Domain of a Function

Assignment

$\frac{25a+4b}{7y+4\cdot3+2}=9b$

What is the domain of the equation?

Solution

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

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 $0$. Therefore, we set the denominator to zero to determine which value $y$ cannot take:

$7y+12+2=0$

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

$7y+14=0$

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

$7y=-14$

We divide by: $7$

$y=-2$

If $y$ is equal to: $-2$ then the denominator is equal to $0$ and the exercise has no solution

Answer

$y\ne-2$

Assignment

What is the domain of the equation?

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

Solution

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

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

Multiply by $2$ in both elements inside the parentheses

$6+2y+4=0$

Add them up

$10+2y=0$

Move the $10$ to the right side

$2y=-10$

Divide by $2$

$y=-5$

$y\ne-5$

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

Answer

$y\ne-5$

Assignment

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

What is the domain of the equation?

Solution

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

$4y-12+8:2=0$

$4y-12+4=0$

We move the terms while keeping the corresponding signs

$4y=12-4$

$4y=8$

Divide by $4$

$y=2$

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

$y\ne2$

Answer

$y\ne2$

Assignment

Solve the following equation:

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

Solution

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

Multiply by:

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

The domain of definition is

$x\ne-1$

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

Reduce by: $3$

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

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

$2x+x^2+2x+1=0$

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

$x_{1,2}=\frac{-4\pm\sqrt{4^2-4}}{2}$

$\frac{-4\pm\sqrt{12}}{2}=$

$-2\pm\frac{\sqrt{12}}{2}$

$-2\pm\frac{2\sqrt{3}}{2}=-2\pm\sqrt{3}$

Answer

$x=\sqrt{3}-2,-\sqrt{3}-2$

Assignment

Solve the following equation

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

Solution

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

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

The domain is $x\ne-2,-\frac{1}{2}$

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

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

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

Combine like terms

$9x^3+18x^2+18x+9=9x^3+22.5x^2+9x$

Divide by: $9$

$x^3+2x^2+2x+1=x^3+2.5x^2+x$

$0.5x^2-x-1=0$

Divide by: $0.5$

$x^2-2x-2=0$

$x_{1,2}=\frac{2\pm\sqrt{\left(-2\right)^2-4\cdot\left(-2\right)}}{2}$

$\frac{2\pm\sqrt{4+8}}{2}$

$\frac{2\pm\sqrt{12}}{2}$

$\frac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$

Answer

$x=1±\sqrt{3}$

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 $X$ (independent variable), which is called the domain of the function, corresponds a unique value from the set $Y$ (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 $X$ 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 $Y$ takes, which depend on the set of numbers from the domain, hence the name dependent variable to the set $Y$.

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:

$\frac{5x}{16-4x}=0$

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

$16-4x=0$

And we proceed to solve this equation:

$-4x=-16$

We divide by $-4$ on the right side

$x=\frac{-16}{-4}$

$x=4$

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

Answer:

$x\ne4$

Example 2:

Determine the domain of the following equation:

$\sqrt{x+5}=0$

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 $X$ this is satisfied:

$x+5>0$

We solve

$x>-5$

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

Answer:

$x>-5$