Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{853 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{853 x_{n}^{3}}{1000} + 9 + e^{- x_{n}}}{- \frac{2559 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{853 (3.0000000000)^{3}}{1000} + 9 + e^{- (3.0000000000)}}{- \frac{2559 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3942488664$ $LaTeX: x_{2} = (2.3942488664) - \frac{- \frac{853 (2.3942488664)^{3}}{1000} + 9 + e^{- (2.3942488664)}}{- \frac{2559 (2.3942488664)^{2}}{1000} - e^{- (2.3942488664)}} = 2.2170151120$ $LaTeX: x_{3} = (2.2170151120) - \frac{- \frac{853 (2.2170151120)^{3}}{1000} + 9 + e^{- (2.2170151120)}}{- \frac{2559 (2.2170151120)^{2}}{1000} - e^{- (2.2170151120)}} = 2.2023395630$ $LaTeX: x_{4} = (2.2023395630) - \frac{- \frac{853 (2.2023395630)^{3}}{1000} + 9 + e^{- (2.2023395630)}}{- \frac{2559 (2.2023395630)^{2}}{1000} - e^{- (2.2023395630)}} = 2.2022431448$ $LaTeX: x_{5} = (2.2022431448) - \frac{- \frac{853 (2.2022431448)^{3}}{1000} + 9 + e^{- (2.2022431448)}}{- \frac{2559 (2.2022431448)^{2}}{1000} - e^{- (2.2022431448)}} = 2.2022431406$