Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{153 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{153 x_{n}^{3}}{1000} + 3 + e^{- x_{n}}}{- \frac{459 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{153 (3.0000000000)^{3}}{1000} + 3 + e^{- (3.0000000000)}}{- \frac{459 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.7413853148$ $LaTeX: x_{2} = (2.7413853148) - \frac{- \frac{153 (2.7413853148)^{3}}{1000} + 3 + e^{- (2.7413853148)}}{- \frac{459 (2.7413853148)^{2}}{1000} - e^{- (2.7413853148)}} = 2.7164472287$ $LaTeX: x_{3} = (2.7164472287) - \frac{- \frac{153 (2.7164472287)^{3}}{1000} + 3 + e^{- (2.7164472287)}}{- \frac{459 (2.7164472287)^{2}}{1000} - e^{- (2.7164472287)}} = 2.7162271508$ $LaTeX: x_{4} = (2.7162271508) - \frac{- \frac{153 (2.7162271508)^{3}}{1000} + 3 + e^{- (2.7162271508)}}{- \frac{459 (2.7162271508)^{2}}{1000} - e^{- (2.7162271508)}} = 2.7162271337$ $LaTeX: x_{5} = (2.7162271337) - \frac{- \frac{153 (2.7162271337)^{3}}{1000} + 3 + e^{- (2.7162271337)}}{- \frac{459 (2.7162271337)^{2}}{1000} - e^{- (2.7162271337)}} = 2.7162271337$