题目
已知函数f(x)=ln
−f′(1)x
ex |
2 |
提问时间:2020-08-02
答案
(I)∵f(x)=lnex2-f′(1)x,∴f′(x)=2ex×e2-f′(1),令x=1,可得f′(1)=1-f′(1),解得f′(1)=12;(II)由(I)知:f′(x)=1x-12=2−x2x,∵x>0,∴当0<x<2时,f′(x)>0,当x>2时,f′(...
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