Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{83 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{83 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{249 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{83 (3.0000000000)^{3}}{1000} + 4 + e^{- (3.0000000000)}}{- \frac{249 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.7895919675$ $LaTeX: x_{2} = (3.7895919675) - \frac{- \frac{83 (3.7895919675)^{3}}{1000} + 4 + e^{- (3.7895919675)}}{- \frac{249 (3.7895919675)^{2}}{1000} - e^{- (3.7895919675)}} = 3.6521870831$ $LaTeX: x_{3} = (3.6521870831) - \frac{- \frac{83 (3.6521870831)^{3}}{1000} + 4 + e^{- (3.6521870831)}}{- \frac{249 (3.6521870831)^{2}}{1000} - e^{- (3.6521870831)}} = 3.6469957222$ $LaTeX: x_{4} = (3.6469957222) - \frac{- \frac{83 (3.6469957222)^{3}}{1000} + 4 + e^{- (3.6469957222)}}{- \frac{249 (3.6469957222)^{2}}{1000} - e^{- (3.6469957222)}} = 3.6469884881$ $LaTeX: x_{5} = (3.6469884881) - \frac{- \frac{83 (3.6469884881)^{3}}{1000} + 4 + e^{- (3.6469884881)}}{- \frac{249 (3.6469884881)^{2}}{1000} - e^{- (3.6469884881)}} = 3.6469884881$