Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{119 x^{3}}{250} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{119 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{357 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{119 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{357 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.6102859551$ $LaTeX: x_{2} = (2.6102859551) - \frac{- \frac{119 (2.6102859551)^{3}}{250} + 3 + e^{- (2.6102859551)}}{- \frac{357 (2.6102859551)^{2}}{250} - e^{- (2.6102859551)}} = 2.0602328276$ $LaTeX: x_{3} = (2.0602328276) - \frac{- \frac{119 (2.0602328276)^{3}}{250} + 3 + e^{- (2.0602328276)}}{- \frac{357 (2.0602328276)^{2}}{250} - e^{- (2.0602328276)}} = 1.8929765850$ $LaTeX: x_{4} = (1.8929765850) - \frac{- \frac{119 (1.8929765850)^{3}}{250} + 3 + e^{- (1.8929765850)}}{- \frac{357 (1.8929765850)^{2}}{250} - e^{- (1.8929765850)}} = 1.8781334503$ $LaTeX: x_{5} = (1.8781334503) - \frac{- \frac{119 (1.8781334503)^{3}}{250} + 3 + e^{- (1.8781334503)}}{- \frac{357 (1.8781334503)^{2}}{250} - e^{- (1.8781334503)}} = 1.8780222115$