Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{189 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{189 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{567 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{189 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{567 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.1837427116$ $LaTeX: x_{2} = (3.1837427116) - \frac{- \frac{189 (3.1837427116)^{3}}{1000} + 6 + e^{- (3.1837427116)}}{- \frac{567 (3.1837427116)^{2}}{1000} - e^{- (3.1837427116)}} = 3.1737560729$ $LaTeX: x_{3} = (3.1737560729) - \frac{- \frac{189 (3.1737560729)^{3}}{1000} + 6 + e^{- (3.1737560729)}}{- \frac{567 (3.1737560729)^{2}}{1000} - e^{- (3.1737560729)}} = 3.1737251721$ $LaTeX: x_{4} = (3.1737251721) - \frac{- \frac{189 (3.1737251721)^{3}}{1000} + 6 + e^{- (3.1737251721)}}{- \frac{567 (3.1737251721)^{2}}{1000} - e^{- (3.1737251721)}} = 3.1737251718$ $LaTeX: x_{5} = (3.1737251718) - \frac{- \frac{189 (3.1737251718)^{3}}{1000} + 6 + e^{- (3.1737251718)}}{- \frac{567 (3.1737251718)^{2}}{1000} - e^{- (3.1737251718)}} = 3.1737251718$