Question

已知二次函数f(x)满足f(x+1)-f(x)=2x(x∈R),且f(0)=1

Answer 1

f(0)=1

设f(x)解析式f(x)=ax²+bx+1

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

a(x+1)²+b(x+1)+1-ax²-bx-1

=2ax+(a+b)=-2x+1

∴2a=-2

a+b=1

∴a=-1

b=2

解析式f(x)=-x²+2x+1

(2)

当x属于（-2,1）时,

f(x)在y=-x+m的图像上方

∴-x²+2x+1>-x+m

-x²+3x+1>m

g(x)=-x²+3x+1对称轴是x=3/2

∴g(x)最小值=g(-2)=-4-12+1=-15>m

m