How to determine the biggest interval where f(x) is invertible and f(x)^-1 passes through P

Question

I need some help here.

$$f(x) = \sin{(3x)} - 1$$ $$P = (-1,0)$$

Answer 1

In order to make $\;\sin x\;$ we restrict its definition domain to $\;\left[-\frac\pi2,\,\frac\pi2\right]\;$ (though this is not the only interval that can be taken. It is just the most usual), and thus in this case we can takt

$$-\frac\pi2\le3x\le\frac\pi2\implies-\frac\pi6\le x\le\frac\pi6$$

and here $\;y=\sin3x-1\;$ is injective and onto $\;[-1,1]\;$,and its inverse is

$$\sin 3x=y+1\implies x=\frac13\arcsin(y+1)\implies f^{-1}=\frac13\arcsin(x+1)$$

Now just check that we actually have $\;P=\left(-1,\,f^{-1}(-1)\right)\;\ldots$ and justify all the above.