已知f(x+1)=x2-5x+4
求f(x)
设x+1=t===>x=t-1
f(x+1)=f(t)=(t-1)^2-5(t-1)+4=t^2-7t+10
即f(x)=x^2-7x+10
f(2x+1)=x2-2x求f(根号2)
设2x+1=t==>x=[t-1]/2
f(t)=[(t-1)/2]^2-(t-1)=1/4t^2-3/2t+5/4
f(根2)=1/4*2-3/2*根2+5/4=7/4-3/2根2