f（x）=1/2x²-ax+（a-1）lnx,a＞1,x>0
求导
f'(x)=x-a+(a-1)/x=[x-(a-1)](x-1)/x
I）当1=0,且f'(x)不恒为0,f(x)在x∈(0,+∞),单调递增
III）当a>2,时
x∈(0,1),f'(x)>0,f(x)单调递增
x∈(1,a-1),f'(x)12是怎么定出来的?是二个根a-1和1比较得到的当a-1>1时有a>2,当a-1<1时有a<2