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

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{4} + \cos{\left(x_{n} \right)} + 4}{- \frac{3 x_{n}^{2}}{4} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{(3.0000000000)^{3}}{4} + \cos{\left((3.0000000000) \right)} + 4}{- \frac{3 (3.0000000000)^{2}}{4} - \sin{\left((3.0000000000) \right)}} = 2.4572736373$ $LaTeX: x_{2} = (2.4572736373) - \frac{- \frac{(2.4572736373)^{3}}{4} + \cos{\left((2.4572736373) \right)} + 4}{- \frac{3 (2.4572736373)^{2}}{4} - \sin{\left((2.4572736373) \right)}} = 2.3634462970$ $LaTeX: x_{3} = (2.3634462970) - \frac{- \frac{(2.3634462970)^{3}}{4} + \cos{\left((2.3634462970) \right)} + 4}{- \frac{3 (2.3634462970)^{2}}{4} - \sin{\left((2.3634462970) \right)}} = 2.3608505311$ $LaTeX: x_{4} = (2.3608505311) - \frac{- \frac{(2.3608505311)^{3}}{4} + \cos{\left((2.3608505311) \right)} + 4}{- \frac{3 (2.3608505311)^{2}}{4} - \sin{\left((2.3608505311) \right)}} = 2.3608485774$ $LaTeX: x_{5} = (2.3608485774) - \frac{- \frac{(2.3608485774)^{3}}{4} + \cos{\left((2.3608485774) \right)} + 4}{- \frac{3 (2.3608485774)^{2}}{4} - \sin{\left((2.3608485774) \right)}} = 2.3608485774$