Find the domain and range of each of the following real valued functions:

Solution:

(i)

Given:

$f(x)=\frac{a x+b}{b x-a}$

Domain of $f$ : Clearly, $f(x)$ is a rational function of $x$ as $\frac{a x+b}{b x-a}$ is a rational expression.

Clearly, $f(x)$ assumes real values for all $x$ except for all those values of $x$ for which $(b x-a)=0$, i.e. $b x=a$.

$\Rightarrow x=\frac{a}{b}$

Hence, domain $(f)=R-\left\{\frac{a}{b}\right\}$

Range of f :

Let f (x) = y

$\Rightarrow \frac{a x+b}{b x-a}=y$

$\Rightarrow(a x+b)=y(b x-a)$

$\Rightarrow(a x+b)=(b x y-a y)$

$\Rightarrow b+a y=b x y-a x$

$\Rightarrow b+a y=x(b y-a)$

$\Rightarrow x=\frac{b+a y}{b y-a}$

Clearly, $f(x)$ assumes real values for all $x$ except for all those values of $x$ for which $(b y-a)=0$, i.e. $b y=a$.

$\Rightarrow y=\frac{a}{b}$

Hence, range $(f)=R-\left\{\frac{a}{b}\right\}$

(ii)

Given:

$f(x)=\frac{a x-b}{c x-d}$

Domain of $f$ : Clearly, $f(x)$ is a rational function of $x$ as $\frac{a x-b}{c x-d}$ is a rational expression.

Clearly, $f(x)$ assumes real values for all $x$ except for all those values of $x$ for which $(c x-d)=0$, i.e. $c x=d$.

$\Rightarrow x=\frac{d}{c}$

Hence, domain $(f)=R-\left\{\frac{d}{c}\right\}$

Range of f :

Let f (x) = y

$\Rightarrow \frac{a x-b}{c x-d}=y$

$\Rightarrow(a x-b)=y(c x-d)$

$\Rightarrow(a x-b)=(c x y-d y)$

$\Rightarrow d y-b=c x y-a x$

$\Rightarrow d y-b=x(c y-a)$

$\Rightarrow x=\frac{d y-b}{c y-a}$

Clearly, $f(x)$ assumes real values for all $x$ except for all those values of $x$ for which $(c y-a)=0$, i.e. $c y=a$.

$\Rightarrow y=\frac{a}{c}$

Hence, range $(f)=R-\left\{\frac{a}{c}\right\}$.

(iii) 

Given:

$f(x)=\sqrt{x-1}$

Domain $(f)$ : Clearly, $f(x)$ assumes real values if $x-1 \geq 0 \Rightarrow x \geq 1 \Rightarrow x \in[1, \infty)$.

Hence, domain $(f)=[1, \infty)$

Range of $f$ : For $x \geq 1$, we have:

$x-1 \geq 0$

$\Rightarrow \sqrt{x-1} \geq 0$

$\Rightarrow f(x) \geq 0$

Thus, f (x) takes all real values greater than zero.

Hence, range (f) = [0, ∞) .

(iv)

Given:

$f(x)=\sqrt{x-3}$

Domain $(f):$ : Clearly, $f(x)$ assumes real values if $x-3 \geq 0 \Rightarrow x \geq 3 \Rightarrow x \in[3, \infty)$.

Hence, domain $(f)=[3, \infty)$

Range of $f:$ For $x \geq 3$, we have:

$x-3 \geq 0$

$\Rightarrow \sqrt{x-3} \geq 0$

$\Rightarrow f(x) \geq 0$

Thus, $f(x)$ takes all real values greater than zero.

Hence, range $(f)=[0, \infty)$.

(v)

Given:

$f(x)=\frac{x-2}{2-x}$

Domain $(f)$ :

Clearly, $f(x)$ is defined for all $x$ satisfying: if $2-x \neq 0 \Rightarrow x \neq 2$.

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

Range of $f$ :

Let $f(x)=y$

$\Rightarrow \frac{x-2}{2-x}=y$

$\Rightarrow x-2=y(2-x)$

$\Rightarrow x-2=-y(x-2)$

$\Rightarrow y=-1$

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

(vi)

The given real function is f (x) = |x – 1|.

It is clear that |x – 1| is defined for all real numbers.

Hence, domain of f = R.

Also, for x ∈ R, (x – 1) assumes all real numbers.

Thus, the range of f is the set of all non-negative real numbers.

Hence, range of f = [0, ∞) .

(vii)

f (x) = – | x |, x ∈ R

We know that

$\therefore f(x)=-|x|= \begin{cases}-x, & x \geq 0 \\ x, & x<0\end{cases}$

It can be observed that the range of f (x) = – | x | is all real numbers except positive real numbers.

∴ The range of f is (– ∞, 0).

(viii) Given:

$f(x)=\sqrt{9-x^{2}}$

$\left(9-x^{2}\right) \geq 0$

$\Rightarrow 9 \geq x^{2}$

$\Rightarrow x \in[-3,3]$

$\sqrt{9-x^{2}}$ is defined for all real numbers that are greater than or equal to $-3$ and less than or equal to $3 .$

Thus, domain of $f(x)$ is $\{x:-3 \leq x \leq 3\}$ or $[-3,3]$.

For any value of x such that – 3 ≤ x ≤ 3, the value of f (x) will lie between 0 and 3.

Hence, the range of f (x) is {x: 0 ≤ x ≤ 3} or [0, 3].

(ix) Given:

$f(x)=\frac{1}{\sqrt{16-x^{2}}}$

$\left(16-x^{2}\right)>0$

$\Rightarrow 16>x^{2}$

$\Rightarrow x \in(-4,4)$

$\frac{1}{\sqrt{16-x^{2}}}$ is defined for all real numbers that are greater than $-4$ and less than $4 .$

Thus, domain of f (x) is {x : – 4 < x < 4} or (– 4, 4).

Range of f :

Let f (x) = y

$\Rightarrow \frac{1}{\sqrt{16-x^{2}}}=y$

$\Rightarrow \frac{1}{16-x^{2}}=y^{2}$

$\Rightarrow \frac{1}{y^{2}}=16-x^{2}$

 

$\Rightarrow x^{2}=16-\frac{1}{y^{2}}$

Since, $-4

$\Rightarrow 0 \leq x^{2}<16$

$\Rightarrow 0 \leq 16-\frac{1}{y^{2}}<16$

$\Rightarrow-16 \leq-\frac{1}{y^{2}}<0$

$\Rightarrow 16 \geq \frac{1}{y^{2}}>0$

$\Rightarrow \frac{1}{16} \leq y^{2}<\infty$

$\Rightarrow \frac{1}{4} \leq y<\infty \quad(\because y \geq 0)$

Hence, range $(f)=\left[\frac{1}{4}, \infty\right)$

(x) Given:

$f(x)=\sqrt{x^{2}-16}$

$\left(x^{2}-16\right) \geq 0$

$\Rightarrow x^{2} \geq 16$

$\Rightarrow x \in(-\infty,-4] \cup[4, \infty)$

$\sqrt{x^{2}-16}$ is defined for all real numbers that are greater than or equal to 4 and less than or equal to $-4$.

Thus, domain of $f(x)$ is $\{x: x \leq-4$ or $x \geq 4\}$ or $(-\infty,-4] \cup[4, \infty)$.

Range of $f$ :

For $x \geq 4$, we have:

$x^{2}-16 \geq 0$

$\Rightarrow \sqrt{x^{2}-16} \geq 0$

$\Rightarrow f(x) \geq 0$

For $x \leq-4$, we have:

$x^{2}-16 \geq 0$

$\Rightarrow \sqrt{x^{2}-16} \geq 0$

$\Rightarrow f(x) \geq 0$

Thus, f (x) takes all real values greater than zero.

Hence, range (f) = [0, ∞).