Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{229 x^{3}}{250} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{229 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{687 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{229 (1.0000000000)^{3}}{250} + 4 + e^{- (1.0000000000)}}{- \frac{687 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.1078347241$ $LaTeX: x_{2} = (2.1078347241) - \frac{- \frac{229 (2.1078347241)^{3}}{250} + 4 + e^{- (2.1078347241)}}{- \frac{687 (2.1078347241)^{2}}{250} - e^{- (2.1078347241)}} = 1.7463913819$ $LaTeX: x_{3} = (1.7463913819) - \frac{- \frac{229 (1.7463913819)^{3}}{250} + 4 + e^{- (1.7463913819)}}{- \frac{687 (1.7463913819)^{2}}{250} - e^{- (1.7463913819)}} = 1.6640489918$ $LaTeX: x_{4} = (1.6640489918) - \frac{- \frac{229 (1.6640489918)^{3}}{250} + 4 + e^{- (1.6640489918)}}{- \frac{687 (1.6640489918)^{2}}{250} - e^{- (1.6640489918)}} = 1.6600201567$ $LaTeX: x_{5} = (1.6600201567) - \frac{- \frac{229 (1.6600201567)^{3}}{250} + 4 + e^{- (1.6600201567)}}{- \frac{687 (1.6600201567)^{2}}{250} - e^{- (1.6600201567)}} = 1.6600108011$