题目内容
函数
【答案】分析:用导数判断函数的单调性,由单调性可求最大值.
解答:解:y′=1+2cosx,当x∈[-
,
]时,y′>0,
所以y=x+2sinx在[-
,
]上单调递增,
所以当x=
时,y=x+2sinx取得最大值为:
+2sin
=
+2.
故答案为:
+2.
点评:本题考查函数的单调性,对于由不同类型的函数构成的函数最值问题,常用函数的性质解决.
解答:解:y′=1+2cosx,当x∈[-


所以y=x+2sinx在[-
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所以当x=
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
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故答案为:

点评:本题考查函数的单调性,对于由不同类型的函数构成的函数最值问题,常用函数的性质解决.

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