Find the range of the function $f(x) = \sqrt{\tan^{-1}x+1}+\sqrt{1-\tan^{-1}x}$ Problem : 
Find the range of the function $f(x) = \sqrt{\tan^{-1}x+1}+\sqrt{1-\tan^{-1}x}$ 
My approach : 
Let $\tan^{-1}x =t $ 
$y = \sqrt{t+1}+\sqrt{1-t}$ 
Squaring both sides we get : 
$y^2= t+1+1-t +2\sqrt{(t+1)(1-t)}$
$\Rightarrow y^2= 2+2\sqrt{(t+1)(1-t)}$
Now how to get the range of this function, please guide, will be of great help , thanks. 
 A: Note that $f(x)$ is even, so we may find extrema in either $x\geq 0$ or $x \leq 0$.
Also, we can find the domain of the function to be $[\tan(1), \tan(1)]$.
Now, consider $f'(x)$ for x $\geq0$:
$$f'(x)=\left(\dfrac{-1}{\sqrt{1-\arctan(x)}}+\dfrac{1}{\sqrt{1+\arctan(x)}}\right)\times\dfrac{1}{2(1+x^2)}$$
$f'(x)< 0$ for $x > 0$. Therefore f is decreasing for $x > 0$.
Extrema occurs at endpoints of the interval.
Therefore the range is $[\sqrt{2},2]$
A: Hint.
Answer the following: 
*Range of $\tan^{-1}(x)$ is ________________
*Range of a positive square root is ________________ 
*Finally, when the two square roots combined, what are the smallest and largest values you get ?
A: HINT:
$g(t)=2+2\sqrt{1-t^2}$
We need $-1\le t\le1$ to keep $g(t)$ real
So, $1\ge\sqrt{1-t^2}\ge0$
Now as $\sqrt{\tan^{-1}x\pm1}\ge0,$
$\sqrt{g(t)_{\text{min}}}\le f(x)\le \sqrt{g(t)_{\text{max}}}$
A: $$f(x)=\sqrt{1+\arctan x}+\sqrt{1-\arctan x}$$
Note:


*

*$f$ is only defined when $\arctan x \in [-1,1]$, i.e. $x\in[-\tan 1,\tan 1]$

*We can write $f(x)=g(h(x))$, where $g(x)=\sqrt{1+x}+\sqrt{1-x},h(x)=\arctan x$

*The image of $[-\tan 1,\tan 1]$ under $h$ is $[-1,1]$

*$g$ is an even function which decreases away from $0$, i.e. for $x\in[-1,1]$, we have $2=g(0)\ge g(x) \ge g(\pm1)=\sqrt{2}$.

*So, the image of $[-1,1]$ under $g$ is $[\sqrt{2},2]$


Thus, the range of $f$ is $[\sqrt{2},2]$.
