Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{457 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{457 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 9}{- \frac{1371 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{457 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{1371 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.6531289368$ $LaTeX: x_{2} = (2.6531289368) - \frac{- \frac{457 (2.6531289368)^{3}}{1000} + \cos{\left((2.6531289368) \right)} + 9}{- \frac{1371 (2.6531289368)^{2}}{1000} - \sin{\left((2.6531289368) \right)}} = 2.6118424272$ $LaTeX: x_{3} = (2.6118424272) - \frac{- \frac{457 (2.6118424272)^{3}}{1000} + \cos{\left((2.6118424272) \right)} + 9}{- \frac{1371 (2.6118424272)^{2}}{1000} - \sin{\left((2.6118424272) \right)}} = 2.6112925002$ $LaTeX: x_{4} = (2.6112925002) - \frac{- \frac{457 (2.6112925002)^{3}}{1000} + \cos{\left((2.6112925002) \right)} + 9}{- \frac{1371 (2.6112925002)^{2}}{1000} - \sin{\left((2.6112925002) \right)}} = 2.6112924036$ $LaTeX: x_{5} = (2.6112924036) - \frac{- \frac{457 (2.6112924036)^{3}}{1000} + \cos{\left((2.6112924036) \right)} + 9}{- \frac{1371 (2.6112924036)^{2}}{1000} - \sin{\left((2.6112924036) \right)}} = 2.6112924036$