Question

Find the domain of each of the following real valued functions of real variable:

(i) $f\left(x\right)=\sqrt{x-2}$
(ii) $f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$
(iii) $f\left(x\right)=\sqrt{9-x^2}$
(iv) $f\left(x\right)=\frac{\sqrt{x-2}}{3-x}$

Answer 1

(i) Given: $f\left(x\right)=\sqrt{x-2}$
Clearly, f (x) assumes real values if x $-$ 2 ≥ 0.
⇒ x ≥ 2
⇒ x ∈ [2, ∞)
Hence, domain (f) = [2, ∞) .

(ii) Given: $f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$
Clearly, f (x) is defined for x² $-$ 1 > 0 .
(x + 1)(x $-$ 1) > 0 [ Since a² $-$ b² = ( a + b)(a - b)]
x < $-$1 and x > 1
x ∈ ($-$∞ , $-$ 1) ∪ (1, ∞)
Hence, domain (f) = ($-$ ∞ , $-$ 1) ∪ (1, ∞)

(iii) Given: $f\left(x\right)=\sqrt{9-x^2}$
We observe that f (x) is defined for all satisfying
9 $-$ x² ≥ 0 .
⇒ x² $-$ 9 ≤ 0
⇒ (x + 3)(x $-$ 3) ≤ 0
⇒ $-$3 ≤ x ≤ 3
x ∈ [ $-$ 3, 3]
Hence, domain ( f ) = [ $-$3, 3]

(iv) Given: $f\left(x\right)=\sqrt{\frac{x-2}{3-x}}$
Clearly, f (x) assumes real values if
x $-$ 2 ≥ 0 and 3 $-$ x > 0
⇒ x ≥ 2 and 3 > x
⇒ x ∈ [2, 3)
Hence, domain ( f ) = [2, 3) .