Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{451 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{451 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1353 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{451 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1353 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4170821090$ $LaTeX: x_{2} = (2.4170821090) - \frac{- \frac{451 (2.4170821090)^{3}}{1000} + 5 + e^{- (2.4170821090)}}{- \frac{1353 (2.4170821090)^{2}}{1000} - e^{- (2.4170821090)}} = 2.2570181989$ $LaTeX: x_{3} = (2.2570181989) - \frac{- \frac{451 (2.2570181989)^{3}}{1000} + 5 + e^{- (2.2570181989)}}{- \frac{1353 (2.2570181989)^{2}}{1000} - e^{- (2.2570181989)}} = 2.2454802230$ $LaTeX: x_{4} = (2.2454802230) - \frac{- \frac{451 (2.2454802230)^{3}}{1000} + 5 + e^{- (2.2454802230)}}{- \frac{1353 (2.2454802230)^{2}}{1000} - e^{- (2.2454802230)}} = 2.2454226528$ $LaTeX: x_{5} = (2.2454226528) - \frac{- \frac{451 (2.2454226528)^{3}}{1000} + 5 + e^{- (2.2454226528)}}{- \frac{1353 (2.2454226528)^{2}}{1000} - e^{- (2.2454226528)}} = 2.2454226513$