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

Dec 16, 2017

Domain: $\left\{x | - 1 \le x \le 1\right\}$
Range: $\left\{y | 0 \le y \le \frac{\pi}{2}\right\}$

#### Explanation:

Domain of the function
We know the inverse sin function is only defined for $- 1 \le x \le 1$ (since $\sin$ only takes on those values), so we know that our function is only defined when ${x}^{2}$ obeys that same interval. We figure out when this is true by solving the following inequality:
$- 1 \le {x}^{2} \le 1$

The first one, $- 1 \le {x}^{2}$, is true for all real numbers, $x \in \mathbb{R}$

And we can solve the second one by taking the square root on both sides:
${x}^{2} \le 1$

$\sqrt{{x}^{2}} \le \sqrt{1}$

$x \le 1$

Now we combine this with our original interval, which we do by taking the most restrictive of the bounds:
$\left\{x | - 1 \le x \le 1\right\}$

And this is when our function is defined, so it must be the domain.

Range of the function
The inverse sine function usually ranges from $- \frac{\pi}{2}$ to $\frac{\pi}{2}$, but since negative values are only produced by negative inputs to the function, the ${x}^{2}$ makes it so that we will only ever get positive (or $0$) values. This means our range will be:
$\left\{y | 0 \le y \le \frac{\pi}{2}\right\}$