Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{19 x^{3}}{125} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{19 x_{n}^{3}}{125} + 5 + e^{- x_{n}}}{- \frac{57 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{19 (3.0000000000)^{3}}{125} + 5 + e^{- (3.0000000000)}}{- \frac{57 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 3.2276927182$ $LaTeX: x_{2} = (3.2276927182) - \frac{- \frac{19 (3.2276927182)^{3}}{125} + 5 + e^{- (3.2276927182)}}{- \frac{57 (3.2276927182)^{2}}{125} - e^{- (3.2276927182)}} = 3.2127626465$ $LaTeX: x_{3} = (3.2127626465) - \frac{- \frac{19 (3.2127626465)^{3}}{125} + 5 + e^{- (3.2127626465)}}{- \frac{57 (3.2127626465)^{2}}{125} - e^{- (3.2127626465)}} = 3.2126945754$ $LaTeX: x_{4} = (3.2126945754) - \frac{- \frac{19 (3.2126945754)^{3}}{125} + 5 + e^{- (3.2126945754)}}{- \frac{57 (3.2126945754)^{2}}{125} - e^{- (3.2126945754)}} = 3.2126945740$ $LaTeX: x_{5} = (3.2126945740) - \frac{- \frac{19 (3.2126945740)^{3}}{125} + 5 + e^{- (3.2126945740)}}{- \frac{57 (3.2126945740)^{2}}{125} - e^{- (3.2126945740)}} = 3.2126945740$