二次函数f（x）=3x^2+bx+1满足f(-1/3-x)=f(x-1/3),求b

二次函数f（x）=3x^2+bx+1满足f(-1/3-x)=f(x-1/3),求b

带入F(-1/3-x)=3(-1/3-x)^2+b(-1/3-3)+1
=3x^2+(2-b)x+4/3-b/3
带入F(x-1/3)=3(x-1/3)^2+b(x-1/3)+1
=3x^2+(2-b)x+4/3-b/3
因为 F(-1/3-x)= F(x-1/3)
所以 3x^2+(2-b)x+4/3-b/3=3x^2+(2-b)x+4/3-b/3
得到 2-b=b-2
解出b=2