Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{761 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{761 x_{n}^{3}}{1000} + 7 + e^{- x_{n}}}{- \frac{2283 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{761 (3.0000000000)^{3}}{1000} + 7 + e^{- (3.0000000000)}}{- \frac{2283 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3446932822$ $LaTeX: x_{2} = (2.3446932822) - \frac{- \frac{761 (2.3446932822)^{3}}{1000} + 7 + e^{- (2.3446932822)}}{- \frac{2283 (2.3446932822)^{2}}{1000} - e^{- (2.3446932822)}} = 2.1301317711$ $LaTeX: x_{3} = (2.1301317711) - \frac{- \frac{761 (2.1301317711)^{3}}{1000} + 7 + e^{- (2.1301317711)}}{- \frac{2283 (2.1301317711)^{2}}{1000} - e^{- (2.1301317711)}} = 2.1075564614$ $LaTeX: x_{4} = (2.1075564614) - \frac{- \frac{761 (2.1075564614)^{3}}{1000} + 7 + e^{- (2.1075564614)}}{- \frac{2283 (2.1075564614)^{2}}{1000} - e^{- (2.1075564614)}} = 2.1073187739$ $LaTeX: x_{5} = (2.1073187739) - \frac{- \frac{761 (2.1073187739)^{3}}{1000} + 7 + e^{- (2.1073187739)}}{- \frac{2283 (2.1073187739)^{2}}{1000} - e^{- (2.1073187739)}} = 2.1073187477$