Question 1: Find the domain of each of the following real valued functions of real variable:

i) $f(x) =$ $\frac{1}{x}$     ii) $f(x) =$ $\frac{1}{x-7}$     iii) $f(x) =$ $\frac{3x-2}{x-1}$    iv) $f(x) =$ $\frac{2x+1}{x^2-9}$                            v) $f(x) =$ $\frac{x^2 + 2x +1}{x^2 - 8x + 12}$

i) Given $f(x) =$ $\frac{1}{x}$

$\therefore f(x)$ assumes real values for all $x$ except for $x = 0$

Hence Domain $(f) = R - \{ 0 \}$

ii) Given $f(x) =$ $\frac{1}{x-7}$

$\therefore f(x)$ assumes real values for all $x$ except for $x = 7$

Hence Domain $(f) = R - \{ 7 \}$

iii) Given $f(x) =$ $\frac{3x-2}{x-1}$

$\therefore f(x)$ assumes real values for all $x$ except for $x = 1$

Hence Domain $(f) = R - \{ 1 \}$

iv) Given $f(x) =$ $\frac{2x+1}{x^2-9}$ $=$ $\frac{2x+1}{(x-3)(x+3)}$

$\therefore f(x)$ assumes real values for all $x$ except for $x = -3, 3$

Hence Domain $(f) = R - \{ -3, 3 \}$

v) Given $f(x) =$ $\frac{x^2 + 2x +1}{x^2 - 8x + 12}$ $=$ $\frac{x^2 + 2x +1}{(x-6)(x-2)}$

$\therefore f(x)$ assumes real values for all $x$ except for $x = 6, 2$

Hence Domain $(f) = R - \{ 6, 2 \}$

$\\$

Question 2: Find the domain of each of the following real valued functions of real variable:

i) $f(x) = \sqrt{x-2}$     ii) $f(x) =$ $\frac{1}{ \sqrt{x^2 - 1} }$     iii) $f(x) = \sqrt{9-x^2}$     iv) $f(x) =$ $\sqrt{ \frac{x-2}{3-x} }$

i) Given $f(x) = \sqrt{x-2}$

$\therefore f(x)$ assumes real values for all $x \geq 2$

$\therefore x \in [ 2, \infty )$

Hence Domain $(f) = [ 2, \infty )$

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

$f(x)$ assumes real values for $\sqrt{x^2-1} > 0$

$\Rightarrow (x-1)(x+1) > 0$

$\Rightarrow x > 1 \ or \ x < -1$

$\Rightarrow x \in (- \infty , -1) \cup (1, \infty)$

Hence the Domain $(f) = (- \infty , -1) \cup (1, \infty)$

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

$f(x)$ assumes real value for $9-x^2 \geq 0$

$\Rightarrow (3-x)(3+x) \geq 0$

$\Rightarrow x \geq -3 \ and \ x \leq 3$

$\therefore -3 \leq x \leq 3$

Hence Domain $(f) = [ -3 , 3 ]$

iv) Given $f(x) =$ $\sqrt{ \frac{x-2}{3-x} }$

$f(x)$ assumes real value if

$x-2 \geq 2 \ and \ 3-x > 0$

$\Rightarrow x \geq 2 \ and \ x < 3$

$\therefore x \in [2, 3)$

Hence the Domain $(f) = [2, 3)$

$\\$

Question 3: Find the domain of each of the following real valued functions of real variable:

i) $f(x) =$ $\frac{ax+b}{bx-a}$          ii) $f(x) =$ $\frac{ax-b}{cx-d}$         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 $f(x) =$ $\frac{ax+b}{bx-a}$

