已知函数f(x)=2x3+3ax2+3bx+8在x=1及x=2处取得极值. (1)求a、b的值; (2)求f(x)的单调区间.
问题描述:
已知函数f(x)=2x3+3ax2+3bx+8在x=1及x=2处取得极值.
(1)求a、b的值;
(2)求f(x)的单调区间.
答
(1)∵函数f(x)=2x3+3ax2+3bx+8,∴f′(x)=6x2+6ax+3b,∵f(x)在x=1及x=2处取得极值,∴f′(1)=6+6a+3b=0f′(2)=24+12a+3b=0,解得a=-3,b=4.(2)∵a=-3,b=4,∴f′(x)=6x2-18x+12,由f′(x)=6x2-...