Show that the equation 3x3−2x2−4=03x^3 - 2x^2 - 4 = 03x3−2x2−4=0 has a solution between x=1x = 1x=1 and x=2x = 2x=2.
Show that the equation 3x3−2x2−4=03x^3 - 2x^2 - 4 = 03x3−2x2−4=0 can be rearranged to give: x=43x−2\displaystyle x = \sqrt{\frac{4}{3x - 2}}x=3x−24
Starting with x0=1.5x_0 = 1.5x0=1.5, use the iteration formula xn+1=43xn−2\displaystyle x_{n+1} = \sqrt{\frac{4}{3x_n - 2}}xn+1=3xn−24 twice to find an estimate for the solution to 3x3−2x2−4=03x^3 - 2x^2 - 4 = 03x3−2x2−4=0.