Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{933 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{933 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{2799 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{933 (1.0000000000)^{3}}{1000} + 5 + e^{- (1.0000000000)}}{- \frac{2799 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.4003941494$ $LaTeX: x_{2} = (2.4003941494) - \frac{- \frac{933 (2.4003941494)^{3}}{1000} + 5 + e^{- (2.4003941494)}}{- \frac{2799 (2.4003941494)^{2}}{1000} - e^{- (2.4003941494)}} = 1.9186232572$ $LaTeX: x_{3} = (1.9186232572) - \frac{- \frac{933 (1.9186232572)^{3}}{1000} + 5 + e^{- (1.9186232572)}}{- \frac{2799 (1.9186232572)^{2}}{1000} - e^{- (1.9186232572)}} = 1.7805724796$ $LaTeX: x_{4} = (1.7805724796) - \frac{- \frac{933 (1.7805724796)^{3}}{1000} + 5 + e^{- (1.7805724796)}}{- \frac{2799 (1.7805724796)^{2}}{1000} - e^{- (1.7805724796)}} = 1.7696878246$ $LaTeX: x_{5} = (1.7696878246) - \frac{- \frac{933 (1.7696878246)^{3}}{1000} + 5 + e^{- (1.7696878246)}}{- \frac{2799 (1.7696878246)^{2}}{1000} - e^{- (1.7696878246)}} = 1.7696230058$