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) $f\left(x\right)=\sqrt{\frac{x-2}{3-x}}$

Answer 1

We have,
$f\left(x\right)=\sqrt{x-2}$
Clearly, $f\left(x\right)$ assumes real values, if
$x-2\ge 0$
$\Rightarrow x\ge 2$
$\Rightarrow xϵ[2,\infty ]$

(ii) We have,
$f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$
Clearly, $f\left(x\right)$ assumes real values, if
$x^2-1>0$
$\Rightarrow (x-1)(x+1)>0$ $[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$
$\Rightarrow x<-1$ or $x>1$
$\Rightarrow xϵ(-\infty ,-1)\cup (1,\infty )$
Hence, domain $\left(f\right)=(-\infty ,-1)\cup (1,\infty )$

(iii) We have,
$f\left(x\right)=\sqrt{9-x^2}$
Clearly, f(x) assumes real values, if
$9-x^2\ge 0$
$\Rightarrow 9\ge x^2$
$\Rightarrow x^2\le 9$
$\Rightarrow -3\le x\le 3$
$\Rightarrow xϵ[-3,3]$
Hence, domain (f) = [-3, 3]

(iv) We have,
$f\left(x\right)=\sqrt{\frac{x-2}{3-x}}$
Clearly, $f\left(x\right)$ assumes real values, if
$9-x^2\ge 0$
$\Rightarrow 9\ge x^2$
$\Rightarrow x^2\le 9$
$\Rightarrow -3\le x\le 3$
$\Rightarrow xϵ[-3,3]$
Hence, domain (f) = [2, 3]