Question

1.若f(x+4)+f(x-4)=x²-2x,求二次函数f(x)的解析式.

Answer 1

因为 f(x+4)+f(x-4)=x²-2x

设x+4=U,x-4=V

则 x²-2x=(x+4)*(x-4)+(x+4)+(x-4)+16

即 f(U)+f(V)=U*V+U+V+16

再令U=V=x,则2f(x)=x²+x+16,

f(x)=(x²+x)/2+8