Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{249 x^{3}}{500} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{249 x_{n}^{3}}{500} + 6 + e^{- x_{n}}}{- \frac{747 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{249 (3.0000000000)^{3}}{500} + 6 + e^{- (3.0000000000)}}{- \frac{747 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 2.4519613496$ $LaTeX: x_{2} = (2.4519613496) - \frac{- \frac{249 (2.4519613496)^{3}}{500} + 6 + e^{- (2.4519613496)}}{- \frac{747 (2.4519613496)^{2}}{500} - e^{- (2.4519613496)}} = 2.3135518037$ $LaTeX: x_{3} = (2.3135518037) - \frac{- \frac{249 (2.3135518037)^{3}}{500} + 6 + e^{- (2.3135518037)}}{- \frac{747 (2.3135518037)^{2}}{500} - e^{- (2.3135518037)}} = 2.3051530930$ $LaTeX: x_{4} = (2.3051530930) - \frac{- \frac{249 (2.3051530930)^{3}}{500} + 6 + e^{- (2.3051530930)}}{- \frac{747 (2.3051530930)^{2}}{500} - e^{- (2.3051530930)}} = 2.3051232342$ $LaTeX: x_{5} = (2.3051232342) - \frac{- \frac{249 (2.3051232342)^{3}}{500} + 6 + e^{- (2.3051232342)}}{- \frac{747 (2.3051232342)^{2}}{500} - e^{- (2.3051232342)}} = 2.3051232338$