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

Solution:

(i) Given: $f(x)=\sqrt{x-2}$

Clearly, f (x) assumes real values if x -">−2 ≥ 0.

$\Rightarrow x \geq 2$

$\Rightarrow x \in[2, \infty)$

Hence, domain $(f)=[2, \infty)$.

(ii) Given: $f(x)=\frac{1}{\sqrt{x^{2}-1}}$

Clearly, $f(x)$ is defined for $x^{2}-1>0$.

$(x+1)(x-1)>0 \quad\left[\right.$ Since $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$

$x<-1$ and $x>1$

$x \in(-\infty,-1) \cup(1, \infty)$

Hence, domain $(f)=(-\infty,-1) \cup(1, \infty)$

(iii) Given: $f(x)=\sqrt{9-x^{2}}$

We observe that f (x) is defined for all satisfying

$9-x^{2} \geq 0$

$\Rightarrow x^{2}-9 \leq 0$

$\Rightarrow(x+3)(x-3) \leq 0$

$\Rightarrow-3 \leq x \leq 3$

$x \in[-3,3]$

Hence, domain $(f)=[-3,3]$

(iv) Given: $f(x)=\sqrt{\frac{x-2}{3-x}}$

Clearly, f (x) assumes real values if

$x-2 \geq 0$ and $3-x>0$

$\Rightarrow x \geq 2$ and $3>x$

$\Rightarrow x \in[2,3)$

Hence, domain ( f ) = [2, 3) .