Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{809 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{809 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{2427 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{809 (1.0000000000)^{3}}{1000} + 6 + e^{- (1.0000000000)}}{- \frac{2427 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.9889514228$ $LaTeX: x_{2} = (2.9889514228) - \frac{- \frac{809 (2.9889514228)^{3}}{1000} + 6 + e^{- (2.9889514228)}}{- \frac{2427 (2.9889514228)^{2}}{1000} - e^{- (2.9889514228)}} = 2.2733394763$ $LaTeX: x_{3} = (2.2733394763) - \frac{- \frac{809 (2.2733394763)^{3}}{1000} + 6 + e^{- (2.2733394763)}}{- \frac{2427 (2.2733394763)^{2}}{1000} - e^{- (2.2733394763)}} = 2.0043350403$ $LaTeX: x_{4} = (2.0043350403) - \frac{- \frac{809 (2.0043350403)^{3}}{1000} + 6 + e^{- (2.0043350403)}}{- \frac{2427 (2.0043350403)^{2}}{1000} - e^{- (2.0043350403)}} = 1.9659505596$ $LaTeX: x_{5} = (1.9659505596) - \frac{- \frac{809 (1.9659505596)^{3}}{1000} + 6 + e^{- (1.9659505596)}}{- \frac{2427 (1.9659505596)^{2}}{1000} - e^{- (1.9659505596)}} = 1.9652130892$