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

Question

Problem :

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

Solution : Since, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$

Since range of $\tan^{-1}x$ is $ (\frac{-\pi}{2}, \frac{\pi}{2})$

$\therefore, \frac{-\pi}{2} \leq \tan^{-1}x \leq \frac{\pi}{2}$

= $ \frac{-\pi}{2} + \frac{\pi}{2} \leq \tan^{-1}x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$.

= $0 \leq \tan^{-1}x+ \frac{\pi}{2} \leq \pi $

Is it correct.. please suggest thanks....

Answer 1

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

First we will calculate domain of function $f(x)$

function $\sin^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$ Similarly function $\cos^{-1}(x)$ is defined in $\displaystyle x\in \left[-1,1\right]$

and function $\tan^{-1}(x)$ is defined in $\displaystyle x\in \left(-\infty,+\infty\right)$

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

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

So $\displaystyle f(x) = \frac{\pi}{2}+\tan^{-1}(x)$

Now $\displaystyle f^{'}(x) = \frac{1}{1+x^2}>0\;\forall x\in [-1,1]$

So $f(x)$ is Strictly Increasing function.

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

and $\displaystyle f(+1) = \frac{\pi}{2}+\tan^{-1}(1) = \frac{\pi}{2}+\frac{\pi}{4} = \frac{3\pi}{4}$

So $\displaystyle f(x)\in \left[\frac{\pi}{4}\;,\frac{3\pi}{4}\right]$

Answer 2

This is not correct because you need to restrict $\tan^{-1}$ to the same domain as the other two functions: $[-1,1]$. The extreme vales that $\tan^{-1}(x)$ takes on $[-1,1]$ are $\pm\frac{\pi}{4}$. So the range of your $f$ is $\left[\frac{\pi}{2}-\frac{\pi}{4},\frac{\pi}{2}+\frac{\pi}{4}\right]=\left[\frac{\pi}{4},\frac{3\pi}{4}\right]$ (and its domain is $[-1,1]$).