Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{11 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=9$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{11 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 9}{- \frac{33 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 9$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (9.0000000000) - \frac{- \frac{11 (9.0000000000)^{3}}{1000} + \cos{\left((9.0000000000) \right)} + 9}{- \frac{33 (9.0000000000)^{2}}{1000} - \sin{\left((9.0000000000) \right)}} = 9.0226473435$ $LaTeX: x_{2} = (9.0226473435) - \frac{- \frac{11 (9.0226473435)^{3}}{1000} + \cos{\left((9.0226473435) \right)} + 9}{- \frac{33 (9.0226473435)^{2}}{1000} - \sin{\left((9.0226473435) \right)}} = 9.0226739817$ $LaTeX: x_{3} = (9.0226739817) - \frac{- \frac{11 (9.0226739817)^{3}}{1000} + \cos{\left((9.0226739817) \right)} + 9}{- \frac{33 (9.0226739817)^{2}}{1000} - \sin{\left((9.0226739817) \right)}} = 9.0226739818$ $LaTeX: x_{4} = (9.0226739818) - \frac{- \frac{11 (9.0226739818)^{3}}{1000} + \cos{\left((9.0226739818) \right)} + 9}{- \frac{33 (9.0226739818)^{2}}{1000} - \sin{\left((9.0226739818) \right)}} = 9.0226739818$ $LaTeX: x_{5} = (9.0226739818) - \frac{- \frac{11 (9.0226739818)^{3}}{1000} + \cos{\left((9.0226739818) \right)} + 9}{- \frac{33 (9.0226739818)^{2}}{1000} - \sin{\left((9.0226739818) \right)}} = 9.0226739818$