Find the domain and range of the function f(x) = sqrt(9 − x2).

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For a square root function, the expression under the square root must be non-negative.
So, we must have:

$$9 - x^2 \ge 0$$

Add $x^2$ to both sides:

$$9 \ge x^2$$ $$x^2 \le 9$$

Take the square root of both sides. Remember to consider both positive and negative roots:

$$\sqrt{x^2} \le \sqrt{9}$$ $$|x| \le 3$$

This inequality means that $x$ must be between $-3$ and 3, inclusive.

$$-3 \le x \le 3$$

In interval notation, the domain is $\boxed{[-3, 3]}$.

The range of a function is the set of all possible output values (y-values). Let $y = f(x)$.

$$y = \sqrt{9 - x^2}$$

Since $y$ is the result of a square root, $y$ must be non-negative:

$$y \ge 0$$

Now, consider the expression $9 - x^2$.
The maximum value of $9 - x^2$ occurs when $x^2$ is at its minimum. Since $x^2 \ge 0$, the minimum value of $x^2$ is 0, which happens when $x=0$.
When $x=0$:

$$y = \sqrt{9 - 0^2} = \sqrt{9} = 3$$

So, the maximum value of $y$ is 3.
The minimum value of $9 - x^2$ occurs when $x^2$ is at its maximum. From the domain, the maximum value of $x^2$ is $3^2 = 9$ (when $x=3$ or $x=-3$).
When $x=3$ or $x=-3$:

$$y = \sqrt{9 - 3^2} = \sqrt{9 - 9} = \sqrt{0} = 0$$

So, the minimum value of $y$ is 0.
Combining these, the range of $y$ is from 0 to 3, inclusive.

$$0 \le y \le 3$$

In interval notation, the range is $\boxed{[0, 3]}$.