已知f(x+1)=x²-x,x∈【1,2】,求f(x)

已知f(x+1)=x²-x,x∈【1,2】,求f(x)
已知f(x+1)=x²-x,x∈【1,2】,求f(x)

已知f(x+1)=x²-x,x∈【1,2】,求f(x)
用x-1代替x得
f(x-1+1)=(x-1)²-(x-1)
∴f(x)=x²-2x+1-x+1
=x²-3x+2

令a=x+1
x=a-1
1<=x<=2则2<=x+1=a<=3
f(a)=(a-1)²-(a-1)=a²-3a+2
所以f(x)=x²-3x+2，x∈[2，3]