Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{63 x^{3}}{500} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{63 x_{n}^{3}}{500} + 3 + e^{- x_{n}}}{- \frac{189 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{63 (3.0000000000)^{3}}{500} + 3 + e^{- (3.0000000000)}}{- \frac{189 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 2.8979621498$ $LaTeX: x_{2} = (2.8979621498) - \frac{- \frac{63 (2.8979621498)^{3}}{500} + 3 + e^{- (2.8979621498)}}{- \frac{189 (2.8979621498)^{2}}{500} - e^{- (2.8979621498)}} = 2.8944308676$ $LaTeX: x_{3} = (2.8944308676) - \frac{- \frac{63 (2.8944308676)^{3}}{500} + 3 + e^{- (2.8944308676)}}{- \frac{189 (2.8944308676)^{2}}{500} - e^{- (2.8944308676)}} = 2.8944267367$ $LaTeX: x_{4} = (2.8944267367) - \frac{- \frac{63 (2.8944267367)^{3}}{500} + 3 + e^{- (2.8944267367)}}{- \frac{189 (2.8944267367)^{2}}{500} - e^{- (2.8944267367)}} = 2.8944267367$ $LaTeX: x_{5} = (2.8944267367) - \frac{- \frac{63 (2.8944267367)^{3}}{500} + 3 + e^{- (2.8944267367)}}{- \frac{189 (2.8944267367)^{2}}{500} - e^{- (2.8944267367)}} = 2.8944267367$