题目内容

函数 y=|x+1|+2 的最小值点是( )。

A. 0
B. 1
C. - 1
D. 2

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函数f(x)=ex-x-1的驻点为()。

A. x=0
B. x=2
C. x=0,y=0
D. x=1,e–2

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