Question

What is the domain and range of `f(x) = sqrt (9 - x^2)`?

Answer 1

Domain: `[-3,3]`
Range: `[0,3]`

Explanation:

The value under a square root cannot be negative, or else the solution is imaginary.

So, we need `9-x^2\geq0`, or `9\geqx^2`, so `x\leq3` and `x\geq-3`, or `[-3.3]`.

As `x` takes on these values, we see that the smallest value of the range is `0`, or when `x=pm3` (so `sqrt(9-9)=sqrt(0)=0`), and a max when `x=0`, where `y=\sqrt(9-0)=sqrt(9)=3`