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 Find the Domain of a Function Algebraically*

## π What is the Domain of a Function?

Let f(x) be a real-valued function. Then the *domain of a function* 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 you to read how to find zeros of a function and zeros of quadratic function first.

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

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

- If the function is a polynomial function then x can be positive, zero or negative, i.e., x>0, x=0 or x<0,
- If f(x)=\frac{g(x)}{h(x)}, then always h(x)\neq0,
- If f(x)=\sqrt{g(x)}, then always g(x)\geq0,
- If f(x)=\frac{g(x)}{\sqrt{h(x)}}, then always h(x)>0,
- If f(x)=\frac{\sqrt{g(x)}}{h(x)}, then always g(x)\geq 0 and h(x)\neq 0,
- If f(x)=\sqrt{\frac{g(x)}{h(x)}}, then always g(x)\geq0 and h(x)>0,
- If f(x)=\ln \left ( g(x) \right ), then always g(x)>0.

The 7 rules mentioned above will make our work easy when we find the domain of a function.

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

Table of Contents - What you will learn

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

There are different ways to ** find the domain of a function**.

Here we will discuss 9 best ways 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

Degree of polynomial Function | Polynomial Function | Name of Polynomial Function | Domain |
---|---|---|---|

1 | 3x+5 | Linear Function | \mathbb{R}=(-\infty,\infty) |

2 | x^{2}+1 | Quadratic Function | \mathbb{R}=(-\infty,\infty) |

3 | 2x^{3}+x^{2}+3 | Cubic Function | \mathbb{R}=(-\infty,\infty) |

4 | 7x^{4}+2 | Quartic Function | \mathbb{R}=(-\infty,\infty) |

5 | 6x^{5}+2x^{2}-3 | Quintic Function | \mathbb{R}=(-\infty,\infty) |

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

\therefore the domain of any polynomial function is \mathbb{R}=(-\infty,\infty).

### #2. Find the Domain of a Rational Function

A Rational Function is a fraction of functions denoted by

f(x)=\frac{g(x)}{h(x)}, h(x)\neq 0

Rational function is also called Quotient Function.

Example:

Let f(x)=\frac{x+2}{x^{2}+3x+2}. Find the domain of f(x).

Solution:

See that x+2 is defined for all x\epsilon \mathbb{R}.

So we do not have to worry for this part.

From Rule 2 we know that the function f(x)=\frac{g(x)}{h(x)} is defined when h(x)\neq0.

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

Now x^{2}+3x+2\neq 0

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

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

i.e., either x\neq -2 or x\neq -1

Therefore f(x)=\frac{x+2}{x^{2}+3x+2} exists for all x\epsilon\mathbb{R} except x\neq -2 and x\neq -1

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

On the Real axis, the green lines are the domain of f(x).

We can write this as,

Domain of f(x)=(-\infty,-2)\cup(-2,-1)\cup(-1,\infty).

Example:

How do you find the domain of the rational function given below

f(x)=\frac{x}{x^{2}+2}

Solution:

For f(x) to be defined,

x^{2}+2\neq 0or, x^{2}\neq -2

or, x\neq \pm \sqrt{-2}

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

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

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

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

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

Example:

Find the domain

f(x)=\sqrt{x+2}

Solution:

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

i.e.,

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

x+2\geq 0

or, x\geq -2

Putting this result on real line we get

\therefore domain of f(x)=\sqrt{x+2} is

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

Example:

Find the domain of the function

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

Solution:

For f(x) to be defined,

x^{2}+3x+2\geq 0i.e., (x+2)(x+1)\geq 0

i.e., either x+2\leq 0 or x+1\geq 0

i.e., either x\leq -2 or x\geq -1

Why we write x\leq -2 and x\geq -1?

See the table given below to understand this

Value of x | Sign of (x+2) | Sign of (x+1) | Sign of (x+2)(x+1) | (x+2)(x+1)\geq 0 satisfied or not |
---|---|---|---|---|

x=-3 i.e., x\epsilon (-\infty,-2) | -ve | -ve | >0 i.e., +ve | β |

x=-2 | 0 | -ve | =0 | β |

x=-1.5 i.e., x\epsilon (-2,-1) | +ve | -ve | <0 i.e., -ve | β |

x=-1 | +ve | 0 | =0 | β |

x=0 i.e., x\epsilon (-1,\infty) | +ve | +ve | >0 i.e., +ve | β |

From the table we can see that the relation (x+2)(x+1)\geq 0 is satisfied when

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

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

i.e., x\epsilon (-\infty,-2] \cup [-1,\infty)

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

D(f) = \: x \: \epsilon \: (-\infty,-2] \cup [-1,\infty)

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

### #4. Finding Domain of a Function 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 find the domain of the function given below

f(x)=\frac{1}{\sqrt{1-x}}

Solution:

For f(x) to be defined,

1-x>0

i.e., 1>x

i.e., x<1

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

Example:

Find the domain of

f(x)=\frac{x^{2}+2x+3}{\sqrt{x+1}}

Solution:

For f(x) to be defined,

x+1>0

i.e., x>-1

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

### #5. Finding Domain of a Function with a Square root in the numerator

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

Example:

Find the domain of

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

Solution:

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

- x+1\geq 0
- x^{2}-4\neq 0

Now

i.e., x\geq -1

and

i.e., (x+2)(x-2)\neq 0

i.e., (x+2)\neq 0 and (x-2)\neq 0

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

\therefore domain of f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is

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

We can also express the domain of the function in interval notation.

