Domain of a Function

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The domain of a function includes all those values of X X (independent variable) that, when substituted into the function, keep the function valid and defined.

The domain of a function is an integral part of function analysis. Moreover, a definition set is required to create a graphical representation of the function.

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Test yourself on domain of a function!

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Consider the following function:

\( \frac{3x+4}{2x-1} \)

What is the domain of the function?

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Exercises on the Domain of a Function

Exercise 1

Assignment

25a+4b7y+4⋅3+2=9b \frac{25a+4b}{7y+4\cdot3+2}=9b

What is the domain of the equation?

Solution

We must calculate for which values of y y it is forbidden to be equal to 25a+4b7y+4⋅3+2=9b \frac{25a+4b}{7y+4\cdot3+2}=9b

For this equation, we can see that we have a rational function, so to calculate the domain, we have a restriction, which is that the denominator cannot be 0 0 . Therefore, we set the denominator to zero to determine which value y y cannot take:

7y+12+2=0 7y+12+2=0

We proceed to solve the previous equation by isolating the variable y y

7y+14=0 7y+14=0

We move the 14 14 to the right side and keep the corresponding sign

7y=−14 7y=-14

We divide by: 7 7

y=−2 y=-2

If y y is equal to: −2 -2 then the denominator is equal to 00 and the exercise has no solution

Answer

y≠−2 y\ne-2


Exercise 2

Assignment

What is the domain of the equation?

xyz2(3+y)+4=8 \frac{xyz}{2(3+y)+4}=8

Solution

We need to calculate for which values of y y it is forbidden to be equal to zero

2(3+y)+4=0 2\left(3+y\right)+4=0

Multiply by 2 2 in both elements inside the parentheses

6+2y+4=0 6+2y+4=0

Add them up

10+2y=0 10+2y=0

Move the 10 10 to the right side

2y=−10 2y=-10

Divide by 2 2

y=−5 y=-5

y≠−5 y\ne-5

If y y is equal to negative 5 5 , then the denominator equals 0 0 and the exercise has no solution

Answer

y≠−5 y\ne-5


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Exercise 3

Assignment

15+34:z4y−12+8:2=5 \frac{\sqrt{15}+34:z}{4y-12+8:2}=5

What is the domain of the equation?

Solution

We must calculate for which Y it is forbidden to be equal to

4y−12+8:2=0 4y-12+8:2=0

4y−12+4=0 4y-12+4=0

We move the terms while keeping the corresponding signs

4y=12−4 4y=12-4

4y=8 4y=8

Divide by 4 4

y=2 y=2

If Y Y is equal to: 2 2 then the denominator is equal to: 0 0 and the exercise has no solution

y≠2 y\ne2

Answer

y≠2 y\ne2


Exercise 4

Assignment

Solve the following equation:

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

Solution

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

Multiply by:

(x+1)2 \left(x+1\right)^2

The domain of definition is

x≠−1 x\ne-1

3+2x(x+1)+(x+1)3=3 3+2x\left(x+1\right)+\left(x+1\right)^3=3

Reduce by: 3 3

(x+1)[2x+(x+1)2]=0 \left(x+1\right)\left\lbrack2x+\left(x+1\right)^2\right\rbrack=0

2x+(x+1)2=0 2x+\left(x+1\right)^2=0

2x+x2+2x+1=0 2x+x^2+2x+1=0

x2+4x+1=0 x^2+4x+1=0

x1,2=−4±42−42 x_{1,2}=\frac{-4\pm\sqrt{4^2-4}}{2}

−4±122= \frac{-4\pm\sqrt{12}}{2}=

−2±122 -2\pm\frac{\sqrt{12}}{2}

−2±232=−2±3 -2\pm\frac{2\sqrt{3}}{2}=-2\pm\sqrt{3}

Answer

x=3−2,−3−2 x=\sqrt{3}-2,-\sqrt{3}-2


Do you know what the answer is?

Exercise 5

Assignment

Solve the following equation

(2x+1)2x+2+(x+2)22x+1=4.5x \frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x

Solution

(2x+1)2x+2+(x+2)22x+1=4.5x \frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x

Multiply by: (x+2)(2x+1) \left(x+2\right)\left(2x+1\right)

The domain is x≠−2,−12 x\ne-2,-\frac{1}{2}

(2x+1)3+(x+2)3=4.5x(x+2)(2x+1) \left(2x+1\right)^3+\left(x+2\right)^3=4.5x\left(x+2\right)\left(2x+1\right)

(2x+1)(2x+1)2+(x+2)(x+2)2=4.5x(2x2+5x+2) \left(2x+1\right)\left(2x+1\right)^2+\left(x+2\right)\left(x+2\right)^2=4.5x(2x^2+5x+2)

(2x+1)(4x2+4x+1)+(x+2)(x2+4x+4)=9x3+22.5x2+9x \left(2x+1\right)\left(4x^2+4x+1\right)+\left(x+2\right)\left(x^2+4x+4\right)=9x^3+22.5x^2+9x

