Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{743 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{743 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{2229 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{743 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{2229 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3032986285$ $LaTeX: x_{2} = (2.3032986285) - \frac{- \frac{743 (2.3032986285)^{3}}{1000} + 6 + e^{- (2.3032986285)}}{- \frac{2229 (2.3032986285)^{2}}{1000} - e^{- (2.3032986285)}} = 2.0534824958$ $LaTeX: x_{3} = (2.0534824958) - \frac{- \frac{743 (2.0534824958)^{3}}{1000} + 6 + e^{- (2.0534824958)}}{- \frac{2229 (2.0534824958)^{2}}{1000} - e^{- (2.0534824958)}} = 2.0214250437$ $LaTeX: x_{4} = (2.0214250437) - \frac{- \frac{743 (2.0214250437)^{3}}{1000} + 6 + e^{- (2.0214250437)}}{- \frac{2229 (2.0214250437)^{2}}{1000} - e^{- (2.0214250437)}} = 2.0209258503$ $LaTeX: x_{5} = (2.0209258503) - \frac{- \frac{743 (2.0209258503)^{3}}{1000} + 6 + e^{- (2.0209258503)}}{- \frac{2229 (2.0209258503)^{2}}{1000} - e^{- (2.0209258503)}} = 2.0209257306$