Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{63 x^{3}}{125} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{63 x_{n}^{3}}{125} + 3 + e^{- x_{n}}}{- \frac{189 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{63 (1.0000000000)^{3}}{125} + 3 + e^{- (1.0000000000)}}{- \frac{189 (1.0000000000)^{2}}{125} - e^{- (1.0000000000)}} = 2.5234378218$ $LaTeX: x_{2} = (2.5234378218) - \frac{- \frac{63 (2.5234378218)^{3}}{125} + 3 + e^{- (2.5234378218)}}{- \frac{189 (2.5234378218)^{2}}{125} - e^{- (2.5234378218)}} = 2.0065155462$ $LaTeX: x_{3} = (2.0065155462) - \frac{- \frac{63 (2.0065155462)^{3}}{125} + 3 + e^{- (2.0065155462)}}{- \frac{189 (2.0065155462)^{2}}{125} - e^{- (2.0065155462)}} = 1.8559065410$ $LaTeX: x_{4} = (1.8559065410) - \frac{- \frac{63 (1.8559065410)^{3}}{125} + 3 + e^{- (1.8559065410)}}{- \frac{189 (1.8559065410)^{2}}{125} - e^{- (1.8559065410)}} = 1.8436977372$ $LaTeX: x_{5} = (1.8436977372) - \frac{- \frac{63 (1.8436977372)^{3}}{125} + 3 + e^{- (1.8436977372)}}{- \frac{189 (1.8436977372)^{2}}{125} - e^{- (1.8436977372)}} = 1.8436211678$