All Tasks - Asymptote

Consider $f(x)= 2 \arcsin\left(\dfrac{x}{3}\right)$ with domain $D_f=[-3, 3]$. The zeros of $f(x)$ are:

Consider $f(x)= \dfrac{\pi}{2} -2 \arcsin(1-2x)$ with domain $D_f=[0, 1]$. The solution of the equation $f(x)=f(0) + 2 \arcsin\left( \dfrac{\sqrt{3}}{2}\right)$ is:

Consider $f(x)= 2 \arcsin\left(\dfrac{x}{3}\right)$ with domain $D_f=[-3, 3]$ and range $D'_f = \left[ -\pi, \pi \right] $. The analytic expression of the inverse function of $f$, $f^{-1}(x)$, and $f^{-1}(\pi)$ are, respectively,

Consider $f(x)= \dfrac{\pi}{2} -2 \arcsin(1-2x)$ with domain $D_f=[0,1]$ and range $D'_f = \left[ -\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] $. The analytic expression of the inverse function of $f$, $f^{-1}(x)$, its domain ($D_{f^{-1}}$) and range ($D'_{f^{-1}}$) are, respectively,

The domain ($D_f$) and the range ($D'_f$) of function $f(x)= 2 \left|-\pi + \arcsin(1-2x)\right| $ are, respectively,

Applying the chain rule, calculate $\frac{dy}{dx}(1)$ where $y(v)=arccot\left(v^2+v\right)$, and $v(u)=\ln(u^2-3)$, and $u(x)=\dfrac{x+1}{x}$