Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{9 x^{3}}{125} - 6$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{9 x_{n}^{3}}{125} + 6 + e^{- x_{n}}}{- \frac{27 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{9 (5.0000000000)^{3}}{125} + 6 + e^{- (5.0000000000)}}{- \frac{27 (5.0000000000)^{2}}{125} - e^{- (5.0000000000)}} = 4.4463829980$ $LaTeX: x_{2} = (4.4463829980) - \frac{- \frac{9 (4.4463829980)^{3}}{125} + 6 + e^{- (4.4463829980)}}{- \frac{27 (4.4463829980)^{2}}{125} - e^{- (4.4463829980)}} = 4.3722276141$ $LaTeX: x_{3} = (4.3722276141) - \frac{- \frac{9 (4.3722276141)^{3}}{125} + 6 + e^{- (4.3722276141)}}{- \frac{27 (4.3722276141)^{2}}{125} - e^{- (4.3722276141)}} = 4.3709675304$ $LaTeX: x_{4} = (4.3709675304) - \frac{- \frac{9 (4.3709675304)^{3}}{125} + 6 + e^{- (4.3709675304)}}{- \frac{27 (4.3709675304)^{2}}{125} - e^{- (4.3709675304)}} = 4.3709671706$ $LaTeX: x_{5} = (4.3709671706) - \frac{- \frac{9 (4.3709671706)^{3}}{125} + 6 + e^{- (4.3709671706)}}{- \frac{27 (4.3709671706)^{2}}{125} - e^{- (4.3709671706)}} = 4.3709671706$