Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{89 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{89 x_{n}^{3}}{125} + 7 + e^{- x_{n}}}{- \frac{267 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{89 (3.0000000000)^{3}}{125} + 7 + e^{- (3.0000000000)}}{- \frac{267 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.3683538742$ $LaTeX: x_{2} = (2.3683538742) - \frac{- \frac{89 (2.3683538742)^{3}}{125} + 7 + e^{- (2.3683538742)}}{- \frac{267 (2.3683538742)^{2}}{125} - e^{- (2.3683538742)}} = 2.1725050015$ $LaTeX: x_{3} = (2.1725050015) - \frac{- \frac{89 (2.1725050015)^{3}}{125} + 7 + e^{- (2.1725050015)}}{- \frac{267 (2.1725050015)^{2}}{125} - e^{- (2.1725050015)}} = 2.1541856648$ $LaTeX: x_{4} = (2.1541856648) - \frac{- \frac{89 (2.1541856648)^{3}}{125} + 7 + e^{- (2.1541856648)}}{- \frac{267 (2.1541856648)^{2}}{125} - e^{- (2.1541856648)}} = 2.1540327224$ $LaTeX: x_{5} = (2.1540327224) - \frac{- \frac{89 (2.1540327224)^{3}}{125} + 7 + e^{- (2.1540327224)}}{- \frac{267 (2.1540327224)^{2}}{125} - e^{- (2.1540327224)}} = 2.1540327118$