Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{301 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{301 x_{n}^{3}}{500} + 8 + e^{- x_{n}}}{- \frac{903 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{301 (3.0000000000)^{3}}{500} + 8 + e^{- (3.0000000000)}}{- \frac{903 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 2.4967909666$ $LaTeX: x_{2} = (2.4967909666) - \frac{- \frac{301 (2.4967909666)^{3}}{500} + 8 + e^{- (2.4967909666)}}{- \frac{903 (2.4967909666)^{2}}{500} - e^{- (2.4967909666)}} = 2.3832438207$ $LaTeX: x_{3} = (2.3832438207) - \frac{- \frac{301 (2.3832438207)^{3}}{500} + 8 + e^{- (2.3832438207)}}{- \frac{903 (2.3832438207)^{2}}{500} - e^{- (2.3832438207)}} = 2.3777651937$ $LaTeX: x_{4} = (2.3777651937) - \frac{- \frac{301 (2.3777651937)^{3}}{500} + 8 + e^{- (2.3777651937)}}{- \frac{903 (2.3777651937)^{2}}{500} - e^{- (2.3777651937)}} = 2.3777527994$ $LaTeX: x_{5} = (2.3777527994) - \frac{- \frac{301 (2.3777527994)^{3}}{500} + 8 + e^{- (2.3777527994)}}{- \frac{903 (2.3777527994)^{2}}{500} - e^{- (2.3777527994)}} = 2.3777527994$