Question

Find the domain and range of each of the following real valued functions:
(i) $f\left(x\right)=\frac{ax+b}{bx-a}$
(ii) $f\left(x\right)=\frac{ax-b}{cx-d}$
(iii) $f\left(x\right)=\sqrt{x-1}$
(iv) $f\left(x\right)=\sqrt{x-3}$
(v) $f\left(x\right)=\frac{x-2}{2-x}$
(vi) $f\left(x\right)=|x-1|$
(vii) $f\left(x\right)=|x|$
(viii) $f\left(x\right)=\sqrt{9-x^2}$

Answer 1

We have,
$f\left(x\right)=\frac{ax+b}{bx-a}$
We observe that f(x) is a rational function of $x$ as $\frac{ax+b}{bx-a}$ is a rational expression.
Clearly, f(x) assumes real values for all x except for the value of x for which bx - a = 0 i.e., bx = a.
$\Rightarrow x=\frac{a}{b}$
$\therefore$ Domain $\left(f\right)=R-\left\{\frac{a}{b}\right\}$
Range of f : Let f(x) = y
$\Rightarrow \frac{ax+b}{bx-a}=y$

$\Rightarrow ax+b=y(bx-a)$
$\Rightarrow ax+b=bxy-ax$
$\Rightarrow b+ay=bxy-ax$
$\Rightarrow b+ay=x(by-a)$
$\Rightarrow \frac{b+ay}{b-ay}=x$
$\Rightarrow x=\frac{b+ay}{by-a}$
Clearly, x will take real value for all $xϵR$ except for
by - a = 0
$\Rightarrow by=a$
$\Rightarrow y=\frac{a}{b}$
$\therefore \text{Range}\left(f\right)=R-\left\{\frac{a}{b}\right\}$

(ii) We have,
$f\left(x\right)=\frac{ax-b}{cx-d}$
We observe that f(x) is a rational function of x as $\frac{ax-b}{cx-d}$ is a rational expression.
Clearly, f(x) assumes real values for all x except for all those values of x for which cx - d = 0 i.e., cx = d.
$x=\frac{d}{c}$
$\Rightarrow \text{Domain}\left(f\right)=R-\left\{\frac{d}{c}\right\}$
Range : Let f(x) = y
$\Rightarrow \frac{ax-b}{cx-d}=y$
$\Rightarrow ax-b=y(cx-d)$
$\Rightarrow ax-b=cxy-dy$
$\Rightarrow dy-b=cxy-9x$
$\Rightarrow dy-b=x(cy-a)$
$\Rightarrow \frac{dy-b}{cy-a}=x$
Clearly, x assumes real values for all y except
$cy-a=0$ i.e., $y=\frac{a}{c}$
Hence, range (f) $=R-\left\{\frac{a}{c}\right\}$

(iii) We have,
$f\left(x\right)=\sqrt{x-1}$
Clearly, f(x) assumes real values, if
$x-1\ge 0$
$\Rightarrow x\ge 1$
$\Rightarrow xϵ[1,\infty )$
Hence, domain (f) $=[1,\infty ]$
Range: For $x\ge 1$, we have,
$x-1\ge 0$
$\Rightarrow \sqrt{x-1}\ge 0$
$\Rightarrow f\left(x\right)\ge 0$
Thus, f(x) takes all real values greater than zero.
Hence, range (f) $=[0,\infty )$

(v) We have,
$f\left(x\right)=\frac{x-2}{2-x}$
Domain of f : Clearly, f(x) is defined for all $xϵR$ except for which $2-x\ne 0$ i.e, $x\ne 2$.
Hence, domain (f) = R - {2}
Range of f : Let f(x) = y
$\Rightarrow \frac{x-2}{2-x}=y$
$\Rightarrow \frac{-1(2-x)}{2-x}=y$
$\Rightarrow -1=y$
$\Rightarrow y=-1$
$\therefore$ Range (f) = (-1)

(vi) We have,
f(x) = |x- 1|
Clearly, f(x) is defined for all $xϵR$
$\Rightarrow$ Domain (f) = R
Range : Let f(x) = y
$\Rightarrow [x-1]=y$
$\Rightarrow f\left(x\right)\ge 0\forall xϵR$
It follows from the above relation that y takes all real values greater or equal to zero.
$\therefore$ Range (f) $=(0,\infty )$

(vii) As |x| is defined for all real numbers, its domain is R and range is only negative numbers because, |x| always positive real number for all real numbers and -|x| is always negative real numbers.

(viii) In order to have f(x) has defined value, term inside square root should always be greater than or equal to zero which gives domain as $-3\le x\le 3$
Whereas Range of above function is limited to [0, 3]