Question

已知f（x）为二次函数,且f（x+1）+f（x-1）=2x的平方+4x,求f（1-根号2）的值?如题

Answer 1

(x)=ax^2+bx+c 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+2b+2c =2x^2-4x 所以a=1,b=-2,c=1 f(x)=x^2-2x+1=(x-1)^2 f(1+√2)=(1+√2-1)^2=2