1）當(dāng)a=1時(shí),求函數(shù)f（x）=(x-1)e^x
f'(x)=xe^(x)=0
=>x=0
f''(x)=e^x[1+x]
f''(x)=1>0
函數(shù)f（x）的極大值=f(0)=-1
(2)f'(x)=e^(x)[ax+a-1)
若函數(shù)f（x）在區(qū)間（0,1）上是單調(diào)增函數(shù)
ax+a-1>0
即a>1/(x+1)
0