# find $A,B$ in way that $f:A\rightarrow B$ is bijective, when $f(x)=\sqrt{x^2-1}$

## Problem

find $$A,B$$ in way that $$f:A\rightarrow B$$ is bijective, when $$f(x)=\sqrt{x^2-1}$$.

## Attempt to solve

map $$f:A \rightarrow B$$ is bijective when it's surjective and injective simultaneously. This map is injective when:

$$\forall(x,y) \in A : x \neq y \implies f(x) \neq f(y)$$

This map is surjective when:

$$\forall y \in B \exists x \in A : f(x)=y$$

if function $$f(x)$$ is "truly" monotonic it implies that function has to be injective.

Observation:

$$\forall x \in \mathbb{R} : \frac{d}{dx}f(x) \neq 0$$

$$\frac{d}{dx}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2+1}} = f'(x)$$

It is visible that only way $$f'(x)$$ could be zero is when $$x=0$$ but $$\sqrt{-1}$$ is not defined in $$\mathbb{R}$$

which implies that $$f(x)$$ is injective if we pick $$A$$ in a way:

$$A\in \mathbb{R}\setminus[-1,1]$$

and $$B$$:

$$B \in \mathbb{R}$$

we have map: $$f:\mathbb{R}\setminus[-1,1] \rightarrow \mathbb{R}$$

Inverse function of $$f(x)$$ can be computed by solving y from following equation. Inverse function is map $$f^{-1}:B \rightarrow A$$

$$\sqrt{x^2-1}=y$$

$$f(x)$$ is defined when : $$x^2-1 \ge 0 \implies x^2\ge 1 \implies -1 \ge x \ge 1$$

$$x^2-1=y^2$$

$$x^2=y^2+1 \implies x=\pm \sqrt{x^2+1}$$

$$h^{-1}(x)=\sqrt{x^2+1}$$

meaning $$f(x)$$ is surjective. Which implies $$f(x)$$ is bijective when:

$$A\in \mathbb{R}\setminus[-1,1], B \in \mathbb{R}$$

• Your argument with the derivative does not work. Note that $f(x)=f(-x)$ when $f$ is defined in $x$. But with your reasoning you may choose $A=[1,\infty)$ and then compute the image $f(A)$. – Severin Schraven Sep 20 '18 at 20:45
• $A=B=\varnothing$. – Asaf Karagila Sep 21 '18 at 12:41

You have made a mistake here

$$\frac{d}{dx}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}} = f'(x)$$

therefore $$f(x)$$ is injective for $$x>1$$ or for $$x<-1$$.

then for the surjectivity it suffices to observe that $$\lim_{x\to \pm \infty} f(x) = \infty$$

then $$f(x)$$ is bijective for the following restrictions

• $$f_1:(-\infty,a]\to [0,\infty)$$
• $$f_2:[b,\infty)\to [0,\infty)$$

with $$a\le -1$$ and $$b\ge 1$$

We can take $$A=[1,+\infty)$$

$$(\forall x>1) \; f'(x)=\frac{x}{f(x)}$$

$f$ is continuous at $A$ and strictly increasing at $(1,\infty)$ thus it is strictly increasing at $A$.

$$f(A)=[0,+\infty)=B$$

$f$ is a bijection from $A$ in $B$.

When you only have to give an $$A$$ and $$B$$ such that $$f: A\to B$$, $$f(x)=\sqrt{x^2-1}$$ is bijective, why not take simply $$A=\{1\}$$ and $$B=\{0\}$$.

If you want to finde the maximal intervalls you might go like this:

We construct $$f^{-1}$$ and while we do so, we have to note when the operations we use are defined or equivialent transformations.

$$x=\sqrt{(y-1)(y+1)}$$ First of all $$y\geq 1$$ or $$y\leq -1$$, else the square root is not defined.

We square both sides. Thus $$x\geq 0$$:

$$x^2=y^2-1\Leftrightarrow y^2=x^2+1$$

$$y_{1,2}=\pm\sqrt{x^2+1}$$

This gives us two solutions.

If $$y\leq -1$$, that means $$y\in (-\infty, -1]$$, then $$f^{-1}(x)=-\sqrt{x^2+1}$$.

If $$y\geq 1$$, that means $$y\in [1,\infty)$$, then $$f^{-1}(x)=\sqrt{x^2+1}$$

This gives us of course the intervals which $$f$$ is a bijection on. We had $$x\geq 0$$ which means $$x\in[0,\infty)$$

And the two solutions are $$f:[1,\infty)\to\mathbb{R}_{\geq 0}$$ or $$f:(-\infty,-1]\to\mathbb{R}_{\geq 0}$$

You have two good choices for $$A$$ and for each case you have a choice for $$B$$

First choice: let $$A= (-\infty,-1]$$ while $$B= [0,\infty)$$ where your function is strictly decreasing.

Second choice: Let $$A= [1,\infty)$$ while $$B= [0,\infty)$$ where your function is strictly increasing.

• There are many many choices. – hamam_Abdallah Sep 20 '18 at 20:54
• You know what I mean by these choices. – Mohammad Riazi-Kermani Sep 21 '18 at 18:24

Since $$f(x)=f(-x)$$, for $$|x|\ge1$$, your set $$A$$ can only contain one among a number and its negative.

The most natural choice is $$A=[1,\infty)$$, but obviously not the only one; also the set $$A'$$ consisting of rationals $$\ge1$$ and the irrationals $$<-1$$ would do.

Both $$A$$ and $$A'$$ are “maximal”, in the sense that you cannot add any number in the domain of $$\sqrt{x^2-1}$$ to either $$A$$ or $$A'$$ and still have an injective function. There are infinitely many choices of such “maximal domains of injectivity”, most notably $$(-\infty,-1]$$; which one to choose is a matter of preferences and simplicity.

Let's stick with $$A=[1,\infty)$$. The equation $$\sqrt{x^2-1}=y$$ can only have a solution if $$y\ge0$$. In this case it becomes $$x^2-1=y^2$$, so $$x=\sqrt{y^2+1}$$, which exists (and is $$\ge1$$) for all $$y\ge0$$.

Hence your set $$B$$ is $$[0,\infty)$$.

• @N.F.Taussig I tried to expand it – egreg Sep 21 '18 at 12:24