8楼: Originally posted by peterflyer at 2015-12-02 10:47:18
令F(X)=10.396x-lnx-3.662
F'(x)=10.396 - 1/x
F"(x)=1/x^2  >0
令F'(x)=0解得唯一的驻点为x*=1/10.396=0.096191
故F(X)=10.396x-lnx-3.662在驻点处取得极小值，为：
F(x*)=-0.32058<0
而F(0)=∞& ...