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 $y> 1$
 A: To determine the range you must study the variations of $y$ and apply the Intermediate value theorem (it's a continuous function).
Now $\;y'=\dfrac x{\sqrt{x^2+1}}-1=\dfrac {x-\sqrt{x^2+1}}{\sqrt{x^2+1}}<0,\;$ since $\;\lvert x\rvert<\sqrt{x^2+1} $. Hence the function is decreasing. 
On the other hand,


*

*$\lim_{x\to-\infty}\sqrt{x^2+1}-x=+\infty+\infty=+\infty$,

*$\lim_{x\to+\infty}\sqrt{x^2+1}-x=\lim_{x\to+\infty}\dfrac{(x^2+1)-x^2}{\sqrt{x^2+1}+x}=0$.


Hence the range of $y$ is $(0,+\infty)$.
A purely computational solution:
As noted in another answer, $y=\dfrac1{x+\sqrt{x^2+1}}$. Hence the range is contained in $(0,+\infty)$. Lets show  for any $m>0$, the equation  $\dfrac1{x+\sqrt{x^2+1}}=m$ has a solution. Indeed, it is equivalent to
$$\sqrt{x^2+1}=\frac1m-x\iff x^2+1=\Bigl(\frac1m-x\Bigr)^2\enspace\textbf{and}\enspace x \le\frac 1m.$$
Now $$x^2+1=\Bigl(\frac1m-x\Bigr)^2=\frac1{m^2}-\frac 2mx+x^2\iff x=\frac{1-m^2}{2m}$$
and the condition $\;x\le\dfrac1m$ is satisfied iff $\;1-m^2\le 2$, i.e. $m^2\ge -1$.
A: Well, there is probably an easy algebraic method:
$$y=\sqrt{x^2+1}-x$$$$\implies y^2+x^2+2xy=x^2+1$$$$\implies y^2+2xy-1=0$$
to find range of $y$, we reform this equation as
$$0x^2+2xy+y^2-1=0$$
this is quadratic in $x$ and since $x$ is real , $y$ should also be real so discriminent should be greater than zero 
$$D=(2y)^2-4(y^2-1)(0)>0$$
and this implies $$y^2>0$$$$\implies y>0$$or$$y<0$$ but $y<0$ is ruled out since 
$|x|=\sqrt{x^2}<\sqrt{x^2+1}$, so $x<\sqrt{x^2+1}$ ,therefore $y>0$ for all values of $x$ so $$y\in (0,\infty)$$
A: Hint: Expand the fraction with $\sqrt{x^2+1}+x$ to see that
$$y=\frac{1}{\sqrt{x^2+1}+x}$$
A: Let's see:
$$\frac1{\sqrt{x^2+1}+x}=\frac1{-x\sqrt{1+\frac1{x^2}}+x}=\frac{-\frac1x}{\sqrt{1+\frac1{x^2}}-1}$$
With l'Hospital now (we have a $\;\frac00\;$ indeterminate);
$$\lim_{x\to-\infty}\frac{-\frac1x}{\sqrt{1+\frac1{x^2}}-1}=\lim_{x\to-\infty}\frac{\frac1{x^2}}{-\frac1{x^3\sqrt{1+\frac1{x^2}}}}=\lim_{x\to-\infty}-x\sqrt{1+\frac1{x^2}}=+\infty$$
And now:
$$\frac1{\sqrt{x^2+1}+x}=\frac1{x\sqrt{1+\frac1{x^2}}+x}=\frac{\frac1x}{\sqrt{1+\frac1{x^2}}+1}\xrightarrow[x\to\infty]{}\frac02=0$$
So I'd say the range is $\;(0,\infty)\;$ as this is a continuous function everywhere.
Other way: calculate any of the two limits above, and then check that
$$\left(\sqrt{x^2+1}-x\right)'=\frac x{\sqrt{x^2+1}}-1\le0\;\;\;\forall\,x\in\Bbb R$$
so the function is monotonic decreasing. Now, if you take the limit when $\;x\to-\infty\;$ then you use this and the fact that the function's always non-negative, and if you took the limit when $\;x\to\infty\;$ then use this and the fact the function's never constant in any interval.
A: 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.
