Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{623 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{623 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1869 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{623 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1869 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3022724498$ $LaTeX: x_{2} = (2.3022724498) - \frac{- \frac{623 (2.3022724498)^{3}}{1000} + 5 + e^{- (2.3022724498)}}{- \frac{1869 (2.3022724498)^{2}}{1000} - e^{- (2.3022724498)}} = 2.0521872409$ $LaTeX: x_{3} = (2.0521872409) - \frac{- \frac{623 (2.0521872409)^{3}}{1000} + 5 + e^{- (2.0521872409)}}{- \frac{1869 (2.0521872409)^{2}}{1000} - e^{- (2.0521872409)}} = 2.0201901240$ $LaTeX: x_{4} = (2.0201901240) - \frac{- \frac{623 (2.0201901240)^{3}}{1000} + 5 + e^{- (2.0201901240)}}{- \frac{1869 (2.0201901240)^{2}}{1000} - e^{- (2.0201901240)}} = 2.0196952986$ $LaTeX: x_{5} = (2.0196952986) - \frac{- \frac{623 (2.0196952986)^{3}}{1000} + 5 + e^{- (2.0196952986)}}{- \frac{1869 (2.0196952986)^{2}}{1000} - e^{- (2.0196952986)}} = 2.0196951815$