Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{381 x^{3}}{500} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{381 x_{n}^{3}}{500} + 9 + e^{- x_{n}}}{- \frac{1143 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{381 (3.0000000000)^{3}}{500} + 9 + e^{- (3.0000000000)}}{- \frac{1143 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 2.4412174208$ $LaTeX: x_{2} = (2.4412174208) - \frac{- \frac{381 (2.4412174208)^{3}}{500} + 9 + e^{- (2.4412174208)}}{- \frac{1143 (2.4412174208)^{2}}{500} - e^{- (2.4412174208)}} = 2.2954224585$ $LaTeX: x_{3} = (2.2954224585) - \frac{- \frac{381 (2.2954224585)^{3}}{500} + 9 + e^{- (2.2954224585)}}{- \frac{1143 (2.2954224585)^{2}}{500} - e^{- (2.2954224585)}} = 2.2859301648$ $LaTeX: x_{4} = (2.2859301648) - \frac{- \frac{381 (2.2859301648)^{3}}{500} + 9 + e^{- (2.2859301648)}}{- \frac{1143 (2.2859301648)^{2}}{500} - e^{- (2.2859301648)}} = 2.2858913505$ $LaTeX: x_{5} = (2.2858913505) - \frac{- \frac{381 (2.2858913505)^{3}}{500} + 9 + e^{- (2.2858913505)}}{- \frac{1143 (2.2858913505)^{2}}{500} - e^{- (2.2858913505)}} = 2.2858913498$