Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{63 x^{3}}{500} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{63 x_{n}^{3}}{500} + 6 + e^{- x_{n}}}{- \frac{189 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{63 (3.0000000000)^{3}}{500} + 6 + e^{- (3.0000000000)}}{- \frac{189 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 3.7670771736$ $LaTeX: x_{2} = (3.7670771736) - \frac{- \frac{63 (3.7670771736)^{3}}{500} + 6 + e^{- (3.7670771736)}}{- \frac{189 (3.7670771736)^{2}}{500} - e^{- (3.7670771736)}} = 3.6348020575$ $LaTeX: x_{3} = (3.6348020575) - \frac{- \frac{63 (3.6348020575)^{3}}{500} + 6 + e^{- (3.6348020575)}}{- \frac{189 (3.6348020575)^{2}}{500} - e^{- (3.6348020575)}} = 3.6299396521$ $LaTeX: x_{4} = (3.6299396521) - \frac{- \frac{63 (3.6299396521)^{3}}{500} + 6 + e^{- (3.6299396521)}}{- \frac{189 (3.6299396521)^{2}}{500} - e^{- (3.6299396521)}} = 3.6299332299$ $LaTeX: x_{5} = (3.6299332299) - \frac{- \frac{63 (3.6299332299)^{3}}{500} + 6 + e^{- (3.6299332299)}}{- \frac{189 (3.6299332299)^{2}}{500} - e^{- (3.6299332299)}} = 3.6299332299$