Question

已知2f(x)+f(x-1)=x²求f(x)的解析式?

Answer 1

由题可知f(x)是二次函数

设f(x)=ax^2+bx+c

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

=ax^2-2ax+a+bx-b+c

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

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

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

=3ax^2+(3b-2a)x+a-b+3c

3a=1

3b-2a=0

a-b+3c=0

a=1/3 b=2/9 c=-1/27

f(x)=(1/3)x^2+(2/9)x-1/27