Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{233 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{233 x_{n}^{3}}{250} + 6 + e^{- x_{n}}}{- \frac{699 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{233 (1.0000000000)^{3}}{250} + 6 + e^{- (1.0000000000)}}{- \frac{699 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.7181057440$ $LaTeX: x_{2} = (2.7181057440) - \frac{- \frac{233 (2.7181057440)^{3}}{250} + 6 + e^{- (2.7181057440)}}{- \frac{699 (2.7181057440)^{2}}{250} - e^{- (2.7181057440)}} = 2.1076725332$ $LaTeX: x_{3} = (2.1076725332) - \frac{- \frac{233 (2.1076725332)^{3}}{250} + 6 + e^{- (2.1076725332)}}{- \frac{699 (2.1076725332)^{2}}{250} - e^{- (2.1076725332)}} = 1.8999981155$ $LaTeX: x_{4} = (1.8999981155) - \frac{- \frac{233 (1.8999981155)^{3}}{250} + 6 + e^{- (1.8999981155)}}{- \frac{699 (1.8999981155)^{2}}{250} - e^{- (1.8999981155)}} = 1.8762748416$ $LaTeX: x_{5} = (1.8762748416) - \frac{- \frac{233 (1.8762748416)^{3}}{250} + 6 + e^{- (1.8762748416)}}{- \frac{699 (1.8762748416)^{2}}{250} - e^{- (1.8762748416)}} = 1.8759812390$