Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{221 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{221 x_{n}^{3}}{1000} + 8 + e^{- x_{n}}}{- \frac{663 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{221 (3.0000000000)^{3}}{1000} + 8 + e^{- (3.0000000000)}}{- \frac{663 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.3461626687$ $LaTeX: x_{2} = (3.3461626687) - \frac{- \frac{221 (3.3461626687)^{3}}{1000} + 8 + e^{- (3.3461626687)}}{- \frac{663 (3.3461626687)^{2}}{1000} - e^{- (3.3461626687)}} = 3.3133366305$ $LaTeX: x_{3} = (3.3133366305) - \frac{- \frac{221 (3.3133366305)^{3}}{1000} + 8 + e^{- (3.3133366305)}}{- \frac{663 (3.3133366305)^{2}}{1000} - e^{- (3.3133366305)}} = 3.3130135183$ $LaTeX: x_{4} = (3.3130135183) - \frac{- \frac{221 (3.3130135183)^{3}}{1000} + 8 + e^{- (3.3130135183)}}{- \frac{663 (3.3130135183)^{2}}{1000} - e^{- (3.3130135183)}} = 3.3130134872$ $LaTeX: x_{5} = (3.3130134872) - \frac{- \frac{221 (3.3130134872)^{3}}{1000} + 8 + e^{- (3.3130134872)}}{- \frac{663 (3.3130134872)^{2}}{1000} - e^{- (3.3130134872)}} = 3.3130134872$