Find the range of the function $y = \sqrt{x^2 + 1}− x$? I have a function $y = \sqrt{x^2 + 1} − x$, where the Domain is $(−\infty,+\infty)$.
Explanation for the domain
I need to make sure the domain of the function does not include values of $x$
that will make the square root negative.
This means that I need:
$x^2+1\ge0$
the discriminant is negative and $x \in\mathbb R$. In fact, $x^2\ge 0$,  $\forall x\in\mathbb R$ and $1 > 0$.
I wish, if it's possible, to explain the value of the range with an algebraic demonstration. I am using an example to solve the function for $x$ using $y$ as parameter:
$\sqrt{x^2 + 1} − x = y$
$\sqrt{x^2 + 1} = y + x$
irrational equation, therefore:
$$
\begin{cases}
& y + x >= 0\\
& x^2 + 1 =(y+x)^2
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& x^2 + 1 =y^2 + x^2 + 2yx
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& 1 =y^2 + 2yx
\end{cases}
$$
$$
\begin{cases}
& x >= -y\\
& x = \frac{1 - y^2}{2y}
\end{cases}
$$
now I need to find for which values is $x >= -y$, therefore:
$ \frac{1 - y^2}{2y} < 0$
solving the numerator and denominator
numerator 
$1 - y^2 > 0$
$y^2 -1 < 0$
solve for $-1 < y < +1$
denominator 
$2y > 0$
solve for $y > 0$
combine the tow solutions the inequality has occurred when
$-1 < y < 0$ or $y> 1$

This is only a partial answer as it only establishes the lower bound for the range.
Multiply by the conjugate
$$y=(\sqrt{x^2+1}-x) \cdot \dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$$
$$y=\dfrac{1}{\sqrt{x^2+1}+x}$$
This function is clearly approaching $0$ as $x$ is approaching positive infinity. Also the function is never negative, since $$1>0$$
$$x^2+1>x^2$$
$$\sqrt{x^2+1}>x$$
Therefore we established a lower bound of $0$ for your range.
