Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{109 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{109 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{327 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{109 (3.0000000000)^{3}}{1000} + 4 + e^{- (3.0000000000)}}{- \frac{327 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.3698181805$ $LaTeX: x_{2} = (3.3698181805) - \frac{- \frac{109 (3.3698181805)^{3}}{1000} + 4 + e^{- (3.3698181805)}}{- \frac{327 (3.3698181805)^{2}}{1000} - e^{- (3.3698181805)}} = 3.3333534302$ $LaTeX: x_{3} = (3.3333534302) - \frac{- \frac{109 (3.3333534302)^{3}}{1000} + 4 + e^{- (3.3333534302)}}{- \frac{327 (3.3333534302)^{2}}{1000} - e^{- (3.3333534302)}} = 3.3329618358$ $LaTeX: x_{4} = (3.3329618358) - \frac{- \frac{109 (3.3329618358)^{3}}{1000} + 4 + e^{- (3.3329618358)}}{- \frac{327 (3.3329618358)^{2}}{1000} - e^{- (3.3329618358)}} = 3.3329617910$ $LaTeX: x_{5} = (3.3329617910) - \frac{- \frac{109 (3.3329617910)^{3}}{1000} + 4 + e^{- (3.3329617910)}}{- \frac{327 (3.3329617910)^{2}}{1000} - e^{- (3.3329617910)}} = 3.3329617910$