Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{81 x^{3}}{100} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{81 x_{n}^{3}}{100} + 6 + e^{- x_{n}}}{- \frac{243 x_{n}^{2}}{100} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{81 (1.0000000000)^{3}}{100} + 6 + e^{- (1.0000000000)}}{- \frac{243 (1.0000000000)^{2}}{100} - e^{- (1.0000000000)}} = 2.9864613748$ $LaTeX: x_{2} = (2.9864613748) - \frac{- \frac{81 (2.9864613748)^{3}}{100} + 6 + e^{- (2.9864613748)}}{- \frac{243 (2.9864613748)^{2}}{100} - e^{- (2.9864613748)}} = 2.2718083058$ $LaTeX: x_{3} = (2.2718083058) - \frac{- \frac{81 (2.2718083058)^{3}}{100} + 6 + e^{- (2.2718083058)}}{- \frac{243 (2.2718083058)^{2}}{100} - e^{- (2.2718083058)}} = 2.0033803240$ $LaTeX: x_{4} = (2.0033803240) - \frac{- \frac{81 (2.0033803240)^{3}}{100} + 6 + e^{- (2.0033803240)}}{- \frac{243 (2.0033803240)^{2}}{100} - e^{- (2.0033803240)}} = 1.9651478409$ $LaTeX: x_{5} = (1.9651478409) - \frac{- \frac{81 (1.9651478409)^{3}}{100} + 6 + e^{- (1.9651478409)}}{- \frac{243 (1.9651478409)^{2}}{100} - e^{- (1.9651478409)}} = 1.9644159395$