Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{613 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{613 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{1839 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{613 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{1839 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3674268040$ $LaTeX: x_{2} = (2.3674268040) - \frac{- \frac{613 (2.3674268040)^{3}}{1000} + 6 + e^{- (2.3674268040)}}{- \frac{1839 (2.3674268040)^{2}}{1000} - e^{- (2.3674268040)}} = 2.1712861819$ $LaTeX: x_{3} = (2.1712861819) - \frac{- \frac{613 (2.1712861819)^{3}}{1000} + 6 + e^{- (2.1712861819)}}{- \frac{1839 (2.1712861819)^{2}}{1000} - e^{- (2.1712861819)}} = 2.1529642173$ $LaTeX: x_{4} = (2.1529642173) - \frac{- \frac{613 (2.1529642173)^{3}}{1000} + 6 + e^{- (2.1529642173)}}{- \frac{1839 (2.1529642173)^{2}}{1000} - e^{- (2.1529642173)}} = 2.1528117472$ $LaTeX: x_{5} = (2.1528117472) - \frac{- \frac{613 (2.1528117472)^{3}}{1000} + 6 + e^{- (2.1528117472)}}{- \frac{1839 (2.1528117472)^{2}}{1000} - e^{- (2.1528117472)}} = 2.1528117367$