Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{89 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{89 x_{n}^{3}}{250} + 8 + e^{- x_{n}}}{- \frac{267 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{89 (3.0000000000)^{3}}{250} + 8 + e^{- (3.0000000000)}}{- \frac{267 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.8383101469$ $LaTeX: x_{2} = (2.8383101469) - \frac{- \frac{89 (2.8383101469)^{3}}{250} + 8 + e^{- (2.8383101469)}}{- \frac{267 (2.8383101469)^{2}}{250} - e^{- (2.8383101469)}} = 2.8288932997$ $LaTeX: x_{3} = (2.8288932997) - \frac{- \frac{89 (2.8288932997)^{3}}{250} + 8 + e^{- (2.8288932997)}}{- \frac{267 (2.8288932997)^{2}}{250} - e^{- (2.8288932997)}} = 2.8288624014$ $LaTeX: x_{4} = (2.8288624014) - \frac{- \frac{89 (2.8288624014)^{3}}{250} + 8 + e^{- (2.8288624014)}}{- \frac{267 (2.8288624014)^{2}}{250} - e^{- (2.8288624014)}} = 2.8288624011$ $LaTeX: x_{5} = (2.8288624011) - \frac{- \frac{89 (2.8288624011)^{3}}{250} + 8 + e^{- (2.8288624011)}}{- \frac{267 (2.8288624011)^{2}}{250} - e^{- (2.8288624011)}} = 2.8288624011$