# 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!

• $\text{Range} [f \circ g] = f( \text{Range}[ g])$. Let $f=\arctan$ – Mark Sep 8 '18 at 7:51
• I see that but apparently there is a $\frac{\pi}{4}$ and I'm not sure how that got there. – Future Math person Sep 8 '18 at 7:54
• Maybe $\text{Range}[g]$ is $(1, \infty)$? That could be the case if $x$ is only allowed to be positive – Mark Sep 8 '18 at 7:57
• Ohh I see it. Thanks! – Future Math person Sep 8 '18 at 8:36

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

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

• Right. I also know that the range of $\arctan(x)$ is from $(-\frac{\pi}{2},\frac{\pi}{2})$. – Future Math person Sep 8 '18 at 7:48
• @FutureMathperson And for the second at first we need to find the range for $x+\sqrt{x^2+1}$. – gimusi Sep 8 '18 at 7:49
• I can't edit your post, but should it not be 0 that is excluded? – Future Math person Sep 8 '18 at 7:53
• @FutureMathperson I've edited, of course we can't reach the value 1, indeed $y=1+1/x \iff x=1/(y-1)$. – gimusi Sep 8 '18 at 7:54
• @FutureMathperson We can reach $0$ for $x=-1$, we can't reach $1$. – gimusi Sep 8 '18 at 7:55

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 $$e^{(x+\sqrt{x^2+1})}>1$$

• How does this imply that the 2nd function is bigger than 1? – Future Math person Sep 8 '18 at 8:35
• $$(x+\sqrt{x^2+1})>0$$ what is $e^0?$ – Deepesh Meena Sep 8 '18 at 8:36
• I see it now. Thank you :) . – Future Math person Sep 8 '18 at 8:37