已知函数f(x)=(k为常数,e是自然对数的底数),曲线y=f(x)在点(1,f(1))

已知函数f(x)=(k为常数,e是自然对数的底数),曲线y=f(x)在点(1,f(1))处的切线与x轴平行.

(1)求实数k的值;

(2)求函数f(x)的单调区间.

答案

 (1) x>0,f'(x)=

又由题知f'(1) ==0,所以k=1.

(2) x>0,f'(x)=.

h(x)=-ln x-1(x>0),则h'(x)=--<0,即h(x)在(0,+∞)上单调递减.

h(1) =0知,当0<x<1时,h(x)>0,所以f'(x)>0;

x>1时,h(x)<0,所以f'(x)<0.

综上,f(x)的单调增区间是(0,1),单调减区间是(1,+∞).

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