Question

Find the domain of each of the following real valued functions of real variable :
(i) $f\left(x\right)=\frac{1}{x}$
(ii) $f\left(x\right)=\frac{1}{x-7}$
(iii) $f\left(x\right)=\frac{3x-2}{x+1}$
(iv) $f\left(x\right)=\frac{2x+1}{x^2-9}$
(v) $f\left(x\right)=\frac{x^2+2x+1}{x^2-8x+12}$

Answer 1

We have,
$f\left(x\right)=\frac{1}{x}$
Clearly, $f\left(x\right)$ assumes real values for all real values for all $x$ except for the values of $x=0$
Hence, Domain (f) = R - {0}

(ii) We have,
$f\left(x\right)=\frac{1}{x-7}$
Clearly, $f\left(x\right)$ assumes real values for all real values for all $x$ except for the values of $x$ satisfying $x-7=0$ i.e., $x=7$
Hence, Domain (f) = R - {7}

(iii) We have,
$f\left(x\right)=\frac{3x-2}{x+1}$
We observe that $f\left(x\right)$ is a rational function of $x$ as $\frac{3x-2}{x+1}$ is a rational expression.
Clearly, $f\left(x\right)$ assumes real values for all $x$ except for the values of $x$ for which $x+1=0$ i.e., $x=-1$
Hence, Domain R - {-1}

(iv) We have,
$f\left(x\right)=\frac{2x+1}{x^2-9}$
$=\frac{2x+1}{(x^2-3^2)}$
$=\frac{2x+1}{(x-3)(x+3)}$
$[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$
We observe that $f\left(x\right)$ is a rational function of $x$ as $\frac{2x+1}{x^2-9}$ is a rational expression.
Clearly, $f\left(x\right)$ assumes real values for all $x$ except for all those values of $x$ for which $x^2-9=0$ i.e., $x=-3,3$

(v) We have,
$f\left(x\right)=\frac{x^2+2x+1}{x^2-8x+12}$
$=\frac{x^2+2x+1}{x^2-6x-2x+12}$
$=\frac{x^2+2x+1}{x(x-6)-2(x-6)}$
$=\frac{x^2+2x+1}{(x-6)(x-2)}$
Clearly, f(x) is a rational function of x as $\frac{x^2+2x+1}{x^2-8x+12}$ is a rational expression in $x$.
We observe that $f\left(x\right)$ assumes real values for all $x$ except for all those values of $x$ for which $x^2-8x+12=0$ i.e., $x=2,6$
$\therefore$ Domain (f) = R - {2, 6}