Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{659 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{659 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{1977 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{659 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{1977 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3418509739$ $LaTeX: x_{2} = (2.3418509739) - \frac{- \frac{659 (2.3418509739)^{3}}{1000} + 6 + e^{- (2.3418509739)}}{- \frac{1977 (2.3418509739)^{2}}{1000} - e^{- (2.3418509739)}} = 2.1254046623$ $LaTeX: x_{3} = (2.1254046623) - \frac{- \frac{659 (2.1254046623)^{3}}{1000} + 6 + e^{- (2.1254046623)}}{- \frac{1977 (2.1254046623)^{2}}{1000} - e^{- (2.1254046623)}} = 2.1024441160$ $LaTeX: x_{4} = (2.1024441160) - \frac{- \frac{659 (2.1024441160)^{3}}{1000} + 6 + e^{- (2.1024441160)}}{- \frac{1977 (2.1024441160)^{2}}{1000} - e^{- (2.1024441160)}} = 2.1021986016$ $LaTeX: x_{5} = (2.1021986016) - \frac{- \frac{659 (2.1021986016)^{3}}{1000} + 6 + e^{- (2.1021986016)}}{- \frac{1977 (2.1021986016)^{2}}{1000} - e^{- (2.1021986016)}} = 2.1021985738$