Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{77 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{77 x_{n}^{3}}{125} + 7 + e^{- x_{n}}}{- \frac{231 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{77 (3.0000000000)^{3}}{125} + 7 + e^{- (3.0000000000)}}{- \frac{231 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.4255883442$ $LaTeX: x_{2} = (2.4255883442) - \frac{- \frac{77 (2.4255883442)^{3}}{125} + 7 + e^{- (2.4255883442)}}{- \frac{231 (2.4255883442)^{2}}{125} - e^{- (2.4255883442)}} = 2.2702711175$ $LaTeX: x_{3} = (2.2702711175) - \frac{- \frac{77 (2.2702711175)^{3}}{125} + 7 + e^{- (2.2702711175)}}{- \frac{231 (2.2702711175)^{2}}{125} - e^{- (2.2702711175)}} = 2.2593966150$ $LaTeX: x_{4} = (2.2593966150) - \frac{- \frac{77 (2.2593966150)^{3}}{125} + 7 + e^{- (2.2593966150)}}{- \frac{231 (2.2593966150)^{2}}{125} - e^{- (2.2593966150)}} = 2.2593453254$ $LaTeX: x_{5} = (2.2593453254) - \frac{- \frac{77 (2.2593453254)^{3}}{125} + 7 + e^{- (2.2593453254)}}{- \frac{231 (2.2593453254)^{2}}{125} - e^{- (2.2593453254)}} = 2.2593453242$