Question

Find the domains of the following real valued functions.
i)
f(x) = sq√(x+2)(x-3)​

Answer 1

(i) As we know that the square of a real number is never negative. Now, f(x) takes real values only when x – 2 ≥ 0 x ≥ 2 ∴ x ∈ [2, ∞) ∴ The domain (f) = [2, ∞) (ii) As we know that the square of a real number is never negative. Now, f(x) takes real values only when x2 – 1 ≥ 0 x2 – 12 ≥ 0 (x + 1) (x – 1) ≥ 0 x ≤ –1 or x ≥ 1 ∴ x ∈ (–∞, –1] ∪ [1, ∞) Here, in addition f(x) is also undefined when x2 – 1 = 0 because denominator will be zero and the result will be indeterminate. x2 – 1 = 0 ⇒ x = ± 1 Therefore, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1} x ∈ (–∞, –1) ∪ (1, ∞) ∴ The domain (f) = (–∞, –1) ∪ (1, ∞) (iii) As we know that the square of a real number is never negative. Now, 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] ∴ The domain (f) = [–3, 3] (iv) As we know the square root of a real number is never negative. Now, f(x) takes real values only when x – 2 and 3 – x are both positive and negative. (a) Here, both x – 2 and 3 – x are positive x – 2 ≥ 0 x ≥ 2 3 – x ≥ 0 x ≤ 3 Thus, x ≥ 2 and x ≤ 3 ∴ x ∈ [2, 3] (b) Here, both x – 2 and 3 – x are negative x – 2 ≤ 0 x ≤ 2 3 – x ≤ 0 x ≥ 3 Thus, x ≤ 2 and x ≥ 3 However, the intersection of these sets is null set. Hence, this case is not possible. Thus, x ∈ [2, 3] – {3} x ∈ [2, 3] ∴ The domain (f) = [2, 3]

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