When the derivative is greater than or equal to 0 in a certain interval, the function increases, and when it is less than or equal to 0, the function decreases. When equal to 0, the function is a constant function in this interval. For your question, when a=-√6/2 and f ′ (x) = 3x? +√ 6x+ 1/2 are all greater than or equal to 0 in the real number field, so the function is incremental. Your math teacher is right.
When f ′ (x) = 0, x=-√6/6 is the only zero. At this time, x=-√6/6 is the equilibrium point of the function f, but it is neither the maximum point nor the minimum point. But f is still a increasing function in the real number field.