Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{52 x^{3}}{125} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{52 x_{n}^{3}}{125} + 9 + e^{- x_{n}}}{- \frac{156 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{52 (3.0000000000)^{3}}{125} + 9 + e^{- (3.0000000000)}}{- \frac{156 (3.0000000000)^{2}}{125} - e^{- (3.0000000000)}} = 2.8065720512$ $LaTeX: x_{2} = (2.8065720512) - \frac{- \frac{52 (2.8065720512)^{3}}{125} + 9 + e^{- (2.8065720512)}}{- \frac{156 (2.8065720512)^{2}}{125} - e^{- (2.8065720512)}} = 2.7928142638$ $LaTeX: x_{3} = (2.7928142638) - \frac{- \frac{52 (2.7928142638)^{3}}{125} + 9 + e^{- (2.7928142638)}}{- \frac{156 (2.7928142638)^{2}}{125} - e^{- (2.7928142638)}} = 2.7927472800$ $LaTeX: x_{4} = (2.7927472800) - \frac{- \frac{52 (2.7927472800)^{3}}{125} + 9 + e^{- (2.7927472800)}}{- \frac{156 (2.7927472800)^{2}}{125} - e^{- (2.7927472800)}} = 2.7927472784$ $LaTeX: x_{5} = (2.7927472784) - \frac{- \frac{52 (2.7927472784)^{3}}{125} + 9 + e^{- (2.7927472784)}}{- \frac{156 (2.7927472784)^{2}}{125} - e^{- (2.7927472784)}} = 2.7927472784$