Question

Find the range of $f\left(x\right)={\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$

Answer 1

Find the range of the given function

Given: $f\left(x\right)={\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$

Domain of ${\sin }^{-1}\left(x\right)=\left[-1,1\right]$

Domain of ${\cos }^{-1}\left(x\right)=\left[-1,1\right]$

Domain of ${\tan }^{-1}\left(x\right)=\left(-\infty ,\infty \right)$

So $f\left(x\right)={\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$ is defined in $\left[-1,1\right]$

So, $f\left(x\right)=\frac{\mathrm{\pi }}{2}+{\tan }^{-1}\left(x\right)\left[\because {\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\frac{\pi }{2},x\in \left[-1,1\right]\right]$

$f\text{'}\left(x\right)=\frac{1}{1+x^2}>0,\forall x\in \left[-1,1\right]$

Hence$f\left(x\right)$ is an increasing function.

$f(-1)=\frac{\mathrm{\pi }}{2}+\tan (-1)=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{4}$

$f(+1)=\frac{\mathrm{\pi }}{2}+\tan (+1)=\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4}$

Therefore, $f\left(x\right)\in \left[\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right]$

Hence, the range of the given function is $\left[\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right]$