1. We are given the function $y = \sin^{-1}(2x)$, which means $y$ is the inverse sine (arcsine) of $2x$. 2. The domain of the arcsine function $\sin^{-1}(z)$ is $-1 \leq z \leq 1$. 3. Since $z = 2x$, we need $-1 \leq 2x \leq 1$. 4. Dividing all parts of the inequality by 2 gives $-\frac{1}{2} \leq x \leq \frac{1}{2}$. 5. This means the domain of $y = \sin^{-1}(2x)$ is $x \in [-\frac{1}{2}, \frac{1}{2}]$. 6. The range of $y = \sin^{-1}(u)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so the range here remains the same but the function input is limited. 7. This function represents the arcsine of the linear function $2x$, scaled horizontally accordingly. Final answer: The domain is $\boxed{-\frac{1}{2} \leq x \leq \frac{1}{2}}$, and the range is $\boxed{-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}}$.