Find the points of local maximum and minimun of the function: $\sin^{-1}(2x\sqrt{1-x^2}), ~x\in (-1,1)$ 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?


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?
