函数f(x)=2x2-lnx的单调递减区间是 _ .
函数f(x)=2x2-lnx的单调递减区间是 ___ .
数学人气:902 ℃时间:2020-03-27 09:05:51
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由f(x)=2x2-lnx,得:f′(x)=(2x2-lnx)′=4x-1x=(2x+1)(2x-1)x.因为函数f(x)=2x2-lnx的定义域为(0,+∞),由f′(x)<0,得:(2x+1)(2x-1)x<0,即(2x+1)(2x-1)<0,解得:0<x<12.所以函数f(x)...
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