Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{271 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{271 x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{813 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{271 (1.0000000000)^{3}}{1000} + 2 + e^{- (1.0000000000)}}{- \frac{813 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.7756930708$ $LaTeX: x_{2} = (2.7756930708) - \frac{- \frac{271 (2.7756930708)^{3}}{1000} + 2 + e^{- (2.7756930708)}}{- \frac{813 (2.7756930708)^{2}}{1000} - e^{- (2.7756930708)}} = 2.1855774274$ $LaTeX: x_{3} = (2.1855774274) - \frac{- \frac{271 (2.1855774274)^{3}}{1000} + 2 + e^{- (2.1855774274)}}{- \frac{813 (2.1855774274)^{2}}{1000} - e^{- (2.1855774274)}} = 2.0061902910$ $LaTeX: x_{4} = (2.0061902910) - \frac{- \frac{271 (2.0061902910)^{3}}{1000} + 2 + e^{- (2.0061902910)}}{- \frac{813 (2.0061902910)^{2}}{1000} - e^{- (2.0061902910)}} = 1.9904290881$ $LaTeX: x_{5} = (1.9904290881) - \frac{- \frac{271 (1.9904290881)^{3}}{1000} + 2 + e^{- (1.9904290881)}}{- \frac{813 (1.9904290881)^{2}}{1000} - e^{- (1.9904290881)}} = 1.9903137318$