<s>[INST] 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. 
 [/INST] 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 $[\sqrt{2},2]$
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