Solved examples on Domain-I

Find the domain of the following :

Students are requested to read all previous posts before reading this.

Here is the quick look

Domain: The Possible point of for which we can get finite value of

To know more about domain and its properties (click here)

1. If $y = \frac{{f(x)}}{{g(x)}}$ then, $g(x) \ne 0$ , i.e. ${\rm{Denominator}} \ne {\rm{0}}$

2. If $y = \sqrt {f(x)}$ then, $f(x) \ge 0$

3. If $y = {\log _{f(x)}}g(x)$ then $f(x) > 0,f(x) \ne 1$ and

4. If $y = {[f(x)]^{g(x)}}$ then $f(x) > 0,f(x) \ne 1$ and g(x) must be defined.

5. If $y = {\sin ^{ - 1}}f(x)$ or $y = {\cos ^{ - 1}}f(x)$ then $- 1 \le f(x) \le 1$

if a function f(x)

is defined in some interval or points( by definition domain), and another function g(x)

is defined in some other interval and points (by definition domain) .... then sum , subtraction , multiplication and division of the functions are defined on the interval or points which are common to both.

1) If domain of y = f(x)

and

are ${D_1}$ and ${D_2}$ respectively then domain of

a) $y = f(x) \pm g(x)$ , $Domain = {D_1}\bigcap {{D_2}}$

b) y = f(x).g(x)

, $Domain = {D_1}\bigcap {{D_2}}$

c) $y = \frac{{f(x)}}{{g(x)}}$ , Since here, $g(x) \ne 0$ , therefore $Domain = {D_1}\bigcap {{D_2}} - \{ g(x) = 0\}$ , i.e. we have to eliminate those points from ${D_1}\bigcap {{D_2}}$ for which

, because $denominator \ne 0$

1. $f(x) = \sqrt {{{\log }_{1/2}}\left( {\frac{{5x - {x^2}}}{4}} \right)}$

We cannot find the root of negative numbers.

therefore,
1. ${\log _{1/2}}\left( {\frac{{5x - {x^2}}}{4}} \right) \ge 0$

2. log is defined when,( i.e. domain of log function = $(0,\infty )$ )

$\left( {\frac{{5x - {x^2}}}{4}} \right) > 0$ ( since log is defined only for positive values )

Take 1st,

${\log _{1/2}}\left( {\frac{{5x - {x^2}}}{4}} \right) \ge 0$

or, ${{{\log }_{1/2}}\left( {\frac{{5x - {x^2}}}{4}} \right) \ge {{\log }_{1/2}}1}$

To know how to solve logarithmic inequality (click here)

since base of log is 1/2 which is less than 1,

therefore, inequality gets reversed

$\left( {\frac{{5x - {x^2}}}{4}} \right) \le 1$

To know how to solve quadratic inequalities (click here)

or, $5x - {x^2} \le 4$

or, ${x^2} - 5x + 4 \ge 0$

or, ${x^2} - 4x - x + 4 \ge 0$

or, $(x - 4)(x - 1) \ge 0$

To know how to solve this (click here)

By wavy-curvy method

$x \in ( - \infty ,1]\bigcup {[4} ,\infty )$ ..................................... (1)

2. $\left( {\frac{{5x - {x^2}}}{4}} \right) > 0$

or, $5x - {x^2} > 0$

or, ${x^2} - 5x < 0$

or,

by wavy-curvy method,

$x \in (0,5)$ ...................... (2)

x must satisfied both the conditions,

therefore, $x \in [1,4]\bigcap {(0,5)}$

..................................................................................................... (Number Line)
-1 0 1 2 3 4 5 6
<-------------------------------------------->
*********************> <*****************

clearly, $x \in (0,1]\bigcup {[4} ,5)$

Therefore, $Domain = (0,1]\bigcup {[4} ,5)$

2. $f(x) = \log (x - [x])$

therefore, x - [x] > 0

=> x > [x]

x is always greater than [x] except x belongs to Integers.

see the definition of greatest integer function ( click here )

when, x = Integers, x = [x]

Also, see from the above graph line y = x touches the graph y = [x] on integers i.e. on integers both the functions are equal.

Therefore,

Domain = R - I

OR you can write $f(x) = \log (x - [x])$

$f(x) = \log \{ x\}$

therefore, {x} > 0

See from the above graph,

{x} is always greater than '0' except on integers,

because, {x} = 0, when x belongs to Integers

Therefore,
Domain = R - I

3. $f(x) = \sin |x| + {\sin ^{ - 1}}(\tan x) + \sin ({\sin ^{ - 1}}x)$

Note: f(x) = g(x) + h(x) + k(x)............

then, $Domain = g(x)\bigcap {h(x)\bigcap {k(x)\bigcap {.....} } }$ .......................

To know more about domain and its properties (click here)

Here are three functions

1. ${f_1}(x) = \sin |x|$

It is defined for all value of 'x'

Domain, ${D_1} = R$

2. ${f_2}(x) = {\sin ^{ - 1}}(\tan x)$

since, ${\sin ^{ - 1}}x$ is defined in the interval [-1,1] i.e. domain of sin inverse x is [-1,1]

therefore, $- 1 \le \tan x \le 1$

( see from the above graph that $- 1 \le \tan x \le 1$ when $- \frac{\pi }{4} \le x \le \frac{\pi }{4}$ )
or, $- \tan \frac{\pi }{4} \le \tan x \le \tan \frac{\pi }{4}$

since, tan x is an increasing function

therefore, $- \frac{\pi }{4} \le x \le \frac{\pi }{4}$

or, $x \in \left[ {\frac{{ - \pi }}{4},\frac{\pi }{4}} \right]$

or, ${D_2} = \left[ {\frac{{ - \pi }}{4},\frac{\pi }{4}} \right]$

3. ${f_3}(x) = \sin ({\sin ^{ - 1}}x)$

since, ${\sin ^{ - 1}}x$ is defined in the interval [-1,1] i.e. domain of sin inverse x is [-1,1]

Therefore, Domain of $\sin ({\sin ^{ - 1}}x)$ = [-1,1]

or, ${D_3} = [ - 1,1]$

also, see the graph

Therefore, domain of $f(x) = \sin |x| + {\sin ^{ - 1}}(\tan x) + \sin ({\sin ^{ - 1}}x)$ ,

$D = {D_1}\bigcap {{D_2}\bigcap {{D_3}} }$

$D = R\bigcap {\left[ {\frac{{ - \pi }}{4},\frac{\pi }{4}} \right]} \bigcap {[ - 1,1]}$

pi/4 = 3.14/4 < 1

and, -pi/4 > -1

therefore, ${D_{}} = \left[ {\frac{{ - \pi }}{4},\frac{\pi }{4}} \right]$

4. $f(x) = {\log _e}|{\log _e}x|$

To know the properties of logarithmic function (click here)

Therefore,

1. for ${\log _e}x$ to be defined

x > 0

or, $x \in (0,\infty )$

2. for ${\log _e}|{\log _e}x|$ to be defined

$|{\log _e}x| > 0$

see the above graph,

at x = 1, ${\log _e}x = 0$

therefore, $x \in (0,\infty ) - \{ 1\}$ or $x \in (0,1)\bigcup {(1,\infty )}$

therefore, take the intersection of (1) and (2)

$Domain = (0,1)\bigcup {(1,\infty )} \bigcap {(0,\infty )}$

or, $Domain = (0,1)\bigcup {(1,\infty )}$

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