Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{109 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{109 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{327 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{109 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{327 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.7039548823$ $LaTeX: x_{2} = (3.7039548823) - \frac{- \frac{109 (3.7039548823)^{3}}{1000} + 5 + e^{- (3.7039548823)}}{- \frac{327 (3.7039548823)^{2}}{1000} - e^{- (3.7039548823)}} = 3.5899460609$ $LaTeX: x_{3} = (3.5899460609) - \frac{- \frac{109 (3.5899460609)^{3}}{1000} + 5 + e^{- (3.5899460609)}}{- \frac{327 (3.5899460609)^{2}}{1000} - e^{- (3.5899460609)}} = 3.5863119960$ $LaTeX: x_{4} = (3.5863119960) - \frac{- \frac{109 (3.5863119960)^{3}}{1000} + 5 + e^{- (3.5863119960)}}{- \frac{327 (3.5863119960)^{2}}{1000} - e^{- (3.5863119960)}} = 3.5863083783$ $LaTeX: x_{5} = (3.5863083783) - \frac{- \frac{109 (3.5863083783)^{3}}{1000} + 5 + e^{- (3.5863083783)}}{- \frac{327 (3.5863083783)^{2}}{1000} - e^{- (3.5863083783)}} = 3.5863083783$