Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{561 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{561 x_{n}^{3}}{1000} + 6 + e^{- x_{n}}}{- \frac{1683 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{561 (3.0000000000)^{3}}{1000} + 6 + e^{- (3.0000000000)}}{- \frac{1683 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4013726131$ $LaTeX: x_{2} = (2.4013726131) - \frac{- \frac{561 (2.4013726131)^{3}}{1000} + 6 + e^{- (2.4013726131)}}{- \frac{1683 (2.4013726131)^{2}}{1000} - e^{- (2.4013726131)}} = 2.2300757027$ $LaTeX: x_{3} = (2.2300757027) - \frac{- \frac{561 (2.2300757027)^{3}}{1000} + 6 + e^{- (2.2300757027)}}{- \frac{1683 (2.2300757027)^{2}}{1000} - e^{- (2.2300757027)}} = 2.2165857960$ $LaTeX: x_{4} = (2.2165857960) - \frac{- \frac{561 (2.2165857960)^{3}}{1000} + 6 + e^{- (2.2165857960)}}{- \frac{1683 (2.2165857960)^{2}}{1000} - e^{- (2.2165857960)}} = 2.2165056100$ $LaTeX: x_{5} = (2.2165056100) - \frac{- \frac{561 (2.2165056100)^{3}}{1000} + 6 + e^{- (2.2165056100)}}{- \frac{1683 (2.2165056100)^{2}}{1000} - e^{- (2.2165056100)}} = 2.2165056072$