Question

已知二次函数f(x)满足f(0)=0,且f(x+1)=f(x)+x+1,g(x)=2f(-x)+x.(1)f(x)的表达

Answer 1

f(x+1)=a(x+1)²+b(x+1)=ax²+bx+x+1

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

2a+b=b+1

a+b=1

a=b=1/2

f(x)的表达式:y=x²/2 + x/2

g(x)=2(x²/2 -x/2) +x=x²

f[g(x)]=f(x²)=(x²)²/2 + x²/2