Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{102 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{102 x_{n}^{3}}{125} + 7 + e^{- x_{n}}}{- \frac{306 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{102 (3.0000000000)^{3}}{125} + 7 + e^{- (3.0000000000)}}{- \frac{306 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.3215126618$ $LaTeX: x_{2} = (2.3215126618) - \frac{- \frac{102 (2.3215126618)^{3}}{125} + 7 + e^{- (2.3215126618)}}{- \frac{306 (2.3215126618)^{2}}{125} - e^{- (2.3215126618)}} = 2.0874258422$ $LaTeX: x_{3} = (2.0874258422) - \frac{- \frac{102 (2.0874258422)^{3}}{125} + 7 + e^{- (2.0874258422)}}{- \frac{306 (2.0874258422)^{2}}{125} - e^{- (2.0874258422)}} = 2.0598064879$ $LaTeX: x_{4} = (2.0598064879) - \frac{- \frac{102 (2.0598064879)^{3}}{125} + 7 + e^{- (2.0598064879)}}{- \frac{306 (2.0598064879)^{2}}{125} - e^{- (2.0598064879)}} = 2.0594419081$ $LaTeX: x_{5} = (2.0594419081) - \frac{- \frac{102 (2.0594419081)^{3}}{125} + 7 + e^{- (2.0594419081)}}{- \frac{306 (2.0594419081)^{2}}{125} - e^{- (2.0594419081)}} = 2.0594418452$