Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{107 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{107 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 4}{- \frac{321 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{107 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 4}{- \frac{321 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.5323641871$ $LaTeX: x_{2} = (2.5323641871) - \frac{- \frac{107 (2.5323641871)^{3}}{500} + \cos{\left((2.5323641871) \right)} + 4}{- \frac{321 (2.5323641871)^{2}}{500} - \sin{\left((2.5323641871) \right)}} = 2.4693719101$ $LaTeX: x_{3} = (2.4693719101) - \frac{- \frac{107 (2.4693719101)^{3}}{500} + \cos{\left((2.4693719101) \right)} + 4}{- \frac{321 (2.4693719101)^{2}}{500} - \sin{\left((2.4693719101) \right)}} = 2.4683151761$ $LaTeX: x_{4} = (2.4683151761) - \frac{- \frac{107 (2.4683151761)^{3}}{500} + \cos{\left((2.4683151761) \right)} + 4}{- \frac{321 (2.4683151761)^{2}}{500} - \sin{\left((2.4683151761) \right)}} = 2.4683148820$ $LaTeX: x_{5} = (2.4683148820) - \frac{- \frac{107 (2.4683148820)^{3}}{500} + \cos{\left((2.4683148820) \right)} + 4}{- \frac{321 (2.4683148820)^{2}}{500} - \sin{\left((2.4683148820) \right)}} = 2.4683148820$