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If $\displaystyle f(x)=\frac{1}{\pi}\left(\arcsin x+\arccos x+\arctan x\right)+\frac{x+1}{x^2+2x+10}\;,$ Then $\max$ value of $f(x)$

$\bf{My\; Try::}$ Here Domain of $\arcsin x\;,\arccos x$ is $\displaystyle x\in \left[-1,1\right]$ and for $\arctan x$ is $(-\infty,\infty)$

So overall domain of function $f(x)$ is $\displaystyle x\in \left[-1,1\right]$

Now Using $\displaystyle \arcsin x+\arccos x= \frac{\pi}{2}\;,\forall x \in \left[-1,1\right]$

So we get $\displaystyle f(x) =\frac1\pi\left(\frac{\pi}{2}+\arctan x\right)+\frac{1}{(x+1)+\frac{9}{(x+1)}}$

How should I proceed?

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  • $\begingroup$ Have you tried taking the derivative and setting this equal to zero to find the extremal points in the interior, and then comparing these to the values at the boudary? I mean... that is how I would proceed in any extremal value problem $\endgroup$ – b00n heT May 31 '16 at 11:28
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    $\begingroup$ Domain of $\arccos$ is $[-\pi/2,\pi/2]$? Do you mean image? $\endgroup$ – Emre May 31 '16 at 11:41
  • $\begingroup$ As user E.Girgin suggests, the domain of $\arcsin$ and $\arccos$ is $\left[-1,1\right]$. Since the roots of $x^2+2x+10$ are not real, you need not add existence conditions for $\frac{x+1}{x^2+2x+10}+\arctan x$. If I were you, I would not divide numerator and denominator by $x+1$, because it unnecessarily alters the domain and it may hinder your possibility of evaluating the derivative of $f$. $\endgroup$ – user228113 May 31 '16 at 12:01
  • $\begingroup$ Actually it was all by mistake, $\endgroup$ – juantheron May 31 '16 at 13:16
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The domain should be $x\in [-1,1]$.

We have $$f(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\arctan x\right)+\frac{x+1}{x^2+2x+10}$$ so $$f'(x)=\frac{1}{\pi(1+x^2)}+\frac{9-(x+1)^2}{(x^2+2x+10)^2}$$ This is positive because of $(x+1)^2\le 4$.

Since $f(x)$ is increasing, the answer is $f(1)=\color{red}{47/52}$.

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For $x\in [-1,1]$, $$f'(x)={8-2x-x^2\over(10+2 x+x^2)^2}+{1\over\pi(1+ x^2)}>0$$ So, $f$ is increasing in $[-1,1]$. Thus, the maxima is taken at $x=1$. Plugging that in, the maxima is $$f(1)=\frac1\pi(\frac\pi2+\frac\pi4)+\frac2{13}=\frac34+\frac2{13}=\frac{47}{52}$$

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