Victor Wall

2022-01-12

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

eninsala06

Beginner2022-01-13Added 37 answers

Step 1

If

$f\left(x\right)=f\left(y\right)$

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).

If

then

Hence

which means

This is true if and only if

Now if

vicki331g8

Beginner2022-01-14Added 37 answers

Step 1

On the on the interval$(2,+\mathrm{\infty})$ , we have

$f\left(x\right)=f\left(y\right)\Rightarrow (x+y)(x-y)=4(x-y)$ ,

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 $(-\mathrm{\infty},2)$ .

On the on the interval

suppose that

alenahelenash

Skilled2022-01-24Added 366 answers

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