Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{69 x^{3}}{125} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{69 x_{n}^{3}}{125} + 8 + e^{- x_{n}}}{- \frac{207 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{69 (3.0000000000)^{3}}{125} + 8 + e^{- (3.0000000000)}}{- \frac{207 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.5416403283$ $LaTeX: x_{2} = (2.5416403283) - \frac{- \frac{69 (2.5416403283)^{3}}{125} + 8 + e^{- (2.5416403283)}}{- \frac{207 (2.5416403283)^{2}}{125} - e^{- (2.5416403283)}} = 2.4502870559$ $LaTeX: x_{3} = (2.4502870559) - \frac{- \frac{69 (2.4502870559)^{3}}{125} + 8 + e^{- (2.4502870559)}}{- \frac{207 (2.4502870559)^{2}}{125} - e^{- (2.4502870559)}} = 2.4468603146$ $LaTeX: x_{4} = (2.4468603146) - \frac{- \frac{69 (2.4468603146)^{3}}{125} + 8 + e^{- (2.4468603146)}}{- \frac{207 (2.4468603146)^{2}}{125} - e^{- (2.4468603146)}} = 2.4468556034$ $LaTeX: x_{5} = (2.4468556034) - \frac{- \frac{69 (2.4468556034)^{3}}{125} + 8 + e^{- (2.4468556034)}}{- \frac{207 (2.4468556034)^{2}}{125} - e^{- (2.4468556034)}} = 2.4468556034$