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8.已知实数a>0,函数f(x)=ax(x-2)2(x∈R)有极大值32.
(1)求实数a的值;
(2)求函数f(x)的单调区间.
解:(1)∵f(x)=ax(x-2)2=ax3-4ax2+4ax,
∴(x)=3ax2-8ax+4a.
由(x)=0,得3ax2-8ax+4a=0.
∵a≠0,∴3x2-8x+4=0.
解得x=2或x=.
∵a>0,∴x<或x>2时,(x)>0;
<x<2时,(x)<0.
∴当x=时,f(x)有极大值32,即
a-a+a=32,∴a=27.
(2)f(x)在(-∞,)和(2,+∞)上是增函数,在(,2)上是减函数.