Question

已知f(x)为多项式函数,且f(x+1)+f(x-1）=2x²-2x+4.求f（x）的解析式.

Answer 1

设f(x)=ax^2+bx+c

则f(x+1)+f(x-1)

=a(x+1)^2+b(x+1)+c+a(x-1)^2+b(x-1)+c

=2ax^2+2bx+2a+2c

所以2a=2,2b=-2,2a+2c=4

a=1,b=-1,c=1

f(x)=x^2-x+1