Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{57 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{57 x_{n}^{3}}{250} + 8 + e^{- x_{n}}}{- \frac{171 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{57 (3.0000000000)^{3}}{250} + 8 + e^{- (3.0000000000)}}{- \frac{171 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 3.3051646870$ $LaTeX: x_{2} = (3.3051646870) - \frac{- \frac{57 (3.3051646870)^{3}}{250} + 8 + e^{- (3.3051646870)}}{- \frac{171 (3.3051646870)^{2}}{250} - e^{- (3.3051646870)}} = 3.2791320417$ $LaTeX: x_{3} = (3.2791320417) - \frac{- \frac{57 (3.2791320417)^{3}}{250} + 8 + e^{- (3.2791320417)}}{- \frac{171 (3.2791320417)^{2}}{250} - e^{- (3.2791320417)}} = 3.2789270328$ $LaTeX: x_{4} = (3.2789270328) - \frac{- \frac{57 (3.2789270328)^{3}}{250} + 8 + e^{- (3.2789270328)}}{- \frac{171 (3.2789270328)^{2}}{250} - e^{- (3.2789270328)}} = 3.2789270201$ $LaTeX: x_{5} = (3.2789270201) - \frac{- \frac{57 (3.2789270201)^{3}}{250} + 8 + e^{- (3.2789270201)}}{- \frac{171 (3.2789270201)^{2}}{250} - e^{- (3.2789270201)}} = 3.2789270201$