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.
问题描述:
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.
答
(1)∵f(x)=ax(x-2)2=ax3-4ax2+4ax,∴f′(x)=3ax2-8ax+4a.由f′(x)=0,得3ax2-8ax+4a=0.∵a≠0,∴3x2-8x+4=0.解得x=2或x=23.∵a>0,∴x<23或x>2时,f′(x)>0;23<x<2时,f′(x)<0.∴当x=...