Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{9 x^{3}}{500} - 9$ using $LaTeX: \displaystyle x_0=7$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{9 x_{n}^{3}}{500} + 9 + e^{- x_{n}}}{- \frac{27 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 7$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (7.0000000000) - \frac{- \frac{9 (7.0000000000)^{3}}{500} + 9 + e^{- (7.0000000000)}}{- \frac{27 (7.0000000000)^{2}}{500} - e^{- (7.0000000000)}} = 8.0680037750$ $LaTeX: x_{2} = (8.0680037750) - \frac{- \frac{9 (8.0680037750)^{3}}{500} + 9 + e^{- (8.0680037750)}}{- \frac{27 (8.0680037750)^{2}}{500} - e^{- (8.0680037750)}} = 7.9392213895$ $LaTeX: x_{3} = (7.9392213895) - \frac{- \frac{9 (7.9392213895)^{3}}{500} + 9 + e^{- (7.9392213895)}}{- \frac{27 (7.9392213895)^{2}}{500} - e^{- (7.9392213895)}} = 7.9371108341$ $LaTeX: x_{4} = (7.9371108341) - \frac{- \frac{9 (7.9371108341)^{3}}{500} + 9 + e^{- (7.9371108341)}}{- \frac{27 (7.9371108341)^{2}}{500} - e^{- (7.9371108341)}} = 7.9371102731$ $LaTeX: x_{5} = (7.9371102731) - \frac{- \frac{9 (7.9371102731)^{3}}{500} + 9 + e^{- (7.9371102731)}}{- \frac{27 (7.9371102731)^{2}}{500} - e^{- (7.9371102731)}} = 7.9371102731$