# Which Best Describes the Domain of a Function

Which Best Describes the Domain of a Function.

A
function
is expressed as

y=f(x)
,

where
x
is the independent variable and
y
is the dependent variable.

First, we learn
what is the Domain
before learning

How to Notice the Domain of a Office Algebraically

## 👉 What is the Domain of a Function?

Let
$f(x)$
be a existent-valued role. Then the

domain of a part

is the set of all possible values of
$x$
for which
$f(x)$
is defined.

The domain of a function
$f(x)$
is expressed as
$D(f)$.

We suggest y’all to read
how to discover zeros of a office
and
zeros of quadratic function
first.

## 👉 Rules to remember when finding the Domain of a Function

We should always remember the following rules when finding the domain of a function:

1. If the function is a polynomial office then x tin be positive, zero or negative, i.east.,
$ten>0$,
$x=0$
or
$ten<0$,
2. If
$f(x)=\frac{chiliad(10)}{h(ten)}$, then always
$h(x)\neq0$,
3. If
$f(10)=\sqrt{k(x)}$, then always
$g(x)\geq0$,
4. If
$f(10)=\frac{chiliad(x)}{\sqrt{h(x)}}$, and then always
$h(ten)>0$,
5. If
$f(x)=\frac{\sqrt{g(x)}}{h(10)}$, then always
$1000(10)\geq 0$
and
$h(x)\neq 0$,
6. If
$f(x)=\sqrt{\frac{chiliad(x)}{h(ten)}}$, then always
$g(x)\geq0$
and
$h(ten)>0$,
7. If
$f(x)=\ln \left ( 1000(x) \right )$, and then ever
$g(x)>0$.

The vii rules mentioned above will make our piece of work like shooting fish in a barrel when we find the domain of a function.

There are 2 other rules. We will learn them at the time of discussion.

## 👉 How to Find the Domain of a Function Algebraically

There are dissimilar ways to

find the domain of a function
.

Here we will discuss 9 best means for different functions.

### #1. Find the Domain of a Polynomial Function

There are different types of
Polynomial Function
based on degree.

Some of them are

See that all the polynomial functions are defined for all
$x\epsilon\mathbb{R}$.

$\therefore$
the domain of whatsoever polynomial office is
$\mathbb{R}=(-\infty,\infty)$.

### #ii. Detect the Domain of a Rational Function

A
Rational Function
is a fraction of functions denoted by

$f(10)=\frac{yard(ten)}{h(ten)}, h(x)\neq 0$

Rational function is also chosen Quotient Function.

Example:

Permit
$f(10)=\frac{x+two}{ten^{2}+3x+ii}$. Observe the domain of
$f(x)$.

Solution:

Run across that 10+2 is defined for all
$x\epsilon \mathbb{R}$.

So nosotros practise not have to worry for this part.

From Rule two we know that the function
$f(x)=\frac{g(ten)}{h(10)}$
is divers when
$h(x)\neq0$.

In this trouble, we have to find at what points
$10^{2}+3x+2\neq 0$.

Now
$ten^{ii}+3x+2\neq 0$

i.e.,
$(10+2)(x+one)\neq 0$

i.e., either
$(x+2)\neq 0$
or
$(x+1)\neq 0$

i.due east., either
$ten\neq -ii$
or
$x\neq -one$

Therefore
$f(x)=\frac{ten+2}{x^{2}+3x+2}$
exists for all
$10\epsilon\mathbb{R}$
except
$x\neq -2$
and
$x\neq -one$

$\therefore$
domain of f(ten) = {$x\epsilon \mathbb{R}:ten\neq -2,-1$}.

On the Real centrality, the
light-green lines
are the domain of f(x).

We can write this equally,

Domain of f(x)=$(-\infty,-two)\cup(-2,-ane)\cup(-ane,\infty)$.

Case:

How practise you find the domain of the rational function given beneath

$f(ten)=\frac{x}{x^{two}+two}$

Solution:

For f(ten) to be defined,

$x^{ii}+2\neq 0$

or,
$ten^{2}\neq -two$

or,
$ten\neq \pm \sqrt{-two}$

or,
$x\neq \pm \sqrt{2}i \epsilon \mathbb{C}$, an imaginary number (i.eastward., not a real number).

This implies that f(10) exists for all
$x\epsilon \mathbb{R}$

$\therefore$
the domain of the function
$f(x)=\frac{x}{x^{2}+ii}$
is

$D(f)=\mathbb{R}=(-\infty,\infty)$.

### #3. Finding Domain of a Function with a Square Root

Case:

Observe the domain

$f(x)=\sqrt{x+ii}$

Solution:

From Rule 3 we know that a function of the form
$f(x)=\sqrt{g(10)}$
is defined when
$one thousand(ten)\geq 0$

i.e.,

$f(x)=\sqrt{10+2}$
is defined when

$10+ii\geq 0$

or,
$x\geq -2$

Putting this result on existent line nosotros get

$\therefore$
domain of
$f(x)=\sqrt{x+ii}$
is

$D(f)$={$x\epsilon \mathbb{R}: ten\geq -2$}=$[-two,\infty )$.

