Question

已知f(x)是二次函数且满足f(0)=1,f(x+1)-f(x)=2x,求f(x,)

Answer 1

设：f(x) = ax^2 + bx + c
f(0) = c = 1
f(x+1) - f(x) = a(x+1)^2 + b(x+1) + c - (ax^2 + bx + c) = 2x
所以：a = 1
当x = 1时：
f(1) - f(0) = 2
所以：1+b+1 -1 = 2
b = -1
f(x) = x^2 -x +1

Answer 2

f(0)=1,∴c=1
f(x+1)-f(x)=2x ∴a(x+1)^2+b(x+1)+1-ax^2-bx-1=2x
整理得：2ax+(a+b)=2x
由待定系数法可以知道 a=1,a+b=0,b=-1
∴f(x)=x^2-x+1