Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{173 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{173 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{519 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{173 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{519 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.2920672016$ $LaTeX: x_{2} = (3.2920672016) - \frac{- \frac{173 (3.2920672016)^{3}}{1000} + 6 + e^{- (3.2920672016)}}{- \frac{519 (3.2920672016)^{2}}{1000} - e^{- (3.2920672016)}} = 3.2681891394$ $LaTeX: x_{3} = (3.2681891394) - \frac{- \frac{173 (3.2681891394)^{3}}{1000} + 6 + e^{- (3.2681891394)}}{- \frac{519 (3.2681891394)^{2}}{1000} - e^{- (3.2681891394)}} = 3.2680169416$ $LaTeX: x_{4} = (3.2680169416) - \frac{- \frac{173 (3.2680169416)^{3}}{1000} + 6 + e^{- (3.2680169416)}}{- \frac{519 (3.2680169416)^{2}}{1000} - e^{- (3.2680169416)}} = 3.2680169326$ $LaTeX: x_{5} = (3.2680169326) - \frac{- \frac{173 (3.2680169326)^{3}}{1000} + 6 + e^{- (3.2680169326)}}{- \frac{519 (3.2680169326)^{2}}{1000} - e^{- (3.2680169326)}} = 3.2680169326$