Finding the inverse of $f(x) = x\sqrt{2+x^2}$ For a function $f(x) = x\sqrt{2+x^2}$ find out if it's bijective and if so, find its inverse.
The function is surjective because $x^2 > 0:\forall x\in\mathbb{R}$. I'm having difficulties proving that the function is injective.
I tried the following: $$f(x) = f(y) \iff x\sqrt{2+x^2} = y\sqrt{2+y^2} \\ x^2(2+x^2) = y^2(2+y^2) \\ 2x^2+x^4 = 2y^2 + y^4$$
And well, I think I just made it more difficult for myself. I couldn't find any thread on MSE where one tried tackling such a specific problem.
Solution appreciated, thanks in advance.
 A: What do you know about monotonous functions? Is this function (strictly) monotonous, and if yes, does this tell you something about injectivity?
A: Hint Let $a=x^2 \geq 0$ and $b=y^2 \geq 0$. The equation reduces to $$2a+a^2=2b+b^2$$ 
A: Hint:
To find the inverse let $y= f^{-1}(x)$ then since $f(f^{-1}(x)) = x$ we have
$$f(f^{-1}(x)) = x = y\sqrt{2+y^2}$$
which gives
$$x^2 = y^2(2+y^2)\to y^4 + 2y^2 - x^2 = 0$$
This is a quadratic equation in $y^2$: $(y^2)^2 + 2y^2 - x^2 = 0$ for which you can use the standard formula on. To determine which root to pick notice that $f^{-1}(0)=0$ and $\text{sign}(f^{-1}(x)) = \text{sign}(x)$.
A: The condition $x^2\ge0$, which implies $2+x^2>0$ just tells you that the function is defined on $\mathbb{R}$, but not that it is surjective. For instance, the function
$$
g(x)=\frac{1}{\sqrt{2+x^2}}
$$
is not surjective if considered as $g\colon\mathbb{R}\to\mathbb{R}$, because $0<g(x)\le\frac{1}{\sqrt{2}}$ for all $x$.
If you're allowed to use calculus, surjectivity and injectivity follow from considering these facts:


*

*$\displaystyle\lim_{x\to-\infty}f(x)=-\infty$

*$\displaystyle\lim_{x\to\infty}f(x)=\infty$

*$\displaystyle f'(x)=\sqrt{2+x^2}+\frac{x^2}{\sqrt{2+x^2}}>0$
Therefore $f$ is increasing and assumes every real value.
Without calculus it's a bit more difficult. Let's try solving the equation
$$
y=x\sqrt{2+x^2}
$$
with respect to $x$. First we note that if $x<0$, then also $y<0$ and conversely. Now we can square:
$$
y^2=2x^2+x^4
$$
or
$$
x^4+2x^2-y^2=0
$$
that furnishes
$$
x^2=-1+\sqrt{1+y^2}
$$
(the other solution of the quadratic equation must be discarded, because it's negative). Thus we can write
$$
|x|=\sqrt{\sqrt{1+y^2}-1}
$$
and, remembering the consideration above, we have
$$
x=\begin{cases}
\sqrt{\sqrt{1+y^2}-1} & \text{if $y>0$}\\[2ex]
0 & \text{if $y=0$}\\[2ex]
-\sqrt{\sqrt{1+y^2}-1} & \text{if $y<0$}\\[1ex]
\end{cases}
$$
and this proves both surjectivity and injectivity, because the function has an inverse.
