2·e的x-e的-x=0,怎么求x?在线等。。。。。
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第1个回答 2016-08-26
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求函数y=(2e^x)+(e^-x)的极值. 请尽量说得详细些答:因为函数处处可导,求一介导数,令f'(x)=(2e^x)-(e^-x)=0,得驻点x=-(ln2/2),当x属于(负无穷,-(ln2/2))时,一介导数恒小于0,那么原函数单调递减,而当x属于(-(ln2/2),正无穷)时,一介导数大于0,那么原函数单调递增,所以当x=-(ln2/2)时,有极小值,Ymin=2√2 ...