Question

已知函数f(x)=x三次方+ax二次方+bx+c在x=-2/3与x=1时都取得极值.若对x属于[-1,2]都有f(x)

Answer 1

f(x)=x^3+ax^2+bx+c

f'(x)=3x^2+2ax+b

由f'(-2/3)=0 f'(1)=0得a=-1/2 b=-2

则f(x)=x^3-1/2x^2-2x+c