Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{747 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{747 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{2241 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{747 (1.0000000000)^{3}}{1000} + 5 + e^{- (1.0000000000)}}{- \frac{2241 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.7712123329$ $LaTeX: x_{2} = (2.7712123329) - \frac{- \frac{747 (2.7712123329)^{3}}{1000} + 5 + e^{- (2.7712123329)}}{- \frac{2241 (2.7712123329)^{2}}{1000} - e^{- (2.7712123329)}} = 2.1439210726$ $LaTeX: x_{3} = (2.1439210726) - \frac{- \frac{747 (2.1439210726)^{3}}{1000} + 5 + e^{- (2.1439210726)}}{- \frac{2241 (2.1439210726)^{2}}{1000} - e^{- (2.1439210726)}} = 1.9285210088$ $LaTeX: x_{4} = (1.9285210088) - \frac{- \frac{747 (1.9285210088)^{3}}{1000} + 5 + e^{- (1.9285210088)}}{- \frac{2241 (1.9285210088)^{2}}{1000} - e^{- (1.9285210088)}} = 1.9034591877$ $LaTeX: x_{5} = (1.9034591877) - \frac{- \frac{747 (1.9034591877)^{3}}{1000} + 5 + e^{- (1.9034591877)}}{- \frac{2241 (1.9034591877)^{2}}{1000} - e^{- (1.9034591877)}} = 1.9031378837$