enter image description here I tried differentiating the function given but the problem is that it changes with intervals.


Taking the usual codomain of $\;\arcsin\;$ , we have that $\;\arcsin:[-1,1]\to\left[-\frac\pi2,\,\frac\pi2\right]\;$ . Now

$$x\in[0,\pi]\implies0\le\frac{x+\sin x}2\le\frac{\pi+\sin\pi}2=\frac\pi2\implies0\le\sin\left(\frac{x+\sin x}2\right)\le1$$

and thus

$$f(x):=\arcsin\sin\left(\frac{x+\sin x}2\right)=\frac{x+\sin x}2$$

Try now to take it from here.

  • $\begingroup$ But what do we do in the intervals$ [\pi,2\pi]~and~[2\pi,4\pi]$ as in the interval $ [\pi,2\pi],~ f(x)=\frac{ x+sinx}{2} $ only till $x+sinx=2$ after which it would become $ f(x)=\pi-\frac{ x+sinx}{2} $ for some x. $\endgroup$ – Gaurav Agarwal May 31 '17 at 13:34
  • $\begingroup$ @GauravAgarwal I can't understand that. The function is explicitly defined for $\;x\in[0,\pi]\;$ . What business have $\;x\in[\pi,2\pi],\,[2\pi,4\pi]\;$ I can't say...but perhaps they mean you "extend" the definition of $\;f\;$ to those intervals...That looks like a mistake to me. Perhaps the definition interval was $\;[0,4\pi]\;$ and you have to figure out what to do in separate intervals... $\endgroup$ – DonAntonio May 31 '17 at 13:43
  • 1
    $\begingroup$ Yup, you were right, I got the answer. I instead stick-ed to my initial approach, one thing that I did different this time was differentiating the function without defining a specific interval for x. Then, getting increase and decrease in the function, which had multiple solutions (trigonometric solutions) which easily gave out all the three answers that too without the use of the mess given in the question. Thanks a lot. $\endgroup$ – Gaurav Agarwal May 31 '17 at 14:09

Observe that for $x \in [0, \pi]$ we have that

$$f(x)=\frac{x+ \sin x}{2}.$$


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