Example:

Detect the domain of the function

$f(x)=\sqrt{ten^{2}+3x+2}$

Solution:

For
$f(x)$
to be defined,

$10^{2}+3x+two\geq 0$

i.e.,
$(x+ii)(x+i)\geq 0$

i.e., either
$x+2\leq 0$
or
$10+i\geq 0$

i.e., either
$x\leq -two$
or
$x\geq -1$

Why nosotros write
$x\leq -2$
and
$x\geq -i$?

See the table given below to understand this

From the table we can see that the relation
$(x+2)(10+i)\geq 0$
is satisfied when

$x\epsilon (-\infty,-2)$,
$10=-2$, and
$10=-ane$,
$x\epsilon (-1,\infty)$

i.e.,
$10\epsilon (-\infty,-2]$
and
$10\epsilon [-1,\infty)$

i.e.,
$x\epsilon (-\infty,-two] \cup [-1,\infty)$

$\therefore$
domain of
$f(ten)=\sqrt{10^{2}+3x+2}$
is

$D(f)$
=
$\: x \: \epsilon \: (-\infty,-ii] \cup [-one,\infty)$

$\: \: \: \: \: \: \: \: \: \:$
= {$ten \: \epsilon \: \mathbb{R}:x\leq -2,ten\geq -1$}

### #4. Finding Domain of a Part with a Square root in the denominator

From Rule 4 we know that a function of the form
$f(x)=\frac{g(x)}{\sqrt{h(x)}}$
is defined when
$h(x)>0$.

Example:

How to discover the domain of the function given below

$f(x)=\frac{one}{\sqrt{i-x}}$

Solution:

For
$f(ten)$
to exist defined,

$1-x>0$

i.due east.,
$ane>10$

i.due east.,
$x<1$

$\therefore$
domain of
$f(ten)$={$x\epsilon \mathbb{R}:x} =
$(-\infty,1)$

Example:

Find the domain of

$f(10)=\frac{ten^{2}+2x+three}{\sqrt{ten+1}}$

Solution:

For
$f(x)$
to be defined,

$x+one>0$

i.e.,
$10>-1$

$\therefore$
domain of
$f(x)=\frac{x^{two}+2x+3}{\sqrt{ten+1}}$
is {$10\epsilon \mathbb{R}:10>-1$} =
$(-1,\infty)$

### #five. Finding Domain of a Part with a Square root in the numerator

From Rule 5 we know that a function of the form
$f(10)=\frac{\sqrt{g(x)}}{h(x)}$
is divers when
$g(x)\geq 0$
and
$h(x)\neq 0$.

Instance:

Find the domain of

$f(x)=\frac{\sqrt{x+1}}{x^{2}-4}$

Solution:

The function
$f(x)=\frac{\sqrt{ten+1}}{10^{2}-4}$
is defined when

• $x+1\geq 0$
• $x^{2}-4\neq 0$

Now

$x+1\geq 0$

i.east.,
$10\geq -ane$

and

$10^{ii}-4\neq 0$

i.eastward.,
$(x+2)(ten-two)\neq 0$

i.e.,
$(x+two)\neq 0$
and
$(x-ii)\neq 0$

i.e.,
$x\neq -2$
and
$x\neq 2$

$\therefore$
domain of
$f(x)=\frac{\sqrt{x+1}}{10^{ii}-four}$
is

{$10\epsilon \mathbb{R}:10\geq -one,x\neq 2,$} (We doesn’t include
$x\neq -2$
because
$x\geq -1$)

We can also express the domain of the office in interval annotation.

Domain of
$f(10)=\frac{\sqrt{x+i}}{x^{2}-4}$
in interval annotation is
$[-1,ii)\cup (2,\infty)$.

### #half dozen. Finding Domain of a Function with a Square root in the numerator and denominator

From Rule half dozen we know that a function of the form
$f(x)=\sqrt{\frac{g(x)}{h(ten)}}$
is divers when
$thou(ten)\geq0$
and
$h(ten)>0$.

Example:

Find the domain of

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

Solution:

For
$f(10)$
to be defined,

$3-x\neq 0$

i.e.,
$iii\neq x$

i.e.,
$x\neq 3$

At present we have to notice the set of values of x so that

$\frac{x-2}{3-x}\geq 0$

Hither we tin can non directly say
$x-2>0$
because we do non know the sign of
$3-x$.

