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8.已知實(shí)數(shù)a>0,函數(shù)f(x)=ax(x-2)2(x∈R)有極大值32.
(1)求實(shí)數(shù)a的值;
(2)求函數(shù)f(x)的單調(diào)區(qū)間.
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時(shí),(x)>0;
<x<2時(shí),(x)<0.
∴當(dāng)x=時(shí),f(x)有極大值32,即
a-a+a=32,∴a=27.
(2)f(x)在(-∞,)和(2,+∞)上是增函數(shù),在(,2)上是減函數(shù).
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