Question

设函数f(x)为二次函数,且f(x+1）+f(x-1)=x∧2-2x+6,求f(x)

Answer 1

f(x)=ax²+bx+c

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

=a(x+1)²+b(x+1)+c+a(x-1)²+b(x-1)+c

=2ax²+2bx+(2a+2c)

=x²-2x+6

2a=1

2b=-2

2a+2c=6

a=1/2,b=-1,c=5/2

f(x)=x²/2-x+5/2