Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{137 x^{3}}{200} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{137 x_{n}^{3}}{200} + 6 + e^{- x_{n}}}{- \frac{411 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{137 (3.0000000000)^{3}}{200} + 6 + e^{- (3.0000000000)}}{- \frac{411 (3.0000000000)^{2}}{200} - e^{- (3.0000000000)}} = 2.3289104434$ $LaTeX: x_{2} = (2.3289104434) - \frac{- \frac{137 (2.3289104434)^{3}}{200} + 6 + e^{- (2.3289104434)}}{- \frac{411 (2.3289104434)^{2}}{200} - e^{- (2.3289104434)}} = 2.1016434845$ $LaTeX: x_{3} = (2.1016434845) - \frac{- \frac{137 (2.1016434845)^{3}}{200} + 6 + e^{- (2.1016434845)}}{- \frac{411 (2.1016434845)^{2}}{200} - e^{- (2.1016434845)}} = 2.0759411702$ $LaTeX: x_{4} = (2.0759411702) - \frac{- \frac{137 (2.0759411702)^{3}}{200} + 6 + e^{- (2.0759411702)}}{- \frac{411 (2.0759411702)^{2}}{200} - e^{- (2.0759411702)}} = 2.0756293380$ $LaTeX: x_{5} = (2.0756293380) - \frac{- \frac{137 (2.0756293380)^{3}}{200} + 6 + e^{- (2.0756293380)}}{- \frac{411 (2.0756293380)^{2}}{200} - e^{- (2.0756293380)}} = 2.0756292925$