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Find the domain and range of each of the following real valued functions : f(x) = √(9-x2) |
Answer» f(x) = √(9-x2) 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, 0 ≤ \(\sqrt{9-x^2}\) ≤ 3 ⇒ 0 ≤ f(x) ≤ 3 ∴ f(x) ∈ [0, 3] Thus, Range of f = [0, 3] |
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