Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{227 x^{3}}{250} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{227 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{681 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{227 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{681 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 1.7955935825$ $LaTeX: x_{2} = (1.7955935825) - \frac{- \frac{227 (1.7955935825)^{3}}{250} + 3 + e^{- (1.7955935825)}}{- \frac{681 (1.7955935825)^{2}}{250} - e^{- (1.7955935825)}} = 1.5619676404$ $LaTeX: x_{3} = (1.5619676404) - \frac{- \frac{227 (1.5619676404)^{3}}{250} + 3 + e^{- (1.5619676404)}}{- \frac{681 (1.5619676404)^{2}}{250} - e^{- (1.5619676404)}} = 1.5254305684$ $LaTeX: x_{4} = (1.5254305684) - \frac{- \frac{227 (1.5254305684)^{3}}{250} + 3 + e^{- (1.5254305684)}}{- \frac{681 (1.5254305684)^{2}}{250} - e^{- (1.5254305684)}} = 1.5245925737$ $LaTeX: x_{5} = (1.5245925737) - \frac{- \frac{227 (1.5245925737)^{3}}{250} + 3 + e^{- (1.5245925737)}}{- \frac{681 (1.5245925737)^{2}}{250} - e^{- (1.5245925737)}} = 1.5245921399$