find the range of function $f(x)=x+\sqrt{x^2+1}$ find the range of function $f(x)=x+\sqrt{x^2+1}$
My Try : $$x^2 \geq 0 \\x^2+1\geq1 \\ \sqrt{x^2+1}\geq1$$
Now what ?
 A: This is a continuous  increasing function with $$\lim _{x\to -\infty}f(x)=0$$ and $$\lim_{x\to \infty} f( x)=\infty$$
Thus the range is $(0,\infty)$
A: WLOG $x=\cot2y,0<2y<\pi$
$f=x+\sqrt{1+x^2}=\cdots=\cot y$
Now $0<y<\dfrac\pi2$
$\cot0>f>\cot\dfrac\pi2$
A: The domain is $\Bbb R$ as there are no restrictions. However, the range of $f(x)$ is not $\Bbb R$. For $x\ge0$, the range is clearly $[1,\infty)$. For $x<0$, let $k=-x>0$. Then the function $$f(x)=x+\sqrt{x^2+1}\implies g(k)=-k+\sqrt{k^2+1}>0$$ for all positive $k$. Thus the range is $(0,1)\cup[1,\infty)=(0,\infty)$.
A: Since $\sqrt{x^2+1}>|x|$, which implies that $f(x)>0$ for all $x\in\mathbb{R}$, the range of $f$ must be a subset of $(0,\infty)$. 
To show that $(0,\infty)$ is the range of $f$, consider any $b>0$. Now you want to show that there exists some $a\in\mathbb{R}$ such that $f(a)=b$. 
Note that if $b=f(a)$, then
$$
(b-a)^2=a^2+1
$$
and thus
$$
b^2-2ab=1.
$$
Consequently, you can check by substitution that $a=\frac{b^2-1}{2b}$ is such that $f(a)=b$. 
A: If we consider only the positive magnitude of the sqrt,then the range of the function is(0,infinity).
As simplifire made it for positive domain,I would like to give you for negative domain.
When you consider the function with negative domain,the difference between x and sqrt(x^2+1) varies from 0 to1.The value is 1 when x is 0 and decreases to 0 when x is -infintiy,but never approaches 0,because the value under sqrt is always greater than x.
If one consider both + and - sqrt then it is not a function at all.Since a function can't have two outputs for single input.
A: $\sqrt{x^2}= |x| \implies \sqrt{x^2+1}> |x|\geq -x$
So, $x+\sqrt{x^2+1}> 0$
Then, the range of $f(x) \subseteq  ]0,+\infty[$
Also since, $f$ is a continuous function and 
$\lim _{x\to -\infty}f(x)=0$ and 
$\lim_{x\to \infty} f( x)=+\infty$
Thus, the range of $f(x)$ is $]0,+\infty[$
