Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{41 x^{3}}{200} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{41 x_{n}^{3}}{200} + 7 + e^{- x_{n}}}{- \frac{123 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{41 (3.0000000000)^{3}}{200} + 7 + e^{- (3.0000000000)}}{- \frac{123 (3.0000000000)^{2}}{200} - e^{- (3.0000000000)}} = 3.2712345251$ $LaTeX: x_{2} = (3.2712345251) - \frac{- \frac{41 (3.2712345251)^{3}}{200} + 7 + e^{- (3.2712345251)}}{- \frac{123 (3.2712345251)^{2}}{200} - e^{- (3.2712345251)}} = 3.2503633652$ $LaTeX: x_{3} = (3.2503633652) - \frac{- \frac{41 (3.2503633652)^{3}}{200} + 7 + e^{- (3.2503633652)}}{- \frac{123 (3.2503633652)^{2}}{200} - e^{- (3.2503633652)}} = 3.2502308460$ $LaTeX: x_{4} = (3.2502308460) - \frac{- \frac{41 (3.2502308460)^{3}}{200} + 7 + e^{- (3.2502308460)}}{- \frac{123 (3.2502308460)^{2}}{200} - e^{- (3.2502308460)}} = 3.2502308407$ $LaTeX: x_{5} = (3.2502308407) - \frac{- \frac{41 (3.2502308407)^{3}}{200} + 7 + e^{- (3.2502308407)}}{- \frac{123 (3.2502308407)^{2}}{200} - e^{- (3.2502308407)}} = 3.2502308407$