f(x)=2x^3+3ax^2+3bx+8在x=1、x=2处得极值.求ab值
问题描述:
f(x)=2x^3+3ax^2+3bx+8在x=1、x=2处得极值.求ab值
求ab值和f(x)单调区间
答
f'(x)=6x^2+6ax+3b=0
x=1时
6+6a+3b=0
x=2时
24+12a+3b=0
两式相减得
a=-3,b=4
f(x)=2x^3-9x^2+12x+8
f'(x)=6x^2-18x+12>0时,为增.
x^2-3x+2>0
x>2 or x