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

Question:

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

(i) $f(x)=\frac{1}{x}$

(ii) $f(x)=\frac{1}{x-7}$

(iii) $f(x)=\frac{3 x-2}{x+1}$

(iv) $f(x)=\frac{2 x+1}{x^{2}-9}$

(v) $f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$

Solution:

(i) Given: $f(x)=\frac{1}{x}$

Domain of f :

We observe that (x) is defined for all x except at x = 0.

At $x=0, f(x)$ takes the intermediate form $\frac{1}{0}$.

Hence, domain ( f ) = R -">{ 0 }

(ii) Given: $f(x)=\frac{1}{(x-7)}$

Domain of f :

Clearly,  (x) is not defined for all (x -"> 7)  = 0 i.e. x = 7.

At $x=7, f(x)$ takes the intermediate form $\frac{1}{0}$.

Hence, domain ( f ) = R -"> { 7 }.

(iii) Given: $f(x)=\frac{3 x-2}{(x+1)}$

Domain of f :

Clearly, $f(x)$ is not defined for all $(x+1)=0$, i.e. $x=-1$.

At $x=-1, f(x)$ takes the intermediate form $\frac{1}{0}$.

Hence, domain $(f)=R-\{-1\}$.

(iv) Given: $f(x)=\frac{2 x+1}{x^{2}-9}$

Domain of $f$ :

Clearly, $f(x)$ is defined for all $x \in R$ except for $x^{2}-9 \neq 0$, i.e. $x=\pm 3$.

At $x=-3,3, f(x)$ takes the intermediate form $\frac{1}{0}$.

Hence, domain $(f)=R-\{-3,3\}$.

(v) Given: $f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$

$=\frac{x^{2}+2 x+1}{x^{2}-6 x-2 x+12}$

$=\frac{x^{2}+2 x+1}{x(x-6)-2(x-6)}$

$=\frac{x^{2}+2 x+1}{(x-6)(x-2)}$

Domain of $f$ : Clearly, $f(x)$ is a rational function of $x$ as $\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$ is a rational expression.

Clearly, $f(x)$ assumes real values for all $x$ except for all those values of $x$ for which $x^{2}-8 x+12=0$, i.e. $x=2,6$.

Hence, domain $(f)=R-\{2,6\}$.