设函数f(x)=-1⼀3x^3+2ax^2-3a^2x+b,0<a<1 (1)求f(x)的单调区间极值 若当x属于[a+1,a+2]时,恒有|f✀(x)|

2025-01-07 03:00:37
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回答1:

f'(x)=-x^2+4ax-3a^2=-(x-3a)(x-a) 负无穷到a单调减,a到3a单调增,3a到正无穷单调减

f'(x)对称轴为2a当a<1/2时,3a当a>1/2时,a+1<3a
恒有|f'(x)| <4a-4