To overcome this problem nosotros will brand the denominator +ve by multiplying the numerator and denominator by (iii-x)

$\frac{x-2}{3-10}\times \frac{{\colour{Magenta} 3-x}}{{\color{Magenta} three-10}}\geq 0$

i.e.,
$\frac{(x-2)(iii-ten)}{(ten-3)^{two}}\geq 0$

i.eastward.,
$(x-2)(3-x)\geq 0$

Next we accept to detect the values of x and so that
$(x-2)(3-10)\geq 0$

At present run across the tabular array given beneath:

Now putting the signs on real axis for each interval and value of x, nosotros become

$\therefore$
the domain of the function
$f(x)=\sqrt{\frac{ten-2}{iii-x}}$
is
$D(f)$
=
$[2,3)$

### #7. Observe Domain Of A Logarithmic Function

From Rule vii we know that a
Logarithmic Part
of the course
$f(ten)=\ln \left ( thou(ten) \right )$
is defined when
$g(x)>0$.

Instance:

Find the domain of

$f(x)= \ln (x-2)$

Solution:

The office
$f(x)= \ln (x-ii)$
is defined when

$x-two>0$

i.due east.,
$ten>2$

Therefore
$f(x)= \ln (10-2)$
is defined for all
$x>2$.

$\therefore$
domain of
$f(10)= \ln (x-2)$
is

$D(f)$
= {$ten\epsilon \mathbb{R}:x>2$} =
$(ii,\infty)$

Example:

Detect the domain of

$f(10)= \ln (x^{2}-3x+2)$

Solution:

The function
$f(x)= \ln (x^{2}-3x+ii)$
is defined when

$x^{2}-3x+2>0$

i.e.,
$(x-1)(x-2)>0$

i.e.,
$10<1$
and
$10>2$

$\therefore$
domain of the part
$f(x)= \ln (x^{2}-3x+2)$
is

$D(f)$
= {$ten\epsilon \mathbb{R}:x<1,x>2$} =
$(-\infty,1)\cup (2,\infty)$

### #viii. Find the Domain of a Part using a Relation

Rules:

• Earlier finding the domain of a role using a relation first nosotros have to bank check that the given relation is a function or not,
• A Relation is the prepare of ordered pairs i.e., the set of (x,y) and the domain of the relation is the set of all ten-coordinates i.e., the ready {ten}.

Case:

Find the domain of the relation

{$(two,5), (3,6), (four,17), (11,8)$}

Solution:

Showtime we check the relation {$(ii,5), (iii,6), (4,17), (eleven,viii)$} is a office or not.

The diagram of the relation is

See that each element of the ready {$2, iii, 4, 11$} is related to a unique element of the set {$five, 6, 8, 17$}.

Therefore the relation is a part.

In the relation

{$(2,5), (iii,six), (4,17), (11,8)$}

the set of x-coordinates is {$2, 3, four, xi$} and the prepare of y-coordinates is {$v, vi, 17, 8$}.

$\therefore$
the domain of the relation is {$2, three, 4, xi$}

Which relation is a Function?

Example:

Discover the domain of the relation

{$(2, iii), (5, 8), (6, 7), (six, fifteen), (11,17)$}

Solution:

The diagram of the given relation is

See that the element half-dozen is related to two different elements 7 and 15

i.due east., 6 is not related to a unique element.

Therefore the given relation is non a function.

Which is relations are non a Function?

### #nine. Find Domain of a Function on a Graph

Finding the domain of a function using a graph is the easiest way to find the domain.

Rule:

The domain of a function on a graph is the ready of all possible values of x on the ten-axis.

For domain, we have to find where the 10 value starts and where the 10 value ends i.due east., the function of x-axis where f(x) is divers.

See the example given below to understand this concept

Case:

Find the domain of the part from graph

$x^{ii}+y^{2}=4$.

Solution:

Step i: Draw the graph

Step 2: Detect the possible values of ten where f(x) is defined

Here the x values showtime from -ii and ends in 2.

Step 3: The possible values of ten is the domain of the function.

$\therefore$
the domain of the circle is {$x\epsilon \mathbb{R}:-ii\leq x\leq 2$} =
$[-ii,2]$

Example:

Notice the domain of the part

$y^{two}=2(x-two)$

Solution:

Step ane: The graph of the given parabola is

Pace 2:

See that the 10 value starts from 2 and extends to infinity (i.due east., it will never stop).

Stride three:

$\therefore$
the domain of the parabola is {$x\epsilon \mathbb{R}:2\leq ten\leq \infty$} =
$[ii,\infty)$

Example:

Find the domain of the direct line
$y=ten$
from the graph

Solution:

From the graph of
$y=x$
we tin can come across that the ten value starts from
$-\infty$
and extends to
$+\infty$.

Therefore the domain of the straight line is
$(-\infty,\infty)$.

Example:

Find the domain of
$x^{2}=2y$
from the graph given below.

Solution: The x values of
$x^{2}=2y$
on the graph are shown by the green line.

See that the
$x$
value starts from
$-\infty$
and extends to
$+\infty$.

Therefore domain of
$10^{2}=2y$
is
$(-\infty,\infty)$.

## Final words

We just learned 9 different ways to
Notice the Domain of a Function Algebraically.

Now it’southward your plough to practice them over again and again and master them.

If you have any doubts or suggestions, please tell us in the comment section. We love to hear from you.