Question

高三数学导函数

高三数学导函数这道题第一小题怎么做呢

Answer 1

（1）f(x)=-e^(x-1)+xlnx+1
f'(x)=-e^(x-1)+lnx+1
f''(x)=-e^(x-1)+1/x
令f''(x)=0,有xe^(x-1)=1,得f''(1)=0
故当0<x<1，f''(x)>0; x>1, f''(x)<0
即f'(x)<=f'(1)=0，故f(x)在（0,+∞）单调递减
（2）f(x)=0 => a=-(xlnx+1)/e^x
令g(x)=-(xlnx+1)/e^x
g'(x)=(x-1)lnx/e^x
由g'(1)=0知当0<x<1,g'(x)>0; x>1,g'(x)>0
g(0)=-1,g(1)=-1/e,g(+∞)=lim(x->+∞) -(xlnx+1)/e^x
=- lim(x->+∞) (lnx+1)/e^x = -lim(x->+∞) 1/x / e^x =0
故当-1/e<=x<0时,f(x)有1个零点
当x>=0时,f(x)无零点