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

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{69 x_{n}^{3}}{125} + 5 + e^{- x_{n}}}{- \frac{207 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{69 (3.0000000000)^{3}}{125} + 5 + e^{- (3.0000000000)}}{- \frac{207 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.3410222516$ $LaTeX: x_{2} = (2.3410222516) - \frac{- \frac{69 (2.3410222516)^{3}}{125} + 5 + e^{- (2.3410222516)}}{- \frac{207 (2.3410222516)^{2}}{125} - e^{- (2.3410222516)}} = 2.1245130974$ $LaTeX: x_{3} = (2.1245130974) - \frac{- \frac{69 (2.1245130974)^{3}}{125} + 5 + e^{- (2.1245130974)}}{- \frac{207 (2.1245130974)^{2}}{125} - e^{- (2.1245130974)}} = 2.1016400125$ $LaTeX: x_{4} = (2.1016400125) - \frac{- \frac{69 (2.1016400125)^{3}}{125} + 5 + e^{- (2.1016400125)}}{- \frac{207 (2.1016400125)^{2}}{125} - e^{- (2.1016400125)}} = 2.1013976259$ $LaTeX: x_{5} = (2.1013976259) - \frac{- \frac{69 (2.1013976259)^{3}}{125} + 5 + e^{- (2.1013976259)}}{- \frac{207 (2.1013976259)^{2}}{125} - e^{- (2.1013976259)}} = 2.1013975989$