Question

设函数f(x)=-1/3x^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)|

Answer 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<a+1<a+2
当a<1/2时，3a<a+1<a+2 |f'(x)| < |f'(a+2)| =4(a-1)
当a>1/2时，a+1<3a<a+2 |f'(x)| <min{|f'(a+2)| ,|f'(a+1)| }=min{4a-4,2a-1}=4a-4

恒有|f'(x)| <4a-4