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}

Answer:

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 \}

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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} }

Answer:

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)

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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}

Answer: 

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)

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