Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{17 x^{3}}{125} - 1$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{125} + 1 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{17 (3.0000000000)^{3}}{125} + 1 + e^{- (3.0000000000)}}{- \frac{51 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.2954425163$ $LaTeX: x_{2} = (2.2954425163) - \frac{- \frac{17 (2.2954425163)^{3}}{125} + 1 + e^{- (2.2954425163)}}{- \frac{51 (2.2954425163)^{2}}{125} - e^{- (2.2954425163)}} = 2.0536384117$ $LaTeX: x_{3} = (2.0536384117) - \frac{- \frac{17 (2.0536384117)^{3}}{125} + 1 + e^{- (2.0536384117)}}{- \frac{51 (2.0536384117)^{2}}{125} - e^{- (2.0536384117)}} = 2.0267915922$ $LaTeX: x_{4} = (2.0267915922) - \frac{- \frac{17 (2.0267915922)^{3}}{125} + 1 + e^{- (2.0267915922)}}{- \frac{51 (2.0267915922)^{2}}{125} - e^{- (2.0267915922)}} = 2.0264847872$ $LaTeX: x_{5} = (2.0264847872) - \frac{- \frac{17 (2.0264847872)^{3}}{125} + 1 + e^{- (2.0264847872)}}{- \frac{51 (2.0264847872)^{2}}{125} - e^{- (2.0264847872)}} = 2.0264847475$