Question

(1/2)这两个函数题怎么解:1、已知函数f(x)=(x-k)e的x次方,(1)求f(x)的单调曲间,(2)求f(x)在

Answer 1

(1)f ' (x)=e^x+(x-k)e^x=(x-k+1)e^x=0

则x-k+1=0 x=k-1

所以在(-无穷大,k-1]上是单调减函数.

在[k-1,+无穷大)上是单调增函数.

(2).i>若k-1若k-1>=1 则最小值=f(0)=-k

iii>若0