Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{69 x^{3}}{250} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{69 x_{n}^{3}}{250} + 8 + e^{- x_{n}}}{- \frac{207 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{69 (3.0000000000)^{3}}{250} + 8 + e^{- (3.0000000000)}}{- \frac{207 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 3.0796859552$ $LaTeX: x_{2} = (3.0796859552) - \frac{- \frac{69 (3.0796859552)^{3}}{250} + 8 + e^{- (3.0796859552)}}{- \frac{207 (3.0796859552)^{2}}{250} - e^{- (3.0796859552)}} = 3.0776909546$ $LaTeX: x_{3} = (3.0776909546) - \frac{- \frac{69 (3.0776909546)^{3}}{250} + 8 + e^{- (3.0776909546)}}{- \frac{207 (3.0776909546)^{2}}{250} - e^{- (3.0776909546)}} = 3.0776896800$ $LaTeX: x_{4} = (3.0776896800) - \frac{- \frac{69 (3.0776896800)^{3}}{250} + 8 + e^{- (3.0776896800)}}{- \frac{207 (3.0776896800)^{2}}{250} - e^{- (3.0776896800)}} = 3.0776896800$ $LaTeX: x_{5} = (3.0776896800) - \frac{- \frac{69 (3.0776896800)^{3}}{250} + 8 + e^{- (3.0776896800)}}{- \frac{207 (3.0776896800)^{2}}{250} - e^{- (3.0776896800)}} = 3.0776896800$