Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{103 x^{3}}{200} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{103 x_{n}^{3}}{200} + 8 + e^{- x_{n}}}{- \frac{309 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{103 (3.0000000000)^{3}}{200} + 8 + e^{- (3.0000000000)}}{- \frac{309 (3.0000000000)^{2}}{200} - e^{- (3.0000000000)}} = 2.5804154587$ $LaTeX: x_{2} = (2.5804154587) - \frac{- \frac{103 (2.5804154587)^{3}}{200} + 8 + e^{- (2.5804154587)}}{- \frac{309 (2.5804154587)^{2}}{200} - e^{- (2.5804154587)}} = 2.5058352131$ $LaTeX: x_{3} = (2.5058352131) - \frac{- \frac{103 (2.5058352131)^{3}}{200} + 8 + e^{- (2.5058352131)}}{- \frac{309 (2.5058352131)^{2}}{200} - e^{- (2.5058352131)}} = 2.5036124272$ $LaTeX: x_{4} = (2.5036124272) - \frac{- \frac{103 (2.5036124272)^{3}}{200} + 8 + e^{- (2.5036124272)}}{- \frac{309 (2.5036124272)^{2}}{200} - e^{- (2.5036124272)}} = 2.5036104897$ $LaTeX: x_{5} = (2.5036104897) - \frac{- \frac{103 (2.5036104897)^{3}}{200} + 8 + e^{- (2.5036104897)}}{- \frac{309 (2.5036104897)^{2}}{200} - e^{- (2.5036104897)}} = 2.5036104897$