Combine like terms

9x3+18x2+18x+9=9x3+22.5x2+9x 9x^3+18x^2+18x+9=9x^3+22.5x^2+9x

Divide by: 9 9

x3+2x2+2x+1=x3+2.5x2+x x^3+2x^2+2x+1=x^3+2.5x^2+x

0.5x2−x−1=0 0.5x^2-x-1=0

Divide by: 0.5 0.5

x2−2x−2=0 x^2-2x-2=0

x1,2=2±(−2)2−4⋅(−2)2 x_{1,2}=\frac{2\pm\sqrt{\left(-2\right)^2-4\cdot\left(-2\right)}}{2}

2±4+82 \frac{2\pm\sqrt{4+8}}{2}

2±122 \frac{2\pm\sqrt{12}}{2}

2±232=1±3 \frac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}

Answer

x=1±3 x=1±\sqrt{3}


Review Questions

What does it mean for a function to be well-defined?

A well-defined function means that it satisfies the definition of a function, which is:

To each element of a set X X (independent variable), which is called the domain of the function, corresponds a unique value from the set Y Y (dependent variable), known as the codomain.


What is the domain of a function?

The domain of a function in mathematics is all the possible values that the independent variable X X can take, such that the function is well-defined when taking these values.


What is the range of a function?

The range, also called the image of a function, are those values that the dependent variable Y Y takes, which depend on the set of numbers from the domain, hence the name dependent variable to the set Y Y .


How is the domain of a function calculated?

The domain of a function depends on the type of function you are working with, as some functions have certain restrictions or ambiguities for the function to exist, that is, to be well-defined.

For example: If we work with a rational function, our restriction for it to be a defined function is that the denominator cannot be equal to zero. Then we must check for which values of the independent variable this restriction is met.

If we work with a radical function in the real numbers, then the restriction is that we cannot have a negative number inside the radical. Similarly, we must observe for which values of the independent variable it is true that it is positive or equal to zero.


How to calculate the domain of the following examples?

Example 1:

Assignment

Determine the domain of the following equation:

5x16−4x=0 \frac{5x}{16-4x}=0

We can see that it is a rational function, then we must determine for which values of X X , our denominator is different from zero, for this we equal the denominator to zero

16−4x=0 16-4x=0

And we proceed to solve this equation:

−4x=−16 -4x=-16

We divide by −4 -4 on the right side

x=−16−4 x=\frac{-16}{-4}

x=4 x=4

Therefore, we conclude that when x=4 x=4 the denominator is equal to 0 0 , so the domain will be any number except 4 4

Answer:

x≠4 x\ne4


Example 2:

Determine the domain of the following equation:

x+5=0 \sqrt{x+5}=0

Here we can see that the equation is a radical function and our restriction is that what's inside the radical be positive or equal to zero, so let's see for which values of X X this is satisfied:

x+5>0 x+5>0

We solve

x>−5 x>-5

We conclude that if the variable takes values equal to or greater than −5 -5 , it will give us the root of a positive number, that is, a defined function.

Answer:

x>−5 x>-5


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Examples with solutions for Domain of a Function

Exercise #1

Does the given function have a domain? If so, what is it?

9x4 \frac{9x}{4}

Video Solution

Step-by-Step Solution

Since the function's denominator equals 4, the domain of the function is all real numbers. This means that any one of the x values could be compatible.

Answer

No, the entire domain

Exercise #2

Given the following function:

5x \frac{5}{x}

Does the function have a domain? If so, what is it?

Video Solution

Step-by-Step Solution

Since the unknown is in the denominator, we should remember that the denominator cannot be equal to 0.

In other words, x≠0 x\ne0

The domain of the function is all those values that, when substituted into the function, will make the function legal and defined.

The domain in this case will be all real numbers that are not equal to 0.

Answer

Yes, x≠0 x\ne0

Exercise #3

Select the field of application of the following fraction:

6x \frac{6}{x}

Video Solution

Step-by-Step Solution

Since the domain of definition depends on the denominator, and X appears in the denominator

All numbers will be suitable except for 0.

In other words, the domain of definition:

x≠0 x\ne0

Answer

All numbers except 0

Exercise #4

Select the field of application of the following fraction:

8+x5 \frac{8+x}{5}

Video Solution

Step-by-Step Solution

Since the domain depends on the denominator, we note that there is no variable in the denominator.

Therefore, the domain is all numbers.

Answer

All numbers

Exercise #5

Given the following function:

5+4x2+x2 \frac{5+4x}{2+x^2}

Does the function have a domain? If so, what is it?

Video Solution

Step-by-Step Solution

Since the denominator is positive for all X, the domain of the function is the entire domain.

That is, all X, therefore there is no domain restriction.

Answer

No, the entire domain

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