We know the square of a real number is never negative.

Clearly, f(x) takes real values only when 9 – x² ≥ 0

⇒ 9 ≥ x²

⇒ x² ≤ 9

⇒ x² – 9 ≤ 0

⇒ x² – 3² ≤ 0

⇒ (x + 3)(x – 3) ≤ 0

⇒ x ≥ –3 and x ≤ 3

∴ x ∈ [–3, 3]

Thus, domain of f = [–3, 3]

When x ∈ [–3, 3], we have 0 ≤ 9 – x² ≤ 9

Hence,

∴ f(x) ∈ [0, 3]

Thus, range of f = [0, 3]