已知f(2x+1) =4x2+2x+1,
f(2x+1)=4x2+2x+1=(2x+1)²-(2x+1)+1;
f(x)=x²-x+1;
或令t=2x+1 x=(t-1)/2
f(2x+1)=f(t)=4[(t-1)/2]^2+2[(t-1)/2]+1
=(t-1)^2+(t-1)+1
=t^2-t+1
所以f(x)=x^2-x+1
已知f(2x+1) =4x2+2x+1,
f(2x+1)=4x2+2x+1=(2x+1)²-(2x+1)+1;
f(x)=x²-x+1;
或令t=2x+1 x=(t-1)/2
f(2x+1)=f(t)=4[(t-1)/2]^2+2[(t-1)/2]+1
=(t-1)^2+(t-1)+1
=t^2-t+1
所以f(x)=x^2-x+1