f(x)=lnx-ax(a>0) (1)当a=2时,求f(x)的单调区间与极值
问题描述:
f(x)=lnx-ax(a>0) (1)当a=2时,求f(x)的单调区间与极值
答
f(x)=lnx-2x(x>0),f'(x)=1/x-2=(1-2x)/x>0,则0
f(x)=lnx-ax(a>0) (1)当a=2时,求f(x)的单调区间与极值
f(x)=lnx-2x(x>0),f'(x)=1/x-2=(1-2x)/x>0,则0