Let f(x) = √(9 - x² )

Clearly, f(x) is a not defined when 9 - x² < 0

So, f(x) is defined only when 9 - x² ≥ 0

Now, 9 - x² ≥ 0

=> x² - 9 ≤ 0

=> (x - 3)*(x + 3) ≤ 0

=> -3 ≤ x ≤ 3

=> x ∈ [-3, 3]

So, domain of f(x) = [-3, 3]

Again let y = √(9 - x² )

squaring on both side, we get

      y² = 9 - x²

=> x² = 9 - y²

=> x = ±√(9 - y² )

Clearly, x is defined, when 9 - y² ≥ 0

Now, 9 - y² ≥ 0

=> y² - 9 ≤ 0

=> (y - 3)*(y + 3) ≤ 0

=> -3 ≤ y ≤ 3

But f(x) attains only non-negative values,

So, range of f(x) = [0, 3]