Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{489 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{489 x_{n}^{3}}{500} + 5 + e^{- x_{n}}}{- \frac{1467 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{489 (1.0000000000)^{3}}{500} + 5 + e^{- (1.0000000000)}}{- \frac{1467 (1.0000000000)^{2}}{500} - e^{- (1.0000000000)}} = 2.3295093050$ $LaTeX: x_{2} = (2.3295093050) - \frac{- \frac{489 (2.3295093050)^{3}}{500} + 5 + e^{- (2.3295093050)}}{- \frac{1467 (2.3295093050)^{2}}{500} - e^{- (2.3295093050)}} = 1.8759303591$ $LaTeX: x_{3} = (1.8759303591) - \frac{- \frac{489 (1.8759303591)^{3}}{500} + 5 + e^{- (1.8759303591)}}{- \frac{1467 (1.8759303591)^{2}}{500} - e^{- (1.8759303591)}} = 1.7515621994$ $LaTeX: x_{4} = (1.7515621994) - \frac{- \frac{489 (1.7515621994)^{3}}{500} + 5 + e^{- (1.7515621994)}}{- \frac{1467 (1.7515621994)^{2}}{500} - e^{- (1.7515621994)}} = 1.7426231043$ $LaTeX: x_{5} = (1.7426231043) - \frac{- \frac{489 (1.7426231043)^{3}}{500} + 5 + e^{- (1.7426231043)}}{- \frac{1467 (1.7426231043)^{2}}{500} - e^{- (1.7426231043)}} = 1.7425787447$