Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{116 x^{3}}{125} - 2$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{116 x_{n}^{3}}{125} + 2 + e^{- x_{n}}}{- \frac{348 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{116 (1.0000000000)^{3}}{125} + 2 + e^{- (1.0000000000)}}{- \frac{348 (1.0000000000)^{2}}{125} - e^{- (1.0000000000)}} = 1.4568320166$ $LaTeX: x_{2} = (1.4568320166) - \frac{- \frac{116 (1.4568320166)^{3}}{125} + 2 + e^{- (1.4568320166)}}{- \frac{348 (1.4568320166)^{2}}{125} - e^{- (1.4568320166)}} = 1.3532226115$ $LaTeX: x_{3} = (1.3532226115) - \frac{- \frac{116 (1.3532226115)^{3}}{125} + 2 + e^{- (1.3532226115)}}{- \frac{348 (1.3532226115)^{2}}{125} - e^{- (1.3532226115)}} = 1.3455287979$ $LaTeX: x_{4} = (1.3455287979) - \frac{- \frac{116 (1.3455287979)^{3}}{125} + 2 + e^{- (1.3455287979)}}{- \frac{348 (1.3455287979)^{2}}{125} - e^{- (1.3455287979)}} = 1.3454882526$ $LaTeX: x_{5} = (1.3454882526) - \frac{- \frac{116 (1.3454882526)^{3}}{125} + 2 + e^{- (1.3454882526)}}{- \frac{348 (1.3454882526)^{2}}{125} - e^{- (1.3454882526)}} = 1.3454882514$