044)^{2}}{1000} - e^{- (1.4043338044)}} = 1.4042180195$ $LaTeX: x_{5} = (1.4042180195) - \frac{- \frac{811 (1.4042180195)^{3}}{1000} + 2 + e^{- (1.4042180195)}}{- \frac{2433 (1.4042180195)^{2}}{1000} - e^{- (1.4042180195)}} = 1.4042180107$