How do I prove that $f(x)=x^2-4x+3$ is injective at $(-\infty, 2)$ and $(2,\infty)$
Solution 1:
If $f(x)=f(y)$ then $x^2-4x=y^2-4y$. Hence $x^2-y^2-4(x-y)=0$, which means $$(x-y)(x+y-4)=0.$$ This is true if and only if $x=y$ or $x+y-4=0$. Now if $x,y > 2$, then $x+y>4$, so the only possibility is $x=y$. Same for $x,y < 2$ (try).