Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{x^{3}}{4} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{4} + 6 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{4} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{(3.0000000000)^{3}}{4} + 6 + e^{- (3.0000000000)}}{- \frac{3 (3.0000000000)^{2}}{4} - e^{- (3.0000000000)}} = 2.8970242855$ $LaTeX: x_{2} = (2.8970242855) - \frac{- \frac{(2.8970242855)^{3}}{4} + 6 + e^{- (2.8970242855)}}{- \frac{3 (2.8970242855)^{2}}{4} - e^{- (2.8970242855)}} = 2.8933528436$ $LaTeX: x_{3} = (2.8933528436) - \frac{- \frac{(2.8933528436)^{3}}{4} + 6 + e^{- (2.8933528436)}}{- \frac{3 (2.8933528436)^{2}}{4} - e^{- (2.8933528436)}} = 2.8933482805$ $LaTeX: x_{4} = (2.8933482805) - \frac{- \frac{(2.8933482805)^{3}}{4} + 6 + e^{- (2.8933482805)}}{- \frac{3 (2.8933482805)^{2}}{4} - e^{- (2.8933482805)}} = 2.8933482805$ $LaTeX: x_{5} = (2.8933482805) - \frac{- \frac{(2.8933482805)^{3}}{4} + 6 + e^{- (2.8933482805)}}{- \frac{3 (2.8933482805)^{2}}{4} - e^{- (2.8933482805)}} = 2.8933482805$