（1）显然,定义域应满足x＞0
对f(x)求导,得
f'(x)=1/x-ax+(a-1)x
=-(1/x)[ax²-(a-1)x-1]
=-(1/x)(ax+1)(x-1)
令f'(x)=0,
可得x=1或x=-1/a
则,若
①a＞0,可得f(x)在[0,1]递增,在[1,+∞)递减；
②-1＜a＜0,可得f(x)在[1,-1/a]递增,在[0,1]及[-1/a,+∞)递减；
③a＜-1,可得f(x)在[-1/a,1]递增,在[0,-1/a]及[1,+∞)递减.
（2）f(x)的图像的切线斜率为
k=f'(x)=-(1/x)(ax+1)(x-1)
……
……