Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{69 x^{3}}{125} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{69 x_{n}^{3}}{125} + 6 + e^{- x_{n}}}{- \frac{207 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{69 (3.0000000000)^{3}}{125} + 6 + e^{- (3.0000000000)}}{- \frac{207 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.4078949439$ $LaTeX: x_{2} = (2.4078949439) - \frac{- \frac{69 (2.4078949439)^{3}}{125} + 6 + e^{- (2.4078949439)}}{- \frac{207 (2.4078949439)^{2}}{125} - e^{- (2.4078949439)}} = 2.2411085169$ $LaTeX: x_{3} = (2.2411085169) - \frac{- \frac{69 (2.2411085169)^{3}}{125} + 6 + e^{- (2.2411085169)}}{- \frac{207 (2.2411085169)^{2}}{125} - e^{- (2.2411085169)}} = 2.2284019008$ $LaTeX: x_{4} = (2.2284019008) - \frac{- \frac{69 (2.2284019008)^{3}}{125} + 6 + e^{- (2.2284019008)}}{- \frac{207 (2.2284019008)^{2}}{125} - e^{- (2.2284019008)}} = 2.2283311458$ $LaTeX: x_{5} = (2.2283311458) - \frac{- \frac{69 (2.2283311458)^{3}}{125} + 6 + e^{- (2.2283311458)}}{- \frac{207 (2.2283311458)^{2}}{125} - e^{- (2.2283311458)}} = 2.2283311436$