Show that $f(x)=x/\sqrt{x^2+1}$ is a bijection of $\mathbb R$ onto $\{ y: -1I am looking for help in regard to a practice question about functions. The question is 

Show that a function $f$, defined by $f(x)=x/\sqrt{x^2+1}$ , $x \in \Bbb R$ is a bijection of $\Bbb R$ onto $\{ y: -1<y<1\}$.

So what I know that for it to be a bijection, it must be an injection and also a surjection.
So to proof this question, do I just need to prove both of those?
So for injection, when $x_1=x_2$ $f(x_1)=f(x_2)$
To do this I wrote $$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}},$$ squared both sides and expanded to solve that $x_1=x_2$.
Next for a surjection, must show that the range is contained,
so the bottom cannot be $0$ or negative because cannot square root a negative and cannot divide by $0$.
I believe in this situation you are supposed to write it as $y=x/\sqrt{x^2+1}$ and solve for $x$ in terms of $y$ but I have trouble doing that. Or can I jsut do it by solving an inequality such as $\sqrt{x^2+1}>0$,
 A: Injective: suppose that
$$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}}\ .\tag{$*$}$$
Squaring both sides and multiplying out denominators,
$$x_1^2(x_2^2+1)=x_2^2(x_1^2+1)\quad\Rightarrow\quad x_1^2=x_2^2\ .$$
Now substituting back into the denominator on the RHS of $(*)$,
$$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_1^2+1}}\quad\Rightarrow\quad
  x_1=x_2\ .$$
Doing it this way avoids having to (explicitly) consider whether or not $x_1$ and $x_2$ have the same sign.
Surjective: we want to solve
$$y=\frac{x}{\sqrt{x^2+1}}\tag{$*\!*$}$$
for any $y\in(-1,1)$.  Easy algebra gives
$$x^2=\frac{y^2}{1-y^2}\ ,$$
and we now have to consider which square root to take in order to get the right $x$.  It is not hard to see that $y$ should have the same sign as $x$, so we guess
$$x=\frac{y}{\sqrt{1-y^2}}\ .$$
Note the word "guess": if you stop here, the solution is logically backwards and therefore incorrect.  We need to actually substitute this expression for $x$ into the RHS of $(**)$ and confirm that it simplifies to $y$.  This is easy, so I leave it to you.
A: since $f$ is differentiable
$$
\frac{df}{dx} = \frac1{(x^2+1)^{\frac32}}
$$
$f$ is monotonic. but it is also continuous...
A: Let $y \in (-1, 1)$. Then $y^2 \in (0, 1)$ . That is  $ 0 \lt  y^2 \lt 1 $. Now take,  $$ t = \frac{y^2}{1 - y^2} \implies y^2 = \frac{t}{t + 1} \;\;; \; t\gt 0 \;  \;\text{by its definition }$$ 
This was inspired by trying to write $y^2$ in the form of $  \dfrac{t}{t + 1}$ and then solving for $t$. Now, 
$$ y = \pm \sqrt {\dfrac{t}{t + 1} } = \pm \dfrac{\sqrt t}{ \sqrt{(\sqrt t)^2 + 1}} $$
Now take $x = \sqrt t$ or $ x = - \sqrt t $. Such an $x$ exists in $\Bbb R$ since $t \gt 0$. hence we have proven that there is $x \in \Bbb R$ such that $ y = \dfrac{x}{ \sqrt {x^2 +1}} $
$ \mathscr Q.E.D.$
Ultimately your $x$ is equal to $ \pm \sqrt{\dfrac{y^2}{1 - y^2}} $
A: I am starting with known facts
$\displaystyle\sin:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to (-1,1) $ and $\displaystyle\tan:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to \mathbb{R}$ $\;$ are bijective
For One one:
Choose $\tan(x),\tan(y)\in \mathbb{R}$ with $x,y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
then $\displaystyle f(\tan(x))=f(\tan(y))\Rightarrow\frac{\tan x}{\sqrt{1+\tan^2 x}}=\frac{\tan y}{\sqrt{1+\tan^2 y}}$
$\Rightarrow \sin x=\sin y\iff x=y\iff\tan x=\tan y \;$
for every $\displaystyle x,y\in\left(-\frac{\pi}
{2},\frac{\pi}{2}\right)$
Now for onto:
for every $a\in (-1,1) $ we can find $\displaystyle\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that $a=\sin\theta$
There exists $b\in\mathbb{R}$ such that $f(b)=a $ where $b=\tan\theta$
Since $a\text{ and } b$ are arbitrary
Hence $\forall \; y\in(-1,1)\;\exists \;x\in\mathbb{R}\text{such that }f(x)=y$
we are done
