Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{193 x^{3}}{500} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{193 x_{n}^{3}}{500} + 6 + e^{- x_{n}}}{- \frac{579 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{193 (3.0000000000)^{3}}{500} + 6 + e^{- (3.0000000000)}}{- \frac{579 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 2.5824769065$ $LaTeX: x_{2} = (2.5824769065) - \frac{- \frac{193 (2.5824769065)^{3}}{500} + 6 + e^{- (2.5824769065)}}{- \frac{579 (2.5824769065)^{2}}{500} - e^{- (2.5824769065)}} = 2.5090654408$ $LaTeX: x_{3} = (2.5090654408) - \frac{- \frac{193 (2.5090654408)^{3}}{500} + 6 + e^{- (2.5090654408)}}{- \frac{579 (2.5090654408)^{2}}{500} - e^{- (2.5090654408)}} = 2.5069281198$ $LaTeX: x_{4} = (2.5069281198) - \frac{- \frac{193 (2.5069281198)^{3}}{500} + 6 + e^{- (2.5069281198)}}{- \frac{579 (2.5069281198)^{2}}{500} - e^{- (2.5069281198)}} = 2.5069263421$ $LaTeX: x_{5} = (2.5069263421) - \frac{- \frac{193 (2.5069263421)^{3}}{500} + 6 + e^{- (2.5069263421)}}{- \frac{579 (2.5069263421)^{2}}{500} - e^{- (2.5069263421)}} = 2.5069263421$