Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{163 x^{3}}{200} - 3$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{163 x_{n}^{3}}{200} + 3 + e^{- x_{n}}}{- \frac{489 x_{n}^{2}}{200} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{163 (1.0000000000)^{3}}{200} + 3 + e^{- (1.0000000000)}}{- \frac{489 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.9075680258$ $LaTeX: x_{2} = (1.9075680258) - \frac{- \frac{163 (1.9075680258)^{3}}{200} + 3 + e^{- (1.9075680258)}}{- \frac{489 (1.9075680258)^{2}}{200} - e^{- (1.9075680258)}} = 1.6302199391$ $LaTeX: x_{3} = (1.6302199391) - \frac{- \frac{163 (1.6302199391)^{3}}{200} + 3 + e^{- (1.6302199391)}}{- \frac{489 (1.6302199391)^{2}}{200} - e^{- (1.6302199391)}} = 1.5801582019$ $LaTeX: x_{4} = (1.5801582019) - \frac{- \frac{163 (1.5801582019)^{3}}{200} + 3 + e^{- (1.5801582019)}}{- \frac{489 (1.5801582019)^{2}}{200} - e^{- (1.5801582019)}} = 1.5786310766$ $LaTeX: x_{5} = (1.5786310766) - \frac{- \frac{163 (1.5786310766)^{3}}{200} + 3 + e^{- (1.5786310766)}}{- \frac{489 (1.5786310766)^{2}}{200} - e^{- (1.5786310766)}} = 1.5786296849$