Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{47 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{47 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 7}{- \frac{141 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{47 (5.0000000000)^{3}}{500} + \cos{\left((5.0000000000) \right)} + 7}{- \frac{141 (5.0000000000)^{2}}{500} - \sin{\left((5.0000000000) \right)}} = 4.2667407177$ $LaTeX: x_{2} = (4.2667407177) - \frac{- \frac{47 (4.2667407177)^{3}}{500} + \cos{\left((4.2667407177) \right)} + 7}{- \frac{141 (4.2667407177)^{2}}{500} - \sin{\left((4.2667407177) \right)}} = 4.0936061529$ $LaTeX: x_{3} = (4.0936061529) - \frac{- \frac{47 (4.0936061529)^{3}}{500} + \cos{\left((4.0936061529) \right)} + 7}{- \frac{141 (4.0936061529)^{2}}{500} - \sin{\left((4.0936061529) \right)}} = 4.0863559926$ $LaTeX: x_{4} = (4.0863559926) - \frac{- \frac{47 (4.0863559926)^{3}}{500} + \cos{\left((4.0863559926) \right)} + 7}{- \frac{141 (4.0863559926)^{2}}{500} - \sin{\left((4.0863559926) \right)}} = 4.0863443606$ $LaTeX: x_{5} = (4.0863443606) - \frac{- \frac{47 (4.0863443606)^{3}}{500} + \cos{\left((4.0863443606) \right)} + 7}{- \frac{141 (4.0863443606)^{2}}{500} - \sin{\left((4.0863443606) \right)}} = 4.0863443605$