二次函数y=x²-4x-1在-1≤x≤1的范围内的最小值
人气：182 ℃ 时间：2019-08-18 14:16:20
解答
y=x²-4x-1
y=x²-4x+4-5
y=(x-2)²-5
-1≤x≤1
x=1时y最小
最小值是：y=-4
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