Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{207 x^{3}}{250} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{207 x_{n}^{3}}{250} + 9 + e^{- x_{n}}}{- \frac{621 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{207 (3.0000000000)^{3}}{250} + 9 + e^{- (3.0000000000)}}{- \frac{621 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.4061260651$ $LaTeX: x_{2} = (2.4061260651) - \frac{- \frac{207 (2.4061260651)^{3}}{250} + 9 + e^{- (2.4061260651)}}{- \frac{621 (2.4061260651)^{2}}{250} - e^{- (2.4061260651)}} = 2.2372393797$ $LaTeX: x_{3} = (2.2372393797) - \frac{- \frac{207 (2.2372393797)^{3}}{250} + 9 + e^{- (2.2372393797)}}{- \frac{621 (2.2372393797)^{2}}{250} - e^{- (2.2372393797)}} = 2.2240712693$ $LaTeX: x_{4} = (2.2240712693) - \frac{- \frac{207 (2.2240712693)^{3}}{250} + 9 + e^{- (2.2240712693)}}{- \frac{621 (2.2240712693)^{2}}{250} - e^{- (2.2240712693)}} = 2.2239944299$ $LaTeX: x_{5} = (2.2239944299) - \frac{- \frac{207 (2.2239944299)^{3}}{250} + 9 + e^{- (2.2239944299)}}{- \frac{621 (2.2239944299)^{2}}{250} - e^{- (2.2239944299)}} = 2.2239944273$