Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{59 x^{3}}{200} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{59 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{177 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{59 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{177 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 2.6544923422$ $LaTeX: x_{2} = (2.6544923422) - \frac{- \frac{59 (2.6544923422)^{3}}{200} + 2 + e^{- (2.6544923422)}}{- \frac{177 (2.6544923422)^{2}}{200} - e^{- (2.6544923422)}} = 2.1078244219$ $LaTeX: x_{3} = (2.1078244219) - \frac{- \frac{59 (2.1078244219)^{3}}{200} + 2 + e^{- (2.1078244219)}}{- \frac{177 (2.1078244219)^{2}}{200} - e^{- (2.1078244219)}} = 1.9496534340$ $LaTeX: x_{4} = (1.9496534340) - \frac{- \frac{59 (1.9496534340)^{3}}{200} + 2 + e^{- (1.9496534340)}}{- \frac{177 (1.9496534340)^{2}}{200} - e^{- (1.9496534340)}} = 1.9371336331$ $LaTeX: x_{5} = (1.9371336331) - \frac{- \frac{59 (1.9371336331)^{3}}{200} + 2 + e^{- (1.9371336331)}}{- \frac{177 (1.9371336331)^{2}}{200} - e^{- (1.9371336331)}} = 1.9370589807$