Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{68 x^{3}}{125} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{68 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{204 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{68 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{204 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.6174315109$ $LaTeX: x_{2} = (2.6174315109) - \frac{- \frac{68 (2.6174315109)^{3}}{125} + 9 + e^{- (2.6174315109)}}{- \frac{204 (2.6174315109)^{2}}{125} - e^{- (2.6174315109)}} = 2.5568333173$ $LaTeX: x_{3} = (2.5568333173) - \frac{- \frac{68 (2.5568333173)^{3}}{125} + 9 + e^{- (2.5568333173)}}{- \frac{204 (2.5568333173)^{2}}{125} - e^{- (2.5568333173)}} = 2.5553976721$ $LaTeX: x_{4} = (2.5553976721) - \frac{- \frac{68 (2.5553976721)^{3}}{125} + 9 + e^{- (2.5553976721)}}{- \frac{204 (2.5553976721)^{2}}{125} - e^{- (2.5553976721)}} = 2.5553968785$ $LaTeX: x_{5} = (2.5553968785) - \frac{- \frac{68 (2.5553968785)^{3}}{125} + 9 + e^{- (2.5553968785)}}{- \frac{204 (2.5553968785)^{2}}{125} - e^{- (2.5553968785)}} = 2.5553968785$