Given that f (x) = x ^ 3-2x ^ 2 + X-2, if the extremum of function g (x) = f (x) + (1 / 3) MX exists, then the value range of real number m is?

Given that f (x) = x ^ 3-2x ^ 2 + X-2, if the extremum of function g (x) = f (x) + (1 / 3) MX exists, then the value range of real number m is?

∵f（x）=x^3-2x^2+x-2
The derivative of F (x) = 3x ^ 2-4x + 1
∵g(x)=f(x)+1/3mx
The derivative of G (x) = 3x ^ 2-4x + 1 + 1 / 3M
If we want g (x) to have an extreme value, then the derivative of G (x) must be equal to 0 and Δ ≥ 0,
The solution m is less than or equal to 1

When x ∈ (0, 2], the function f (x) = AX2 + 4 (a + 1) x-3 gets the maximum value at x = 2, then the value range of a is ()
A. - 12 ≤ a < 0b. A ≥ - 12C. - 12 ≤ a < 0 or & nbsp; a > 0d. A ∈ R

When a = 0, f (x) = 4x-3, x = 2, the maximum value is obtained, which is in line with the meaning of the problem; when a ≠ 0, the axis of symmetry is x = - 2 + 2AA, (1) when a > 0, to make x = 2 get the maximum value, then - 2 + 2AA ≤ 1, the solution is a > 0. (2) when a < 0, to make x = 2 get the maximum value, then - 2 + 2AA ≥ 2, a ≥ - 12, a ≤ 0