已知二次函数 f(x)=ax^+bx(a不等于零),且f(x+1)为偶函数,定义:满足f(x)=x的实数x

已知f(x) = ax^2 + bx,则f(x + 1) = a(x + 1)^2 + b(x + 1) = ax^2 + (2a + b)x + (a + b)
因为f(x + 1) 为偶函数,则f(x + 1) = f(-x + 1)
=> ax^2 + (2a + b)x + (a + b) =a(-x)^2 + (2a + b)(-x) + (a + b)
=> 2a + b = 0,即b = -2a
因此 f(x) = ax^2 -2ax
当 f(x) = x 时,有 ax^2 - 2ax = x,即 ax^2 - (2a + 1)x = 0
解出 x = 0 或 x = (2a + 1) /a