Question

问1道不算难的数学题以知f(x)为二次函数,且f(x+1)+f(x-1)=2x平方－4x,则f(x)的表达式是什么,最好

Answer 1

设 f(x) = ax^2 + bx + c

则：f(x+1) = a(x+1)^2 + b(x+1) + c = ax^2 + (2a+b)x + (a+b+c)

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

所以：

f(x+1) + f(x-1) = 2ax^2 + 2bx + 2(a+c)

= 2x^2 - 4x

对比系数,得：

2a = 2

2b = -4

2(a+c) = 0

解得：a = 1 ,b = -2 ,c = -1

所以：

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