Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{117 x^{3}}{200} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{117 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{351 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.8398401749$ $LaTeX: x_{2} = (1.8398401749) - \frac{- \frac{117 (1.8398401749)^{3}}{200} + 2 + e^{- (1.8398401749)}}{- \frac{351 (1.8398401749)^{2}}{200} - e^{- (1.8398401749)}} = 1.5964664443$ $LaTeX: x_{3} = (1.5964664443) - \frac{- \frac{117 (1.5964664443)^{3}}{200} + 2 + e^{- (1.5964664443)}}{- \frac{351 (1.5964664443)^{2}}{200} - e^{- (1.5964664443)}} = 1.5584587630$ $LaTeX: x_{4} = (1.5584587630) - \frac{- \frac{117 (1.5584587630)^{3}}{200} + 2 + e^{- (1.5584587630)}}{- \frac{351 (1.5584587630)^{2}}{200} - e^{- (1.5584587630)}} = 1.5575942195$ $LaTeX: x_{5} = (1.5575942195) - \frac{- \frac{117 (1.5575942195)^{3}}{200} + 2 + e^{- (1.5575942195)}}{- \frac{351 (1.5575942195)^{2}}{200} - e^{- (1.5575942195)}} = 1.5575937797$