Question

已知函数f(x)=ax的三次方+bx的平方-2x在x=-2,x=1处取得极值,求函数f(x)的解析式及

Answer 1

求导f'(x)=3ax^2+2bx-2
在x=-2,x=1处取得极值,说明f'(x)=0的二个零点是-2和1
韦达定理得:-2+1=-2b/(3a),-2*1=-2/(3a)
解得:a=1/3, b=1/2

Answer 2

f(x) = ax^3+bx^2-2x
f'(x) = 3ax^2+2bx -2
f'(-2) = 12a-4b-2 =0 (1)
f'(1) = 3a+2b -2 =0 (2)
(1) - 4(2)
-4b-2 - 8b+8 =0
12b = 6
b = 1/2
a= 1/3
f(x) = (1/3)x^3 + (1/2)x^2 - 2x

Answer 3

f'(x)=3ax^2+2bx-2
f'(x)=0的二个零点是-2和1
韦达定理得:-2+1=-2b/(3a),-2*1=-2/(3a)
解得:a=1/3, b=1/2