A problem on the maximum value of closed interval of quadratic function I know that there are three possibilities, but when a ≤ 1 ≤ B, how can I judge whether to take a or B as the minimum? I see a solution (3) a

F (x) = - (x-1) ^ 2 + 1, the axis of symmetry is x = 1, the opening downward parabola
1) When a > = 1, the interval [a, b] is on the right side of the symmetry axis, and f (x) is a decreasing function on this interval, so f (a) max = 2-A, f (b) min = 2-b
The solution is a = 1 or 2, B = 1 or 2, because a

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