Question

已知二次函数fx满足f括号x加一括号加f括号x减一等于二x平方减四x,

Answer 1

设f(x)=ax^2+bx+c,则
f(x+1)+f(x-1)
=a[(x+1)^2+(x-1)^2]+b(x+1+x-1)+2c
=2ax^2+2bx+2a+2c
=2x^2-4x,
比较系数得a=1,b=-2,2a+2c=0,c=-1,
∴f(x)=x^2-2x-1.