Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{461 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{461 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1383 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{461 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1383 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4080708192$ $LaTeX: x_{2} = (2.4080708192) - \frac{- \frac{461 (2.4080708192)^{3}}{1000} + 5 + e^{- (2.4080708192)}}{- \frac{1383 (2.4080708192)^{2}}{1000} - e^{- (2.4080708192)}} = 2.2419267567$ $LaTeX: x_{3} = (2.2419267567) - \frac{- \frac{461 (2.2419267567)^{3}}{1000} + 5 + e^{- (2.2419267567)}}{- \frac{1383 (2.2419267567)^{2}}{1000} - e^{- (2.2419267567)}} = 2.2293865742$ $LaTeX: x_{4} = (2.2293865742) - \frac{- \frac{461 (2.2293865742)^{3}}{1000} + 5 + e^{- (2.2293865742)}}{- \frac{1383 (2.2293865742)^{2}}{1000} - e^{- (2.2293865742)}} = 2.2293180647$ $LaTeX: x_{5} = (2.2293180647) - \frac{- \frac{461 (2.2293180647)^{3}}{1000} + 5 + e^{- (2.2293180647)}}{- \frac{1383 (2.2293180647)^{2}}{1000} - e^{- (2.2293180647)}} = 2.2293180626$