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Find the points of local maximum and minimun of the function: $$f(x)=\sin^{-1}(2x\sqrt{1-x^2})~~~~;~~x\in (-1,1)$$ I know $$f'(x)=-\frac{2}{\sqrt{1-x^2}}$$

How to find the local maximum and minimum? I have drawn the fig and seen the points of local maximum and minimum. But how to find then analytically? enter image description here

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    @Mattos I have checked. Simplify your answer, you will get my answer. – 2017-01-30
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    I suggest graphing. – 2017-01-30
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    @JackLam I have graphed. But I need the analytical solution. – 2017-01-30
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    See http://math.stackexchange.com/questions/1764431/solving-arcsin-left2x-sqrt1-x2-right-2-arcsin-x?noredirect=1&lq=1 – 2017-01-30
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    @labbhattacharjee Please elaborate how the link above help me to find the local maximum and minimum? – 2017-01-30

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Hint: We know that $$\arcsin x + \arcsin y = \arcsin (x\sqrt{1-y^2} + y\sqrt {1-x^2}) $$ If $x=y $, then $$2\arcsin x = \arcsin (2x\sqrt {1-x^2}) $$

Hope you can take it from here.

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    But how to find the local maximum and minimum? – 2017-01-30
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    I hope you know the global maxima and minima of $\arcsin x $. They are at $1$ and $-1$ respectively. Hope it helps. – 2017-01-30
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    Please elaborate what you want to say to find the local maximum and minimum. – 2017-01-30
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    Since the maximum and minimum of the OP's function are not at $1$ and $-1$, this seems not the best method. Indeed, the identity you start with is generally false. – 2017-01-30
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As suggested in the comment using my answer here in Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$,

$$ \arcsin(2x\sqrt{1-x^2}) =\begin{cases}2\arcsin x \;\;;-\dfrac1{\sqrt2}\le x\le \dfrac1{\sqrt2}\iff-\dfrac\pi4\le\arcsin x\le\dfrac\pi4\\ \pi - 2\arcsin x\;\;; \dfrac1{\sqrt2}< x\le 1\iff\dfrac\pi4<\arcsin x\le\dfrac\pi2\\ -\pi -2\arcsin x\;\;;-1< x \le-\dfrac1{\sqrt2}\iff-\dfrac\pi2\le\arcsin x<-\dfrac\pi4\end{cases} $$

Can you take it from here?

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    I understand the above from here and the link you have provided earlier, but unable to get the local maximum and minimum. – 2017-01-30
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    @user1942348, Can u find them now? – 2017-01-30
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    No, unable to find the local maximum and minimum – 2017-01-30
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    @user1942348, Let's take the toughest case:) For $-1< x \le-\dfrac1{\sqrt2}\iff-\dfrac\pi2\le\arcsin x<-\dfrac\pi4,$ $$\pi-\pi\ge-\pi-2\arcsin x>-\pi+\dfrac\pi2$$ – 2017-01-30
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    implies $0\ge\arcsin(2x\sqrt{1-x^2})>-\dfrac\pi2$ Please conclude – 2017-01-30
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    @user1942348, For $$-1< x \le-\dfrac1{\sqrt2}$$ – 2017-01-30
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    Yes, $0\ge\arcsin(2x\sqrt{1-x^2})>-\dfrac\pi2$ for $-1< x \le-\dfrac1{\sqrt2}$ – 2017-01-30
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52717/discussion-between-lab-bhattacharjee-and-user1942348). – 2017-01-30
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First of all, a bit of theory.

Suppose you have a function $F$ defined over the interval $(a,b)$ and increasing thereon. Suppose also that $G$ takes on values in $(a,b)$. Then, in order to find the points where $F(G(x))$ has a local maximum or minimum, it's sufficient to find the points where $G$ has a local maximum or minimum. This is because $F(G(x_1))

The arcsine is an increasing function, so you just need to find the local maxima and minima of $g(x)=x\sqrt{1-x^2}$. You should already have worked out that $-1\le 2x\sqrt{1-x^2}\le 1$ for $x\in[-1,1]$ when you have discussed the domain of $f$. With notation as before, $G(x)=2x\sqrt{1-x^2}$ and $F(t)=\arcsin t$.

Since $$ G'(x)=\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=\frac{1-2x^2}{\sqrt{1-x^2}} $$ you find that:

  1. $\lim_{x\to-1^+}G'(x)=-\infty=\lim_{x\to1^-}G'(x)$

  2. $G'(x)$ vanishes at $-1/\sqrt{2}$ and at $1/\sqrt{2}$, being positive in the interval $(-1/\sqrt{2},1/\sqrt{2})$.

Now you should be able to end: $G$ has a local minimum at $-1/\sqrt{2}$ and a local maximum at $1/\sqrt{2}$.

If you include $-1$ and $1$ in the domain of $f$, then $-1$ is a local maximum point and $1$ a local minimum point.

As a side note, $f$ is not differentiable at $\pm1/\sqrt{2}$, but this is irrelevant.

Your computation of the derivative is wrong: it should be $$ f'(x)=\frac{1}{\sqrt{1-(2x\sqrt{1-x^2})^2}}\frac{1-2x^2}{\sqrt{1-x^2}}= \frac{1-2x^2}{|1-2x^2|}\frac{1}{\sqrt{1-x^2}} $$ Note that $\sqrt{(1-2x^2)^2}=|1-2x^2|$.

From this you could derive the same conclusions as before. The function is increasing where $1-2x^2>0$ and decreasing where $1-2x^2<0$.

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    the given function in question is not an increasing function. Also please end up the solution. – 2017-01-30
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    @user1942348 The given function isn't, but the arcsine is. – 2017-01-30
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    ok, please end up the solution. – 2017-01-30
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    You have written: "Now you should be able to end: G has a local minimum at −1/√2 and a local maximum at 1/√2". Have you used 2nd derivate method or else? – 2017-01-30
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    @user1942348 $G$ has negative derivative over $(-1,-1/\sqrt{2})$, positive over $(-1/\sqrt{2},1/\sqrt{2})$ and negative over $(1/\sqrt{2},-1)$. – 2017-01-30