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函数y=log1/2(x²-2x)的递增区间是 递减区间是
人气:330 ℃ 时间:2020-05-19 20:14:30
解答
设g(x)=x^2-2x
=(x-1)^2-1
对称轴x=1的左侧(x1),g(x)为单调递增函数.
y=log1/2 (x^2-2x)
=log1/2 g(x)
∵00
x(x-2)>0
x2
x2在对称轴x=1的右侧,g(x)我单调递增,y=log1/2 g(x)为单调递减,即:y=log1/2 (x^2-2x)的单调递减区间是:(2,+∞)
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