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

(i) Given: $f\left(x\right)=\frac{1}{x}$
Domain of f :
We observe that f (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: $f\left(x\right)=\frac{1}{\left(x-7\right)}$
Domain of f :
Clearly, f (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: $f\left(x\right)=\frac{3x-2}{\left(x+1\right)}$
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: $f\left(x\right)=\frac{2x+1}{x^2-9}$
Domain of f :
Clearly, f (x) is defined for all x ∈ R except for x² $-$ 9 ≠ 0, i.e. x = ± 3.
At x = $-$3, 3, f (x) takes the intermediate form $\frac{1}{0}.$
Hence, domain ( f ) = R $-$ { $-$ 3, 3 }.

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

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

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

$=\frac{x^2+2x+1}{\left(x-6\right)\left(x-2\right)}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{x^2+2x+1}{x^2-8x+12}$ is a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which x² $-$ 8x + 12 = 0, i.e. x = 2, 6.
Hence, domain ( f ) = R $-$ {2,6}.