Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{719 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{719 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{2157 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{719 (1.0000000000)^{3}}{1000} + 4 + e^{- (1.0000000000)}}{- \frac{2157 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.4451697700$ $LaTeX: x_{2} = (2.4451697700) - \frac{- \frac{719 (2.4451697700)^{3}}{1000} + 4 + e^{- (2.4451697700)}}{- \frac{2157 (2.4451697700)^{2}}{1000} - e^{- (2.4451697700)}} = 1.9503283406$ $LaTeX: x_{3} = (1.9503283406) - \frac{- \frac{719 (1.9503283406)^{3}}{1000} + 4 + e^{- (1.9503283406)}}{- \frac{2157 (1.9503283406)^{2}}{1000} - e^{- (1.9503283406)}} = 1.8075508394$ $LaTeX: x_{4} = (1.8075508394) - \frac{- \frac{719 (1.8075508394)^{3}}{1000} + 4 + e^{- (1.8075508394)}}{- \frac{2157 (1.8075508394)^{2}}{1000} - e^{- (1.8075508394)}} = 1.7961600438$ $LaTeX: x_{5} = (1.7961600438) - \frac{- \frac{719 (1.7961600438)^{3}}{1000} + 4 + e^{- (1.7961600438)}}{- \frac{2157 (1.7961600438)^{2}}{1000} - e^{- (1.7961600438)}} = 1.7960906898$