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

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

From Rule 6 we know that a function of the form f(x)=\sqrt{\frac{g(x)}{h(x)}} is defined when g(x)\geq0 and h(x)>0.

Example:

Find the domain of

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

Solution:

For f(x) to be defined,

3-x\neq 0

i.e., 3\neq x

i.e., x\neq 3

Now we have to find the set of values of x so that

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

Here we can not directly say x-2>0 because we do not know the sign of 3-x.

To overcome this problem we will make the denominator +ve by multiplying the numerator and denominator by (3-x)

\frac{x-2}{3-x}\times \frac{{\color{Magenta} 3-x}}{{\color{Magenta} 3-x}}\geq 0i.e., \frac{(x-2)(3-x)}{(x-3)^{2}}\geq 0

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

Next we have to find the values of x so that (x-2)(3-x)\geq 0

Now see the table given below:

Value of x | Sign of (x-2) | Sign of (3-x) | Sign of (x-2)(3-x) | (x-2)(3-x)\geq 0 satisfied or not |
---|---|---|---|---|

x=0 i.e., x\epsilon (-\infty,2) | -ve | +ve | <0 i.e., -ve | β |

x=2 | 0 | +ve | 0 | β |

x=2.5 i.e., x\epsilon (2,3) | +ve | +ve | >0 i.e., +ve | β |

x=4 i.e., x\epsilon (3,\infty) | +ve | -ve | <0 i.e., -ve | β |

Now putting the signs on real axis for each interval and value of x, we get

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

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

From Rule 7 we know that a Logarithmic Function of the form f(x)=\ln \left ( g(x) \right ) is defined when g(x)>0.

Example:

Find the domain of

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

Solution:

The function f(x)= \ln (x-2) is defined when

x-2>0

i.e., x>2

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

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

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

Example:

Find the domain of

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

Solution:

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

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

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

Value of x | Sign of (x-1) | Sign of (x-2) | Sign of (x-1)(x-2) | (x-1)(x-2)>0 satisfied or not |
---|---|---|---|---|

x=.5<1 i.e., x\epsilon (-\infty,1) | -ve | -ve | +ve i.e., >0 | β |

x=1 | 0 | -ve | =0 | β |

x=1.5 i.e., x\epsilon (1,2) | +ve | -ve | -ve i.e., <0 | β |

x=2 | +ve | 0 | =0 | β |

x=3>2 i.e., x\epsilon (2,\infty) | +ve | +ve | +ve i.e., >0 | β |

i.e., x<1 and x>2

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

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

### #8. Find the Domain of a Function using a Relation

Rules:

- Before finding the domain of a function using a relation first we have to check that the given relation is a function or not,
- A Relation is the set of ordered pairs i.e., the set of (x,y) and the domain of the relation is the set of all x-coordinates i.e., the set {x}.

Example:

Find the domain of the relation

{(2,5), (3,6), (4,17), (11,8)}

Solution:

First we check the relation {(2,5), (3,6), (4,17), (11,8)} is a function or not.

The diagram of the relation is

See that each element of the set {2, 3, 4, 11} is related to a unique element of the set {5, 6, 8, 17}.

Therefore the relation is a function.

In the relation

{(2,5), (3,6), (4,17), (11,8)}

the set of x-coordinates is {2, 3, 4, 11} and the set of y-coordinates is {5, 6, 17, 8}.

\therefore the domain of the relation is {2, 3, 4, 11}

Read more: Which relation is a Function?

Example:

Find the domain of the relation

{(2, 3), (5, 8), (6, 7), (6, 15), (11,17)}

Solution:

The diagram of the given relation is

See that the element 6 is related to two different elements 7 and 15

i.e., 6 is not related to a unique element.

Therefore the given relation is not a function.

Additional reading: Which is relations are not a Function?

### #9. 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 set of all possible values of x on the x-axis.

For domain, we have to find where the x value starts and where the x value ends i.e., the part of x-axis where f(x) is defined.

See the example given below to understand this concept

Example:

Find the domain of the function from graph

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

Solution:

Step 1: Draw the graph

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

Here the x values start from -2 and ends in 2.

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

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

Example:

Find the domain of the function

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

Solution:

Step 1: The graph of the given parabola is

Step 2:

See that the x value starts from 2 and extends to infinity (i.e., it will never end).

Step 3:

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

Example:

Find the domain of the straight line y=x from the graph

Solution:

From the graph of y=x we can see that the x 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 x^{2}=2y is (-\infty,\infty).

## Final words

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

Now it’s your turn to practice them 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.

Additionally you can read:

f(x)=-1//x?

how to find its domain?

content-rich lecture. Easy to follow, grasp!

this so helpful. thanks so much