If the quadratic function f (x) = - 4x ^ + 4ax-4a-a ^ has a minimum value of - 5 in [0,1], find the value of real number a, and 0 and 1 are also in the range, the brackets will not be opened Why is a = 1 substituted? I don't know whether the axis of symmetry is in the middle or on both sides of 0,1. It is also possible that (0,1 〉 is within the play range where x is greater than or equal to 0 and less than or equal to 1

If the quadratic function f (x) = - 4x ^ + 4ax-4a-a ^ has a minimum value of - 5 in [0,1], find the value of real number a, and 0 and 1 are also in the range, the brackets will not be opened Why is a = 1 substituted? I don't know whether the axis of symmetry is in the middle or on both sides of 0,1. It is also possible that (0,1 〉 is within the play range where x is greater than or equal to 0 and less than or equal to 1

It doesn't seem difficult
The original formula can be reduced to f (x) = - 4 (x-a / 2) ^ - 4A
(and then it's drawing, but I can't )
Because in [0,1], f (x) min = 5
From the figure: so there are:
X = 1, f (x) = - 5, that is, f (x) = - 4 * (1-A / 2) ^ - 4 * a = - 5
The solution is: A1 = 1, A2 = - 1
After examination, it is consistent with the meaning of the question
A: the value of a is ± 1