已知函数f(2x-1)=x2-3x+2,求f(x-2)

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已知函数f(2x-1)=x2-3x+2,求f(x-2)

已知函数f(2x-1)=x2-3x+2,求f(x-2)
已知函数f(2x-1)=x2-3x+2,求f(x-2)

已知函数f(2x-1)=x2-3x+2,求f(x-2)
f(2x-1)=x^2-3x+2=(x-2)(x-1)
令t=2x-1
x=(t+1)/2
f(t)=[(t+1)/2-1][(t+1)/2-2]
=[(t-1)/2][(t-3)/2]
=(t-1)(t-3)/4
=(t^2-4t+3)/4
即f(x)=(x-1)(x-3)/4
f(x-2)=(x-2-1)(x-2-3)/4
=(x-3)(x-5)/4
=(x^2-8x+15)/4

f(x-1)=x2-3x+2
=(x-1)(x-2)
=(x-1)(x-1-1)
所以f(x)=x(x-1)
所以f(x+1)=(x+1)(x+1-1)=x(x+1)=x²+x

设t-2=2x-1,所以x=(t-1)/2
f(t-2)=((t-1)/2)²-3*(t-1)/2+2