Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{116 x^{3}}{125} - 6$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{116 x_{n}^{3}}{125} + 6 + e^{- x_{n}}}{- \frac{348 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{116 (1.0000000000)^{3}}{125} + 6 + e^{- (1.0000000000)}}{- \frac{348 (1.0000000000)^{2}}{125} - e^{- (1.0000000000)}} = 2.7259160900$ $LaTeX: x_{2} = (2.7259160900) - \frac{- \frac{116 (2.7259160900)^{3}}{125} + 6 + e^{- (2.7259160900)}}{- \frac{348 (2.7259160900)^{2}}{125} - e^{- (2.7259160900)}} = 2.1124244927$ $LaTeX: x_{3} = (2.1124244927) - \frac{- \frac{116 (2.1124244927)^{3}}{125} + 6 + e^{- (2.1124244927)}}{- \frac{348 (2.1124244927)^{2}}{125} - e^{- (2.1124244927)}} = 1.9030263939$ $LaTeX: x_{4} = (1.9030263939) - \frac{- \frac{116 (1.9030263939)^{3}}{125} + 6 + e^{- (1.9030263939)}}{- \frac{348 (1.9030263939)^{2}}{125} - e^{- (1.9030263939)}} = 1.8789339106$ $LaTeX: x_{5} = (1.8789339106) - \frac{- \frac{116 (1.8789339106)^{3}}{125} + 6 + e^{- (1.8789339106)}}{- \frac{348 (1.8789339106)^{2}}{125} - e^{- (1.8789339106)}} = 1.8786314848$