Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{617 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{617 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 9}{- \frac{1851 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{617 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 9}{- \frac{1851 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.4851826956$ $LaTeX: x_{2} = (2.4851826956) - \frac{- \frac{617 (2.4851826956)^{3}}{1000} + \cos{\left((2.4851826956) \right)} + 9}{- \frac{1851 (2.4851826956)^{2}}{1000} - \sin{\left((2.4851826956) \right)}} = 2.3803514230$ $LaTeX: x_{3} = (2.3803514230) - \frac{- \frac{617 (2.3803514230)^{3}}{1000} + \cos{\left((2.3803514230) \right)} + 9}{- \frac{1851 (2.3803514230)^{2}}{1000} - \sin{\left((2.3803514230) \right)}} = 2.3762709571$ $LaTeX: x_{4} = (2.3762709571) - \frac{- \frac{617 (2.3762709571)^{3}}{1000} + \cos{\left((2.3762709571) \right)} + 9}{- \frac{1851 (2.3762709571)^{2}}{1000} - \sin{\left((2.3762709571) \right)}} = 2.3762649184$ $LaTeX: x_{5} = (2.3762649184) - \frac{- \frac{617 (2.3762649184)^{3}}{1000} + \cos{\left((2.3762649184) \right)} + 9}{- \frac{1851 (2.3762649184)^{2}}{1000} - \sin{\left((2.3762649184) \right)}} = 2.3762649184$