Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{x^{3}}{125} - 4$ using $LaTeX: \displaystyle x_0=7$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{125} + 4 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 7$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (7.0000000000) - \frac{- \frac{(7.0000000000)^{3}}{125} + 4 + e^{- (7.0000000000)}}{- \frac{3 (7.0000000000)^{2}}{125} - e^{- (7.0000000000)}} = 8.0679745028$ $LaTeX: x_{2} = (8.0679745028) - \frac{- \frac{(8.0679745028)^{3}}{125} + 4 + e^{- (8.0679745028)}}{- \frac{3 (8.0679745028)^{2}}{125} - e^{- (8.0679745028)}} = 7.9393462387$ $LaTeX: x_{3} = (7.9393462387) - \frac{- \frac{(7.9393462387)^{3}}{125} + 4 + e^{- (7.9393462387)}}{- \frac{3 (7.9393462387)^{2}}{125} - e^{- (7.9393462387)}} = 7.9372420619$ $LaTeX: x_{4} = (7.9372420619) - \frac{- \frac{(7.9372420619)^{3}}{125} + 4 + e^{- (7.9372420619)}}{- \frac{3 (7.9372420619)^{2}}{125} - e^{- (7.9372420619)}} = 7.9372415047$ $LaTeX: x_{5} = (7.9372415047) - \frac{- \frac{(7.9372415047)^{3}}{125} + 4 + e^{- (7.9372415047)}}{- \frac{3 (7.9372415047)^{2}}{125} - e^{- (7.9372415047)}} = 7.9372415047$