<s>[INST] 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.
 [/INST] 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 $\text{sign}(f^{-1}(x)) = \text{sign}(x)$.
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