Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{117 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{117 (3.0000000000)^{3}}{1000} + 2 + e^{- (3.0000000000)}}{- \frac{351 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.6543201814$ $LaTeX: x_{2} = (2.6543201814) - \frac{- \frac{117 (2.6543201814)^{3}}{1000} + 2 + e^{- (2.6543201814)}}{- \frac{351 (2.6543201814)^{2}}{1000} - e^{- (2.6543201814)}} = 2.6080628659$ $LaTeX: x_{3} = (2.6080628659) - \frac{- \frac{117 (2.6080628659)^{3}}{1000} + 2 + e^{- (2.6080628659)}}{- \frac{351 (2.6080628659)^{2}}{1000} - e^{- (2.6080628659)}} = 2.6072886398$ $LaTeX: x_{4} = (2.6072886398) - \frac{- \frac{117 (2.6072886398)^{3}}{1000} + 2 + e^{- (2.6072886398)}}{- \frac{351 (2.6072886398)^{2}}{1000} - e^{- (2.6072886398)}} = 2.6072884257$ $LaTeX: x_{5} = (2.6072884257) - \frac{- \frac{117 (2.6072884257)^{3}}{1000} + 2 + e^{- (2.6072884257)}}{- \frac{351 (2.6072884257)^{2}}{1000} - e^{- (2.6072884257)}} = 2.6072884257$