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) $f\left(x\right)=\left|x-1\right|$

(vii) $f\left(x\right)=-\left|x\right|$

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

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

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

Answer 1

(i)
Given:
$f\left(x\right)=\frac{ax+b}{bx-a}$
Domain of f : Clearly, 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 all those values of x for which ( bx $-$ a) = 0, i.e. bx = a.
$\Rightarrow x=\frac{a}{b}$
Hence, domain ( f ) = $R-\left\{\frac{a}{b}\right\}$
Range of f :
Let f (x) = y
$\Rightarrow \frac{ax+b}{bx-a}=y$
⇒ (ax + b) = y (bx $-$ a)
⇒ (ax + b) = (bxy $-$ ay)
⇒ b + ay = bxy $-$ ax
⇒ b + ay = x(by $-$ a)
$\Rightarrow x=\frac{b+ay}{by-a}$
Clearly, f (x) assumes real values for all x except for all those values of x for which ( by $-$ a) = 0, i.e. by = a.
$\Rightarrow y=\frac{a}{b}$.
Hence, range ( f ) = $R-\left\{\frac{a}{b}\right\}$
(ii)
Given:
$f\left(x\right)=\frac{ax-b}{cx-d}$
Domain of f : Clearly, 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.
$\Rightarrow x=\frac{d}{c}$.
Hence, domain ( f ) = $R-\left\{\frac{d}{c}\right\}$
Range of f :
Let f (x) = y
$\Rightarrow \frac{ax-b}{cx-d}=y$
⇒ (ax $-$ b) = y( cx $-$ d)
⇒ (ax $-$ b) = (cxy $-$ dy)
⇒ dy $-$ b = cxy $-$ ax
⇒ dy $-$ b = x(cy $-$ a)
$\Rightarrow x=\frac{dy-b}{cy-a}$
Clearly, f (x) assumes real values for all x except for all those values of x for which ( cy $-$ a) = 0, i.e. cy = a.
$\Rightarrow y=\frac{a}{c}$.
Hence, range ( f ) = $R-\left\{\frac{a}{c}\right\}$ .

(iii)
Given: $f\left(x\right)=\sqrt{x-1}$
Domain ( f ) : Clearly, f (x) assumes real values if x $-$ 1 ≥ 0 ⇒ x ≥ 1 ⇒ x ∈ [1, ∞) .
Hence, domain (f) = [1, ∞)
Range of f : For x ≥ 1, we have:
x $-$ 1 ≥ 0
$\Rightarrow \sqrt{x-1}\ge 0$
⇒ f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .

(iv)
Given: $f\left(x\right)=\sqrt{x-3}$
Domain ( f ) : Clearly, f (x) assumes real values if x $-$ 3 ≥ 0 ⇒ x ≥ 3 ⇒ x ∈ [3, ∞) .
Hence, domain ( f ) = [3, ∞)
Range of f : For x ≥ 3, we have:
x $-$ 3 ≥ 0
$\Rightarrow \sqrt{x-3}\ge 0$
⇒ f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .

(v)
Given:
$f\left(x\right)=\frac{x-2}{2-x}$
Domain ( f ) :
Clearly, f (x) is defined for all x satisfying: if 2 $-$ x ≠ 0 ⇒ x ≠ 2.
Hence, domain ( f ) = R $-$ {2}.
Range of f :
Let f (x) = y
⇒ $\frac{x-2}{2-x}=y$
⇒ x $-$ 2 = y (2 $-$ x)
⇒ x $-$ 2 = $-$ y (x $-$ 2)
⇒ 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
$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x& x<0\end{array}\right.$
$\therefore f\left(x\right)=-\left|x\right|=\left\{\begin{array}{ll}-x,& x\ge 0\\ x,& x<0\end{array}\right.$
Since f(x) is defined for x ∈ R, domain of f = R.
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:
$f\left(x\right)=\sqrt{9-x^2}$
$$\begin{gathered} (9-x^2)\ge 0 \\ \Rightarrow 9\ge x^2 \\ \Rightarrow x\in \left[-3,3\right] \end{gathered}$$

$\sqrt{9-x^2}$ 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 ≤ x ≤ 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:
$f\left(x\right)=\frac{1}{\sqrt{16-x^2}}$

$$\begin{gathered} (16-x^2)>0 \\ \Rightarrow 16>x^2 \\ \Rightarrow x\in \left(-4,4\right) \end{gathered}$$

$\frac{1}{\sqrt{16-x^2}}$ 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
$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{16-x^2}}=y \\ \Rightarrow \frac{1}{16-x^2}=y^2 \\ \Rightarrow \frac{1}{y^2}=16-x^2 \\ \Rightarrow x^2=16-\frac{1}{y^2} \\ \mathrm{Since},-4<x<4 \\ \Rightarrow 0\le x^2<16 \\ \Rightarrow 0\le 16-\frac{1}{y^2}<16 \\ \Rightarrow -16\le -\frac{1}{y^2}<0 \\ \Rightarrow 16\ge \frac{1}{y^2}>0 \\ \Rightarrow \frac{1}{16}\le y^2<\infty \\ \Rightarrow \frac{1}{4}\le y<\infty \left(\because y\ge 0\right) \end{gathered}$$

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


(x) Given:
$f\left(x\right)=\sqrt{x^2-16}$

$$\begin{gathered} (x^2-16)\ge 0 \\ \Rightarrow x^2\ge 16 \\ \Rightarrow x\in (-\infty ,-4]\cup [4,\infty ) \end{gathered}$$

$\sqrt{x^2-16}$ 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 ≤ – 4 or x ≥ 4} or (–∞, –4] ∪ [4, ∞).

Range of f :
For x ≥ 4, we have:
x² $-$ 16 ≥ 0
$\Rightarrow \sqrt{x^2-16}\ge 0$
⇒ f (x) ≥ 0

For x ≤ – 4, we have:
x² $-$ 16 ≥ 0
$\Rightarrow \sqrt{x^2-16}\ge 0$
⇒ f (x) ≥ 0

Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞).