Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{107 x^{3}}{200} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{107 x_{n}^{3}}{200} + 4 + e^{- x_{n}}}{- \frac{321 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{107 (1.0000000000)^{3}}{200} + 4 + e^{- (1.0000000000)}}{- \frac{321 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 2.9427844202$ $LaTeX: x_{2} = (2.9427844202) - \frac{- \frac{107 (2.9427844202)^{3}}{200} + 4 + e^{- (2.9427844202)}}{- \frac{321 (2.9427844202)^{2}}{200} - e^{- (2.9427844202)}} = 2.2560388981$ $LaTeX: x_{3} = (2.2560388981) - \frac{- \frac{107 (2.2560388981)^{3}}{200} + 4 + e^{- (2.2560388981)}}{- \frac{321 (2.2560388981)^{2}}{200} - e^{- (2.2560388981)}} = 2.0096670502$ $LaTeX: x_{4} = (2.0096670502) - \frac{- \frac{107 (2.0096670502)^{3}}{200} + 4 + e^{- (2.0096670502)}}{- \frac{321 (2.0096670502)^{2}}{200} - e^{- (2.0096670502)}} = 1.9781794603$ $LaTeX: x_{5} = (1.9781794603) - \frac{- \frac{107 (1.9781794603)^{3}}{200} + 4 + e^{- (1.9781794603)}}{- \frac{321 (1.9781794603)^{2}}{200} - e^{- (1.9781794603)}} = 1.9776943150$