Proving a polynomial is injective on restricted domain Show that the following function is injective 
$ f:[2,\infty) \rightarrow \Bbb R : x \mapsto x^2 -4x + 5 $
I've shown that the range is $[1,\infty)$ by $f(2+\sqrt{c-1} )=c$
I know that to show injectivity I need to show $x_{1}\not= x_{2} \implies f(x_{1}) \not= f(x_{2})$. But this leads me to $(x_{1})^2-4(x_{1})=(x_{2})^2-4(x_{2})$. What reasoning can I give for those to be equal? 
 A: Let $x_1,x_2\in [2,+\infty) $ such that
$x_1\neq x_2$ and $f (x_1)=f (x_2) $.
then by Rolle's  Theorem
$\exists c\in (x_1,x_2) :$
$$f'(c)=0=2c-4$$
which is in contradiction with $c>2$.
$f $ is injective.
A: Your approach is good: suppose $c\ge1$; then
$$
x^2-4x+5=c
$$
leads to
$$
x=2-\sqrt{c-1}\qquad\text{or}\qquad x=2+\sqrt{c-1}
$$
and there is a unique solution in $[2,\infty)$.
So you have computed the inverse function from $[1,\infty)$ to $[2,\infty)$. Hence the given function is injective.
The other method can be used as well. Suppose $2\le x_1\le x_2$ and $f(x_1)=f(x_2)$. Then
$$
x_2^2-4x_2+5=x_1^2-4x_1+5
$$
so
$$
(x_2-x_1)(x_2+x_1)-4(x_2-x_1)=0
$$
which becomes
$$
(x_2-x_1)(x_2+x_1-4)=0
$$
Hence either
$$
x_2+x_1=4
$$
which implies $x_1=x_2=2$, or
$$
x_2-x_1=0
$$
which implies $x_1=x_2$.
A: Calculate the maximum point of your parabola, and then you can check if your domain is on one side of the maximum, and thus injective. So just calculate,
$$ \frac{df}{dx} = 0, $$
since you know that $f'$ is a straight line it will differ from zero everywhere except at the maxima and thus the restriction to the left or right side will be monotonic and thus injective.
A: $$f (x_1)=f (x_2)\implies $$
$$x_1^2-4x_1=x_2^2-4x_2 \implies$$
$$x_1^2-x_2^2=4 (x_1-x_2)\implies$$
$$(x_1-x_2)(x_1+x_2-4)=0$$
but
$$x_1>x_2\geq 2$$ then
$$x_1+x_2>2x_2\geq 4$$
or
$$x_1+x_2-4>0$$
thus
$$x_1=x_2$$.
