Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{17 x^{3}}{20} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{20} + 5 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{20} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{17 (1.0000000000)^{3}}{20} + 5 + e^{- (1.0000000000)}}{- \frac{51 (1.0000000000)^{2}}{20} - e^{- (1.0000000000)}} = 2.5483434228$ $LaTeX: x_{2} = (2.5483434228) - \frac{- \frac{17 (2.5483434228)^{3}}{20} + 5 + e^{- (2.5483434228)}}{- \frac{51 (2.5483434228)^{2}}{20} - e^{- (2.5483434228)}} = 2.0081053840$ $LaTeX: x_{3} = (2.0081053840) - \frac{- \frac{17 (2.0081053840)^{3}}{20} + 5 + e^{- (2.0081053840)}}{- \frac{51 (2.0081053840)^{2}}{20} - e^{- (2.0081053840)}} = 1.8402304347$ $LaTeX: x_{4} = (1.8402304347) - \frac{- \frac{17 (1.8402304347)^{3}}{20} + 5 + e^{- (1.8402304347)}}{- \frac{51 (1.8402304347)^{2}}{20} - e^{- (1.8402304347)}} = 1.8245056552$ $LaTeX: x_{5} = (1.8245056552) - \frac{- \frac{17 (1.8245056552)^{3}}{20} + 5 + e^{- (1.8245056552)}}{- \frac{51 (1.8245056552)^{2}}{20} - e^{- (1.8245056552)}} = 1.8243741733$