Victor Wall

2022-01-12

How do I prove that $f\left(x\right)={x}^{2}-4x+3$ is injective at $\left(-\mathrm{\infty },2\right)$ and $\left(2,\mathrm{\infty }\right)$

eninsala06

Step 1
If
$f\left(x\right)=f\left(y\right)$
then
${x}^{2}-4x={y}^{2}-4y$.
Hence
${x}^{2}-{y}^{2}-4\left(x-y\right)=0$,
which means
$\left(x-y\right)\left(x+y-4\right)=0$.
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).

vicki331g8

Step 1
On the on the interval $\left(2,+\mathrm{\infty }\right)$, we have
$f\left(x\right)=f\left(y\right)⇒\left(x+y\right)\left(x-y\right)=4\left(x-y\right)$,
suppose that $x\ne y$, then we have
$x+y=4$, but we know that both $x,y>2$, a contradiction. Similar for the interval $\left(-\mathrm{\infty },2\right)$.

alenahelenash

Step 1This real valued function is obiously continuous and differentiable at every point in $\mathbb{R}$.$\frac{df\left(x\right)}{dx}=2x-4$.So${f}^{\prime }\left(x\right)>0$,$x\in \left(2,+\mathrm{\infty }\right)$and${f}^{\prime }\left(x\right)<0$,$x\in \left(-\mathrm{\infty },2\right)$.So since f is properly monotonous and coninouous in those two specific intervals, is also injective, i.e. if $x\ne y$ then $f\left(x\right)\ne f\left(y\right)$ .