Question

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

Answer 1

Domain: `{x|-1<=x<=1}`
Range: `{y|0<=y<=pi/2}`

Explanation:

Domain of the function
We know the inverse sin function is only defined for `-1<=x<=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<=x^2<=1`

The first one, `-1<=x^2`, is true for all real numbers, `x in RR`

And we can solve the second one by taking the square root on both sides:
`x^2<=1`

`sqrt(x^2)<=sqrt(1)`

`x<=1`

Now we combine this with our original interval, which we do by taking the most restrictive of the bounds:
`{x|-1<=x<=1}`

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 `-pi/2` to `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:
`{y|0<=y<=pi/2}`