The function f (x) = 4x ^ 2-4ax + A ^ 2-2a + 2 has the minimum value of 3 in the interval [0,2], and the value of a is obtained

The function f (x) = 4x ^ 2-4ax + A ^ 2-2a + 2 has the minimum value of 3 in the interval [0,2], and the value of a is obtained

First of all, the formula can be written as y = 4 (x-a / 2) ^ 2-2a + 2
When x = A / 2, the minimum value of F is - 2A + 2
If a / 2 is in [0,2], then - 2A + 2 = 3 = > A = - 1 / 2 is inconsistent with the hypothesis and the possibility is removed
If a / 2 is in (- ∞, 0), then f (0) = a ^ 2-2a + 2 = 3 is the minimum = > A = 1 + √ 2 (rounding off) a = 1 - √ 2
If a / 2 is within (2, + ∞), then f (2) = a ^ 2-10a + 18 = 3 is the minimum = > A = 5 - √ 3 (rounding off) a = 5 + √ 3
So, the value of a can be 1 - √ 2 and 5 + √ 3