Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{249 x^{3}}{250} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{249 x_{n}^{3}}{250} + 5 + e^{- x_{n}}}{- \frac{747 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{249 (1.0000000000)^{3}}{250} + 5 + e^{- (1.0000000000)}}{- \frac{747 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.3027522346$ $LaTeX: x_{2} = (2.3027522346) - \frac{- \frac{249 (2.3027522346)^{3}}{250} + 5 + e^{- (2.3027522346)}}{- \frac{747 (2.3027522346)^{2}}{250} - e^{- (2.3027522346)}} = 1.8598428717$ $LaTeX: x_{3} = (1.8598428717) - \frac{- \frac{249 (1.8598428717)^{3}}{250} + 5 + e^{- (1.8598428717)}}{- \frac{747 (1.8598428717)^{2}}{250} - e^{- (1.8598428717)}} = 1.7405246704$ $LaTeX: x_{4} = (1.7405246704) - \frac{- \frac{249 (1.7405246704)^{3}}{250} + 5 + e^{- (1.7405246704)}}{- \frac{747 (1.7405246704)^{2}}{250} - e^{- (1.7405246704)}} = 1.7322588752$ $LaTeX: x_{5} = (1.7322588752) - \frac{- \frac{249 (1.7322588752)^{3}}{250} + 5 + e^{- (1.7322588752)}}{- \frac{747 (1.7322588752)^{2}}{250} - e^{- (1.7322588752)}} = 1.7322207307$