Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{277 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{277 x_{n}^{3}}{1000} + 9 + e^{- x_{n}}}{- \frac{831 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{277 (3.0000000000)^{3}}{1000} + 9 + e^{- (3.0000000000)}}{- \frac{831 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.2086374676$ $LaTeX: x_{2} = (3.2086374676) - \frac{- \frac{277 (3.2086374676)^{3}}{1000} + 9 + e^{- (3.2086374676)}}{- \frac{831 (3.2086374676)^{2}}{1000} - e^{- (3.2086374676)}} = 3.1958379271$ $LaTeX: x_{3} = (3.1958379271) - \frac{- \frac{277 (3.1958379271)^{3}}{1000} + 9 + e^{- (3.1958379271)}}{- \frac{831 (3.1958379271)^{2}}{1000} - e^{- (3.1958379271)}} = 3.1957871638$ $LaTeX: x_{4} = (3.1957871638) - \frac{- \frac{277 (3.1957871638)^{3}}{1000} + 9 + e^{- (3.1957871638)}}{- \frac{831 (3.1957871638)^{2}}{1000} - e^{- (3.1957871638)}} = 3.1957871630$ $LaTeX: x_{5} = (3.1957871630) - \frac{- \frac{277 (3.1957871630)^{3}}{1000} + 9 + e^{- (3.1957871630)}}{- \frac{831 (3.1957871630)^{2}}{1000} - e^{- (3.1957871630)}} = 3.1957871630$