函数y＝sin²x+sinx-1的值域是什么求过程
人气：489 ℃ 时间：2020-02-15 06:58:27
解答
令sinx =t,则t∈[-1,1]
∴y=t²+t-1,开口向上,对称轴为t=-1/2
∴值域为[-5/4,1]
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