Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{89 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{89 x_{n}^{3}}{1000} + 3 + e^{- x_{n}}}{- \frac{267 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{89 (3.0000000000)^{3}}{1000} + 3 + e^{- (3.0000000000)}}{- \frac{267 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.2636947482$ $LaTeX: x_{2} = (3.2636947482) - \frac{- \frac{89 (3.2636947482)^{3}}{1000} + 3 + e^{- (3.2636947482)}}{- \frac{267 (3.2636947482)^{2}}{1000} - e^{- (3.2636947482)}} = 3.2443553437$ $LaTeX: x_{3} = (3.2443553437) - \frac{- \frac{89 (3.2443553437)^{3}}{1000} + 3 + e^{- (3.2443553437)}}{- \frac{267 (3.2443553437)^{2}}{1000} - e^{- (3.2443553437)}} = 3.2442437149$ $LaTeX: x_{4} = (3.2442437149) - \frac{- \frac{89 (3.2442437149)^{3}}{1000} + 3 + e^{- (3.2442437149)}}{- \frac{267 (3.2442437149)^{2}}{1000} - e^{- (3.2442437149)}} = 3.2442437112$ $LaTeX: x_{5} = (3.2442437112) - \frac{- \frac{89 (3.2442437112)^{3}}{1000} + 3 + e^{- (3.2442437112)}}{- \frac{267 (3.2442437112)^{2}}{1000} - e^{- (3.2442437112)}} = 3.2442437112$