Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{141 x^{3}}{200} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{141 x_{n}^{3}}{200} + 4 + e^{- x_{n}}}{- \frac{423 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{141 (1.0000000000)^{3}}{200} + 4 + e^{- (1.0000000000)}}{- \frac{423 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 2.4752546501$ $LaTeX: x_{2} = (2.4752546501) - \frac{- \frac{141 (2.4752546501)^{3}}{200} + 4 + e^{- (2.4752546501)}}{- \frac{423 (2.4752546501)^{2}}{200} - e^{- (2.4752546501)}} = 1.9686335609$ $LaTeX: x_{3} = (1.9686335609) - \frac{- \frac{141 (1.9686335609)^{3}}{200} + 4 + e^{- (1.9686335609)}}{- \frac{423 (1.9686335609)^{2}}{200} - e^{- (1.9686335609)}} = 1.8199918172$ $LaTeX: x_{4} = (1.8199918172) - \frac{- \frac{141 (1.8199918172)^{3}}{200} + 4 + e^{- (1.8199918172)}}{- \frac{423 (1.8199918172)^{2}}{200} - e^{- (1.8199918172)}} = 1.8077066827$ $LaTeX: x_{5} = (1.8077066827) - \frac{- \frac{141 (1.8077066827)^{3}}{200} + 4 + e^{- (1.8077066827)}}{- \frac{423 (1.8077066827)^{2}}{200} - e^{- (1.8077066827)}} = 1.8076264944$