# 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.

• $f(x) = (x - 1)^2 + 2$. Since the vertex is to the right of $x = 0$, it's clearly injective on the domain $(-\infty, 1]$ or $[1, +\infty)$ (obviously it's not injective on both or any union of any non-empty set of the two). As for being surjective--you need to give the codomain of the function. Because if the codomain is $(-\infty, +\infty)$ then it's not surjective. Apr 3, 2014 at 6:32
• Can you decide your function, whether monotone or not in the given domain ?;(that can be done using derivative test); Apr 3, 2014 at 6:34

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).

• Thank you so much. Also, thank you everyone else who also answered! Apr 3, 2014 at 7:12

The function is strictly decreasing throughout the domain.

Thus if x>y. Then f(y)>f(x).

When xf(y).

Equality of f(x) and f(y) hold only when x=y.

• I need some help. Whenever i post my answer, certain parts are not displayed, please edit the text if you can Apr 3, 2014 at 6:37

If $f$ is strictly decreasing and $x\neq y$, then $f(x)\neq f(y)$. That's because either $x<y$ (so $f(x)>f(y)$), or $y<x$ (so $f(y)>f(x)$).