algebra precalculus - 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!

Answer

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

$$y=x+\sqrt{x^2+1} \iff (y-x)^2=x^2+1 \iff y^2-2xy+x^2=x^2+1 \iff x=\frac{y^2-1}{2y}$$