Question

设函数f(x)=ax^3+bx^2+cx+b(a,b,c,d∈R)的图像关于原点对称且x=1时f(x)去最小值-2

Answer 1

∵函数f（x）=ax^3+bx^2+cx+d的图像关于原点对称

∴ f(x)为奇函数

则f(-x)=-f(x)

即-ax^3 +bx² -cx +d = -(ax^3+bx^2+cx+d) = -ax^3-bx^2 - cx- d

比较系数可得：b = 0,d = 0

∴f(x) = ax^3 +cx

f'(x) = 3ax² + c

∵x=1时f(x)去最小值-2/3

∴f(1)=a+c= -2/3

f'(1)=3a+c=0

解得：a=1/3,c=-1

即f(x) = x^3/3 - x