Question

设函数f（x）=ax（3x上面）+bx（2在x上面）+x在x=1处取得极大值5.求常数a和b.求函数f（x）的极小值

Answer 1

f(x)'=3ax^2+2bx+1,则

3a+2b+1=0

a+b+1=5

a=-9

b=13,即

令f(x)'=3ax^2+2bx+1=-27x^2+26x+1=0

x=1或x=-1/27,即

x=-1/27,f（x）有极小值=-9（-1/27)^3+13(-1/27)^2-1/27=-41/2187