Range of $\arctan(1+\frac{1}{x})$ Pretty self explanatory. I'm trying to find the range of $$f(x)=\arctan(1+\frac{1}{x})$$
but in all honesty, I'm not really sure how to proceed. I feel like there is something very silly and obvious I'm missing. I can find the domain fairly easily but how do I go about finding the range without taking its inverse? Is that even possible?
I have the same problem with:
$$f(x)=e^{x+\sqrt{x^2+1}}$$
In general, how would I go about finding the range? I know I can just take the inverse and find the domain of that, but is that the only way? Thanks!
 A: At first we need to show the range for $f(x)=1+\frac1x$ which is of course $(-\infty,\infty)\setminus\{1\}$, indeed
$$y=1+\frac1x \iff x=\frac 1 {y-1}$$
therefore we can reach any value but not $y=1$.
For the second one we have that $x+\sqrt{x^2+1}>0$ and


*

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

*$\lim_{x\to -\infty} x+\sqrt{x^2+1}=\lim_{u\to \infty} -u+\sqrt{u^2+1}=\lim_{u\to \infty} \frac{-u^2+u^2+1}{u+\sqrt{u^2+1}}=0$
In order to avoid limits note that
$$y=x+\sqrt{x^2+1} \iff (y-x)^2=x^2+1 \iff y^2-2xy+x^2=x^2+1 \iff x=\frac{y^2-1}{2y}$$
A: Second is also the same as the first one you need to minimize $x+\sqrt{x^2+1}$ in order to minimize $e^{x+\sqrt{x^2+1}}$ bcause $e^{x}$ is monotonic increasing
$$\lim_{x\to \infty} x+\sqrt{x^2+1}=\infty$$
$$\lim_{x\to -\infty} x+\sqrt{x^2+1}=0$$
$$\frac{d}{dx}(x+\sqrt{x^2+1})=1+\frac{x}{\sqrt{x^2+1}}=0$$
$$\frac{x}{\sqrt{x^2+1}}=-1$$
When $x>0$ this is not possible because LHS is positive and RHS is negative 
Let's check when $x<0$
$$x^2=x^2+1$$ gives us $0=1$
Thus $(x+\sqrt{x^2+1})$ never becomes 0 it is always greater than 0
Thus $$e^{(x+\sqrt{x^2+1})}>1$$
