Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{421 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{421 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{1263 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{421 (3.0000000000)^{3}}{1000} + 4 + e^{- (3.0000000000)}}{- \frac{1263 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3590829988$ $LaTeX: x_{2} = (2.3590829988) - \frac{- \frac{421 (2.3590829988)^{3}}{1000} + 4 + e^{- (2.3590829988)}}{- \frac{1263 (2.3590829988)^{2}}{1000} - e^{- (2.3590829988)}} = 2.1579476422$ $LaTeX: x_{3} = (2.1579476422) - \frac{- \frac{421 (2.1579476422)^{3}}{1000} + 4 + e^{- (2.1579476422)}}{- \frac{1263 (2.1579476422)^{2}}{1000} - e^{- (2.1579476422)}} = 2.1387605164$ $LaTeX: x_{4} = (2.1387605164) - \frac{- \frac{421 (2.1387605164)^{3}}{1000} + 4 + e^{- (2.1387605164)}}{- \frac{1263 (2.1387605164)^{2}}{1000} - e^{- (2.1387605164)}} = 2.1385944482$ $LaTeX: x_{5} = (2.1385944482) - \frac{- \frac{421 (2.1385944482)^{3}}{1000} + 4 + e^{- (2.1385944482)}}{- \frac{1263 (2.1385944482)^{2}}{1000} - e^{- (2.1385944482)}} = 2.1385944359$