函數及基本微分問題

Suppose f(x) is of degree 1
i.e. Let f(x)=Ax+B
f(x+1)-f(x)=4x-2
[A(x+1)+B]-(Ax+B)=4x-2
A=4x-2
which implies A is not a constant , that is wrong
Now , let f(x) is of degree2
i.e. f(x)=Ax^2+Bx+C
f(x+1)-f(x)=4x-2
[A(x+1)^2+B(x+1)+C]-(Ax^2+Bx+C)=4x-2
(2x+1)A+B=4x-2
2Ax+(A+B)=4x-2
which gives :
2A=4 -> A=2
A+B=-2 -> 2+B=-2 ->B=-4
Also , f(3)=7
that is 2(3)^2-4(3)+C=7 -> C=1
Finally , f(x)=2x^2-4x+1