已知函数F(x)=ax3-3x2+1(a属于r且.a>0),求f`(x)及函数f(x)的极大值与极小值

已知函数F(x)=ax³-3x²+1(a属于r且.a>0),求f`(x)及函数f(x)的极大值与极小值
令f′(x)=3ax²-6x=3ax(x-2/a)=0,故得驻点x₁=0,x₂=2/a.
∵a>0,∴2/a>0,故当x由0的左边变到0的右边时f′(x)由正变负,故x₁=0是极大点；当x由2/a
的左边变到2/a的右边时,f′(x)由负变正,故x₂=2/a是极小点.于是得：
maxf(x)=f(0)=1；minf(x)=f(2/a)=8/a²-12/a²+1=(a²-4)/a²

f(x)=ax^3-3x^2+1
f'(x) = 3ax^2- 6x
f'(x)=0
3x(ax-2)=0
x= 0 or 2/a
f''(x) = 6ax -6
f''(0) = -6 (max)
f''(2/a) = 6 (min)
max f(x) = f(0) =1
minf(x) = f(a/2) = a^4/8 - 3a^2/4 +1

f(x)=ax^3-x^2+1(a>0)
f'(x)=3ax^2-2x
令f'(x)=0
则x=0或x=2/(3a)
f(0)=1
f(2/(3a))=a[2/(3a)]^3-[2/(3a)]^2+1
=-4/(27a^2)+1
所以极大值f(0)=1，极小值f(2/(3a))=-4/(27a^2)+1