当a=0时,
f(x)=-3x^2+6x-3
Δ=36+36>0
f(x)有2个零点
当a≠0时,
f'(x)=3ax^2-3(a+2)x+6
=3a[x^2-(1+2/a)x+2/a]
=3a(x-1)(x-2/a)
当a0,f(x)递增
(1,+∞)上f'(x)2时,2/a0,f(x)递减
(1,+∞)上f'(x)
已知函数f=ax^3-3/2(a+2)x^2+6x-3
当a=0时,
f(x)=-3x^2+6x-3
Δ=36+36>0
f(x)有2个零点
当a≠0时,
f'(x)=3ax^2-3(a+2)x+6
=3a[x^2-(1+2/a)x+2/a]
=3a(x-1)(x-2/a)
当a0,f(x)递增
(1,+∞)上f'(x)2时,2/a0,f(x)递减
(1,+∞)上f'(x)