Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{647 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{647 x_{n}^{3}}{1000} + 1 + e^{- x_{n}}}{- \frac{1941 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{647 (1.0000000000)^{3}}{1000} + 1 + e^{- (1.0000000000)}}{- \frac{1941 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 1.3122204773$ $LaTeX: x_{2} = (1.3122204773) - \frac{- \frac{647 (1.3122204773)^{3}}{1000} + 1 + e^{- (1.3122204773)}}{- \frac{1941 (1.3122204773)^{2}}{1000} - e^{- (1.3122204773)}} = 1.2588621631$ $LaTeX: x_{3} = (1.2588621631) - \frac{- \frac{647 (1.2588621631)^{3}}{1000} + 1 + e^{- (1.2588621631)}}{- \frac{1941 (1.2588621631)^{2}}{1000} - e^{- (1.2588621631)}} = 1.2568492749$ $LaTeX: x_{4} = (1.2568492749) - \frac{- \frac{647 (1.2568492749)^{3}}{1000} + 1 + e^{- (1.2568492749)}}{- \frac{1941 (1.2568492749)^{2}}{1000} - e^{- (1.2568492749)}} = 1.2568464936$ $LaTeX: x_{5} = (1.2568464936) - \frac{- \frac{647 (1.2568464936)^{3}}{1000} + 1 + e^{- (1.2568464936)}}{- \frac{1941 (1.2568464936)^{2}}{1000} - e^{- (1.2568464936)}} = 1.2568464936$