# Maxima and minima of $f(x)=\sin x+\cos x+\tan x+\arcsin x+\arccos x+\arctan x$

Let $f(x)=\sin x+\cos x+\tan x+\arcsin x+\arccos x+\arctan x$ . If $M$ and $m$ are maximum and minimum values of $f(x)$ then their arithmetic mean is equal to?

My Approach:

$\arcsin x + \arccos x =\frac{\pi}{2}$, so the equation becomes $\sin x + \cos x + \tan x + \frac{\pi}{2} + \arctan x$ . After this I differentiated this equation but did not came to any conclusion . Also I tried to solve the equation to make something relevant but the equation did not simplified . Please help...

• Maximum and minimum over what set? All of the functions listed are bounded except for $\tan x$, which is unbounded over $(-\pi/2,\pi/2)$. – Joey Zou Sep 1 '16 at 18:07
• @JoeyZou Obviously, on a subset of $[-1,1]$. – Stop hurting Monica Sep 1 '16 at 18:08
• Since there is a inverse function so domain can be [-1,1]. – saladi Sep 1 '16 at 18:14

Hint: We have $f(x) = \sin x + \cos x + \tan x + \frac{\pi}{2} + \arctan x$ on $[-1,1]$. Now $$f'(x) = \cos x - \sin x + \sec^2 x + \frac{1}{1+x^2}.$$ On $[-1,1]$, we have $\cos x > 0$, $-\sin x > -1$, $\sec^2x\ge 1$, and $\frac{1}{1+x^2}>0$, so for all $x\in[-1,1]$ we have $$f'(x) > 0 - 1 + 1 + 0 = 0.$$ Thus $f$ is strictly increasing. What can you say about where it attains its maximum and minimum?