Question

f(x)=x^3+ax^2+bx+c在x=1处取得极小值-2 1.用c表示啊a、b 2.求函数单调递减区间

Answer 1

1:

f'(x) = 3x^2+2ax+b,当x=1时f'(x) = 0,即3+2a+b = 0.(1)

又：x=1时,f(x) =-2,即：3+a+b+c = -2.(2)

将C看作已知数,联立(1) (2),解二元一次议程组,得

a =c

b = -3-2c

2:

f'(x)