Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{33 x^{3}}{100} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{33 x_{n}^{3}}{100} + 1 + e^{- x_{n}}}{- \frac{99 x_{n}^{2}}{100} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{33 (1.0000000000)^{3}}{100} + 1 + e^{- (1.0000000000)}}{- \frac{99 (1.0000000000)^{2}}{100} - e^{- (1.0000000000)}} = 1.7643384307$ $LaTeX: x_{2} = (1.7643384307) - \frac{- \frac{33 (1.7643384307)^{3}}{100} + 1 + e^{- (1.7643384307)}}{- \frac{99 (1.7643384307)^{2}}{100} - e^{- (1.7643384307)}} = 1.5672554039$ $LaTeX: x_{3} = (1.5672554039) - \frac{- \frac{33 (1.5672554039)^{3}}{100} + 1 + e^{- (1.5672554039)}}{- \frac{99 (1.5672554039)^{2}}{100} - e^{- (1.5672554039)}} = 1.5438637809$ $LaTeX: x_{4} = (1.5438637809) - \frac{- \frac{33 (1.5438637809)^{3}}{100} + 1 + e^{- (1.5438637809)}}{- \frac{99 (1.5438637809)^{2}}{100} - e^{- (1.5438637809)}} = 1.5435578504$ $LaTeX: x_{5} = (1.5435578504) - \frac{- \frac{33 (1.5435578504)^{3}}{100} + 1 + e^{- (1.5435578504)}}{- \frac{99 (1.5435578504)^{2}}{100} - e^{- (1.5435578504)}} = 1.5435577987$