已知函数f(2x+1)=x2-3x+2,求f(x-2)

问题描述:

已知函数f(2x+1)=x2-3x+2,求f(x-2)

f(2x+1)=x2-3x+2
令2x+1=t-2
x=(t-3)/2
f(t-2)=[(t-3)/2]²-3(t-3)/2+2
=1/4(t²-6t+9)-3/2t+13/2
=1/4(t²-12t+35)
所以
f(x-2)=1/4(x²-12x+35)

令2x+1=t
则x=(t-1)/2
所以f(t)=(t-1)²/2²-3×(t-1)/2+2
=t²/4-2t+15/4
故f(x-2)=(x-2)²/4-2(x-2)+15/4
=x²/4-3x+35/4
  祝学习快乐