Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{369 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{369 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{1107 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{369 (3.0000000000)^{3}}{1000} + 4 + e^{- (3.0000000000)}}{- \frac{1107 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4094338678$ $LaTeX: x_{2} = (2.4094338678) - \frac{- \frac{369 (2.4094338678)^{3}}{1000} + 4 + e^{- (2.4094338678)}}{- \frac{1107 (2.4094338678)^{2}}{1000} - e^{- (2.4094338678)}} = 2.2449906305$ $LaTeX: x_{3} = (2.2449906305) - \frac{- \frac{369 (2.2449906305)^{3}}{1000} + 4 + e^{- (2.2449906305)}}{- \frac{1107 (2.2449906305)^{2}}{1000} - e^{- (2.2449906305)}} = 2.2328184592$ $LaTeX: x_{4} = (2.2328184592) - \frac{- \frac{369 (2.2328184592)^{3}}{1000} + 4 + e^{- (2.2328184592)}}{- \frac{1107 (2.2328184592)^{2}}{1000} - e^{- (2.2328184592)}} = 2.2327545314$ $LaTeX: x_{5} = (2.2327545314) - \frac{- \frac{369 (2.2327545314)^{3}}{1000} + 4 + e^{- (2.2327545314)}}{- \frac{1107 (2.2327545314)^{2}}{1000} - e^{- (2.2327545314)}} = 2.2327545296$