Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{91 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{91 x_{n}^{3}}{1000} + 8 + e^{- x_{n}}}{- \frac{273 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{91 (5.0000000000)^{3}}{1000} + 8 + e^{- (5.0000000000)}}{- \frac{273 (5.0000000000)^{2}}{1000} - e^{- (5.0000000000)}} = 4.5069684934$ $LaTeX: x_{2} = (4.5069684934) - \frac{- \frac{91 (4.5069684934)^{3}}{1000} + 8 + e^{- (4.5069684934)}}{- \frac{273 (4.5069684934)^{2}}{1000} - e^{- (4.5069684934)}} = 4.4493906649$ $LaTeX: x_{3} = (4.4493906649) - \frac{- \frac{91 (4.4493906649)^{3}}{1000} + 8 + e^{- (4.4493906649)}}{- \frac{273 (4.4493906649)^{2}}{1000} - e^{- (4.4493906649)}} = 4.4486442081$ $LaTeX: x_{4} = (4.4486442081) - \frac{- \frac{91 (4.4486442081)^{3}}{1000} + 8 + e^{- (4.4486442081)}}{- \frac{273 (4.4486442081)^{2}}{1000} - e^{- (4.4486442081)}} = 4.4486440837$ $LaTeX: x_{5} = (4.4486440837) - \frac{- \frac{91 (4.4486440837)^{3}}{1000} + 8 + e^{- (4.4486440837)}}{- \frac{273 (4.4486440837)^{2}}{1000} - e^{- (4.4486440837)}} = 4.4486440837$