Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{54 x^{3}}{125} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{54 x_{n}^{3}}{125} + 4 + e^{- x_{n}}}{- \frac{162 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{54 (3.0000000000)^{3}}{125} + 4 + e^{- (3.0000000000)}}{- \frac{162 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.3499785435$ $LaTeX: x_{2} = (2.3499785435) - \frac{- \frac{54 (2.3499785435)^{3}}{125} + 4 + e^{- (2.3499785435)}}{- \frac{162 (2.3499785435)^{2}}{125} - e^{- (2.3499785435)}} = 2.1416451479$ $LaTeX: x_{3} = (2.1416451479) - \frac{- \frac{54 (2.1416451479)^{3}}{125} + 4 + e^{- (2.1416451479)}}{- \frac{162 (2.1416451479)^{2}}{125} - e^{- (2.1416451479)}} = 2.1208493627$ $LaTeX: x_{4} = (2.1208493627) - \frac{- \frac{54 (2.1208493627)^{3}}{125} + 4 + e^{- (2.1208493627)}}{- \frac{162 (2.1208493627)^{2}}{125} - e^{- (2.1208493627)}} = 2.1206525550$ $LaTeX: x_{5} = (2.1206525550) - \frac{- \frac{54 (2.1206525550)^{3}}{125} + 4 + e^{- (2.1206525550)}}{- \frac{162 (2.1206525550)^{2}}{125} - e^{- (2.1206525550)}} = 2.1206525375$