Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{889 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{889 x_{n}^{3}}{1000} + 7 + e^{- x_{n}}}{- \frac{2667 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{889 (3.0000000000)^{3}}{1000} + 7 + e^{- (3.0000000000)}}{- \frac{2667 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.2951663820$ $LaTeX: x_{2} = (2.2951663820) - \frac{- \frac{889 (2.2951663820)^{3}}{1000} + 7 + e^{- (2.2951663820)}}{- \frac{2667 (2.2951663820)^{2}}{1000} - e^{- (2.2951663820)}} = 2.0373795956$ $LaTeX: x_{3} = (2.0373795956) - \frac{- \frac{889 (2.0373795956)^{3}}{1000} + 7 + e^{- (2.0373795956)}}{- \frac{2667 (2.0373795956)^{2}}{1000} - e^{- (2.0373795956)}} = 2.0027487856$ $LaTeX: x_{4} = (2.0027487856) - \frac{- \frac{889 (2.0027487856)^{3}}{1000} + 7 + e^{- (2.0027487856)}}{- \frac{2667 (2.0027487856)^{2}}{1000} - e^{- (2.0027487856)}} = 2.0021579069$ $LaTeX: x_{5} = (2.0021579069) - \frac{- \frac{889 (2.0021579069)^{3}}{1000} + 7 + e^{- (2.0021579069)}}{- \frac{2667 (2.0021579069)^{2}}{1000} - e^{- (2.0021579069)}} = 2.0021577369$