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求函数y=sinx/(2+sinx)的值域
人气:432 ℃ 时间:2020-01-31 15:34:21
解答
y=(2+sinx-2)/(2+sinx)
=1-2/(2+sinx)
-1<=sinx<=1
1<=sinx<=3
1/3<=1/(2+sinx)<=1
-2<=-2/(2+sinx)<=-2/3
-1<=1-2/(2+sinx)<=1/3
值域[-1,1/3]看清楚题目啊是(sinx)/(2+sinx)不信拉倒
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