The max and min of a function on a give set can occur at three kinds of points:
1) At a point in the interior of the set where the derivative does not exist
2) At a point in the interior of the set where the derivative is 0
3) On the boundary of the set- which in the case of an interval consists of the two endpoints.
In this problem the derivative always exist so there are no points of type (1).
The derivative of (3x^3-2x^2)-4x is NOT " 6x+4x-3", it is 9x2- 4x- 4. The points at which that is 0 are indeed irrational but you can still calculate them exactly, using the quadratic formula. Since the problem asks only for the value of x at which the function has a minimum value, that should be enough. Even with approximate values for the function value at those points, you might be able to show that one of them gives the minimum value, or that the minimum is at an endpoint.