# How do you find the domain and range of sqrt(sin^-1(2x))?

Jul 31, 2016

Domain is $\left[0 , \frac{1}{2}\right]$ and, Range is $\left[0 , \sqrt{\frac{\pi}{2}}\right]$.

#### Explanation:

Let $f \left(x\right) = \sqrt{{\sin}^{-} 1 \left(2 x\right)}$.

We recall that the Domain of ${\sin}^{-} 1$ function is [-1,1], and, its Range

is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$. So, for our $f$,

$2 x \in \left[0 , 1\right] \Rightarrow - 1 \le 2 x \le 1 \Rightarrow - \frac{1}{2} \le x \le \frac{1}{2}$

But, $- \frac{1}{2} \le x < 0$$\Rightarrow - 1 \le 2 x < 0$, and, since, ${\sin}^{-} 1$ is $\uparrow$,

${\sin}^{-} 1 \left(- 1\right) \le {\sin}^{-} 1 \left(2 x\right) < , {\sin}^{-} 1 0$, i.e.,

$- \frac{\pi}{2} \le {\sin}^{-} 1 2 x < 0$, and, hence $\sqrt{{\sin}^{-} 1 2 x}$ will be undefined.

Therefore, $x$ has to be restricted to $\left[0 , \frac{1}{2}\right]$, and as such,

${\sin}^{-} 1 2 x$ will be in {0,1], so that, $f \left(x\right) = {\sin}^{-} 1 2 x \in \left[0 , \sqrt{\frac{\pi}{2}}\right]$.

Thus, Domain is $\left[0 , \frac{1}{2}\right]$ and, Range is $\left[0 , \sqrt{\frac{\pi}{2}}\right]$.