$f(x)$ will assume real values for all $x$ except for the value of $x$ for

$bx - a = 0 \Rightarrow x =$ $\frac{a}{b}$

Therefore Domain $(f) = R - \{$ $\frac{a}{b}$ $\}$

Let $f(x) = y$

$y =$ $\frac{ax+b}{bx-a}$

$\Rightarrow ax + b = bxy - ay$

$\Rightarrow b + ay = x(by - a)$

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

Clearly $x$ will take real value for all $x \in R$ except for

$by-a = 0$ $\Rightarrow y =$ $\frac{a}{b}$

$\therefore$ Range $(f) = R - \{$ $\frac{a}{b}$ $\}$

ii)   Given $f(x) =$ $\frac{ax-b}{cx-d}$

$f(x)$ will assume real values for all $x$ except for the value of $x$ for

$cx - d = 0 \Rightarrow x =$ $\frac{d}{c}$

Therefore Domain $(f) = R - \{$ $\frac{d}{c}$ $\}$

Let $f(x) = y$

$y =$ $\frac{ax-b}{cx-d}$

$\Rightarrow ax - b = cxy - dy$

$\Rightarrow dy-b = cxy-ax$

$\Rightarrow x =$ $\frac{dy-b}{cy-a}$

Clearly $x$ will take real value for all $y \in R$ except for

$cy-a = 0$ $\Rightarrow y =$ $\frac{a}{c}$

$\therefore$ Range $(f) = R - \{$ $\frac{a}{c}$ $\}$

iii)  Given $f(x) = \sqrt{x-1}$

$\therefore f(x)$ assumes real value if $x- 1 \geq 0 \Rightarrow x \geq 1 \Rightarrow x \in [1, \infty)$

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

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 equal to $0$

Hence Range $(f) = [0, \infty)$

iv)   Given $f(x) = \sqrt{x-3}$

$\therefore f(x)$ assumes real value if $x- 3 \geq 0 \Rightarrow x \geq 3 \Rightarrow x \in [3, \infty)$

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

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 equal to $0$

Hence Range $(f) = [0, \infty)$

v)   Given $f(x) =$ $\frac{x-2}{2-x}$

$\therefore f(x)$ is defined for all $x \in R$ except for which $2-x = 0 \Rightarrow x = 2$

Hence Domain $(f) = R - \{ 2 \}$

Left $f(x) = y$

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

$\Rightarrow x-2 = 2y - xy$

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

Hence Range $(f) = \{ -1 \}$

vi)  Given $f(x) = | x- 1|$

$\therefore f(x)$  is defined for all $x \in R$

Hence Domain $(f) = R$

Let $f(x) = y$

$\Rightarrow |x-1| = y$

$\Rightarrow f(x) \geq 0$ for all $x \in R$

Hence Range $(f) = [0, \infty)$

vii)  Given $f(x) = -| x|$

$\therefore f(x)$  is defined for all $x \in R$

Hence Domain $(f) = R$

Let $f(x) = y$

$\Rightarrow -|x| = y$

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

Hence Range $(f) = (- \infty, 0)$

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

$\therefore f(x)$ will assume real values for all $x$ except when

$9-x^2 < 0 \Rightarrow (3-x)(3+x) < 0$

$\Rightarrow x > 3$ and $x <-3$

Therefore $f(x)$ will be real for $-3 \leq x \leq 3$

Hence Domain $(f) = \{ -3, 3 \}$

Let $f(x) = y$

$\Rightarrow \sqrt{9-x^2} = y$

$y$ will assume real values for all

$9-x^2 \geq 0 \Rightarrow (3-x)(3+x) \geq 0 \Rightarrow -3 \leq x \leq 3$

For any value such that $-3 \leq x \leq 3$, the value of $y$ will lie between $0$ and $3$

Hence Range $(f) = [0, 3]$

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

$\therefore f(x)$ will assume real values when

$16-x^2 > 0 \Rightarrow (4-x)(4+x) > 0 \Rightarrow -4 \leq x \leq 4$

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

Hence Domain $(f) = (-4, 4)$

Let $f(x) = y$

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

Since $-4 \leq x \leq 4$ mean $\frac{1}{4}$ $< y < \infty$

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

x)   Given $f(x) = \sqrt{x^2 - 16}$

$\therefore f(x)$ will assume real values for all $x \geq 4$ or $x \leq -4$

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

Hence Domain $(f) = (-\infty, -4] \cup [4, \infty)$

For $x \geq 4, f(x) \geq 0$

For $x \leq -4 , f(x) \geq 0$

$\therefore f(x)$ will take real values greater that equal to $0$.

Hence Range $(f) = [0, \infty)$

$\\$