Find the domain and range of each of the following real valued functions:
(i) $f(x)=\frac{a x+b}{b x-a}$
(ii) $f(x)=\frac{a x-b}{c x-d}$
(iii) $f(x)=\sqrt{x-1}$
(iv) $f(x)=\sqrt{x-3}$
(v) $f(x)=\frac{x-2}{2-x}$
(vi) $f(x)=|x-1|$
(vii) $f(x)=-|x|$
(viii) $f(x)=\sqrt{9-x^{2}}$
(ix) $f(x)=\frac{1}{\sqrt{16-x^{2}}}$
(x) $f(x)=\sqrt{x^{2}-16}$
(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, ∞).
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