Q:

Prove that a function is a bijection?

I am having trouble with this problem:
Prove that the function $f(x)=x^2-2x+3$ with domain $x\in(-\infty, 0)$, is a bijection from $(-\infty, 0)$ to its range. 
Work: Basically, I try to use the standard contrapositive proof: if f(x)=f(y), then x=y to try and prove the injective. However, I am stuck on this part and I do not know what to do. I know that the function is strictly decreasing along the domain but I do not know how to put this in proof and then say that this means the function is bijective.

A:

The function $f(x) = x^2 - 2x + 3$ is bijective if it maps values from $(-\infty, 0) \mapsto (3, \infty)$.  You can show that the function is monotone on that domain or you can just solve for $x$ given $y$:
$$
0 = x^2 - 2x + 3 - y \\
x = \frac{2 \pm \sqrt{4 - 4(3 - y)}}{2} = \frac{2 \pm \sqrt{4y - 8}}{2}
$$
For $x$ to exist, $4y - 8 \geq 0 \rightarrow y \geq 2$.  All values in the codomain $y \in (3, \infty)$ satisfy that constraint.  Next the question is whether or not values from that codomain can give two different values of $x$.  The minimum value of $4y - 8$ occurs at the minimum value of $y = 3$: $12 - 8 = 4$.  Therefore for all values of $y > 3$, $\sqrt{4y - 8} > \sqrt{4} = 2$.  This means:
$$
x > \frac{2 + 2}{2} = 2 \\
x < \frac{2 - 2}{2} = 0
$$
Since our domain is $x\in (-\infty, 0)$, $x > 2$ is discarded.  This leaves values of $y\in (3, \infty)$ giving values of $x < 0 \rightarrow x \in (-\infty, 0)$ which is the original domain, thus this function is surjective.  It's also injective because we discarded the extra solution (it was outside of the original domain).

