+128407 x^3 ≥ 4x
x^3 - 4x ≥ 0 (1)
x ( x^2 - 4) ≥ 0
x ( x + 2) ( x - 2) ≥ 0
Setting each factor = 0 and solving for x produces three test points
x = 0 x = -2 x = 2
We have the intervals
(-inf, 2] [ -2, 0 ] [ 0, 2] [ 2, inf )
Picking a point in each interval and testing this in (1)
x = -3 (-3)^3 - 4(-3) = -27 + 12 = -15 and this is not ≥ 0....so the first interval doesn't work
x = -1 (-1)^3 - 4(-1) = -1 + 4 = 3 and this is ≥ 0....so the second interval works
x = 1 (1)^3 - 4(1) = -3 and this is not ≥ 0....so this interval does not work
x = 3 (3)^3 - 4(3) = 15 and this is ≥ 0 ...so this interval works
The solution is x = [-2, 0 ] U [ 2, inf )
