Why is not $\,f(x)=\arcsin\left(2x\sqrt{1 - x^2}\right)$ a one to one function?

Question

Why is not $\,f(x)=\arcsin\left(2x\sqrt{1 - x^2}\right)$ a one to one function?

I've always knew inverse trig functions to be one to one on a specific range. But from the graph of the above inverse trig function it shows that it’s not one to one. My question is why? Also what are the necessary restrictions for the function to act as one to one?

Answer 1

Note the function $y=2x\sqrt{1-x^2}$ is not one to one, as shown in the figure. And you can rigorously prove this fact by using the intermediate value theorem and continuity (try it). So the function $f(x)=\arcsin(2x\sqrt{1-x^2})$ is not one to one.

Answer 2

As noticed the function $\arcsin x$ is one to one and also $\arcsin (f(x))$ is one to one when $f(x)$ is one to one, indeed

$$\arcsin (f(x_1))=\arcsin (f(x_2)) \implies f(x_1)=f(x_2)\implies x_1=x_2$$

but in this case $2x\sqrt{1 - x^2}$ is not one to one in the domain of interest, indeed for example

$$2x\sqrt{1 - x^2}=0\iff x_{1,2}=0,1$$

which implies that the composition is not one to one since $\exists x_1$ and $\exists x_2$ such that

$$\arcsin (f(x_1))=\arcsin (f(x_2))$$

with $x_1\neq x_2$.

Answer 3

$f(g(x))$ requires $g(x)$ to be injective and $f(x)$ to be injective to be injective.

Answer 4

Although $\arcsin\left(2x\sqrt{1-x^2}\right)$ is one-to-one on $\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right]$, it is not one-to-one on $[-1,1]$ because $x\sqrt{1-x^2}$ is not one-to-one on that interval. However, $\arcsin\left(2x\sqrt{1-x^2}\right)=2\arcsin(x)$ when $x^2\le\frac12$, and $2\arcsin(x)$ is one-to-one on $[-1,1]$.

Furthermore, for $x^2\ge\frac12$ $$ \arcsin\left(2x\sqrt{1-x^2}\right)+2\arcsin(x)=\pi\operatorname{sgn}(x) $$