Question

f(x)=ax^3+bx^2+cx在x=正负1时取得极值,且f(1)=-1,

Answer 1

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

f(x)=ax^3+bx^2+cx在x=正负1时取得极值

所以f'(1)=f'(-1)=0

3a+2b+c=0 ---1

3a-2b+c=0 ---2

f(1)=-1

a+b+c=-1 ---3

联合1,2,3得

a=1/2 b=0 c=-3/2

f'(x)=3/2x^2-3/2

=3/2(x+1)(x-1)

x < -1时f'(x) > 0 f(x)递增

-1