Given the function f (x) = x 2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant, and the minimum value of a is obtained

Given the function f (x) = x 2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant, and the minimum value of a is obtained

Let the minimum value of F (x) on [- 2, 2] be g (a), then the minimum value of a satisfying g (a) ≥ A is obtained. The formula is f (x) = (x + A2) 2 + 3 − A24 (| x | ≤ 2) (1) when − 2 ≤− A2 ≤ 2, i.e. - 4 ≤ a ≤ 4, G (a) = 3 − A24, from the solution of 3-a24 ≥ a, ■ - 4 ≤ a ≤ 2; (2) when − A2 ≥ 2
For any differentiable function f (x) on R, if (x-1) f ′ (x) ≥ 0, then ()
A. f（0）+f（2）＜2f（1）B. f（0）+f（2）≤2f（1）C. f（0）+f（2）≥2f（1）D. f（0）+f（2）＞2f（1）
According to the meaning of the problem, when x ≥ 1, f '(x) ≥ 0, the function f (x) is an increasing function on (1, + ∞); when x < 1, f' (x) ≤ 0, f (x) is a decreasing function on (- ∞, 1), so when x = 1, the minimum value of F (x) is also the minimum value, that is, f (0) ≥ f (1), f (2) ≥ f (1), f (0) + F (2) ≥ 2F (1)
Given the function f (x) = x 2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant, and the minimum value of a is obtained
Let the minimum value of F (x) on [-2, 2] be g (a), and then satisfy the minimum value of G (a) ≥ a. The formula is f (x) = (x + A2) 2 + 2 + 3 − A24 (|x