Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{753 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{753 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{2259 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{753 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{2259 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.2992805978$ $LaTeX: x_{2} = (2.2992805978) - \frac{- \frac{753 (2.2992805978)^{3}}{1000} + 6 + e^{- (2.2992805978)}}{- \frac{2259 (2.2992805978)^{2}}{1000} - e^{- (2.2992805978)}} = 2.0457861004$ $LaTeX: x_{3} = (2.0457861004) - \frac{- \frac{753 (2.0457861004)^{3}}{1000} + 6 + e^{- (2.0457861004)}}{- \frac{2259 (2.0457861004)^{2}}{1000} - e^{- (2.0457861004)}} = 2.0126061073$ $LaTeX: x_{4} = (2.0126061073) - \frac{- \frac{753 (2.0126061073)^{3}}{1000} + 6 + e^{- (2.0126061073)}}{- \frac{2259 (2.0126061073)^{2}}{1000} - e^{- (2.0126061073)}} = 2.0120687983$ $LaTeX: x_{5} = (2.0120687983) - \frac{- \frac{753 (2.0120687983)^{3}}{1000} + 6 + e^{- (2.0120687983)}}{- \frac{2259 (2.0120687983)^{2}}{1000} - e^{- (2.0120687983)}} = 2.0120686590$