Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{489 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{489 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1467 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{489 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1467 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3847925806$ $LaTeX: x_{2} = (2.3847925806) - \frac{- \frac{489 (2.3847925806)^{3}}{1000} + 5 + e^{- (2.3847925806)}}{- \frac{1467 (2.3847925806)^{2}}{1000} - e^{- (2.3847925806)}} = 2.2022096134$ $LaTeX: x_{3} = (2.2022096134) - \frac{- \frac{489 (2.2022096134)^{3}}{1000} + 5 + e^{- (2.2022096134)}}{- \frac{1467 (2.2022096134)^{2}}{1000} - e^{- (2.2022096134)}} = 2.1867056190$ $LaTeX: x_{4} = (2.1867056190) - \frac{- \frac{489 (2.1867056190)^{3}}{1000} + 5 + e^{- (2.1867056190)}}{- \frac{1467 (2.1867056190)^{2}}{1000} - e^{- (2.1867056190)}} = 2.1865987884$ $LaTeX: x_{5} = (2.1865987884) - \frac{- \frac{489 (2.1865987884)^{3}}{1000} + 5 + e^{- (2.1865987884)}}{- \frac{1467 (2.1865987884)^{2}}{1000} - e^{- (2.1865987884)}} = 2.1865987834$