Inputing Values into a Function

In mathematics, we often assign numerical values to variables in equations or mathematical expressions. A function can be thought of as a machine: you input a value (called the input), the function processes it according to a specific rule, and produces a result (called the output). The input is usually represented by \(x)\, and the output by $f(x)$ or $y$ .

Understanding Variables and Substitution

A variable is a placeholder for an unknown number, often called the "unknown". When we replace a variable with a specific value, we refer to this process as substitution.

If an equation contains only one variable, we can solve for its value by isolating it.
If an equation contains multiple variables, there can be multiple solutions. Each value substituted affects the remaining variables in the equation.

When we talk about inputting values into a function, we are essentially substituting the variables in a mathematical expression or equation with specific numerical values to evaluate the result.
Often "solving an equation" is actually finding the values of the variables inside it.

For example

Suppose we have three variables, two of which have known values:

$X=3$

$Y=2$

$Z=\text{?}$

We are also given the equation:

$X^2+Y=Z$

Remember! when facing this kind of questions, you want to try and find the values of the variables.
To solve the problem, we will first substitute the known values into the equation:

$3^2+2=Z$

Simplify and solve:

$9+2=Z$

$Z=11$

By substituting the known values and performing the necessary operations, we were able to isolate and calculate the value of $Z$

Answer: $Z=11$

X Y Z By assigning the numerical value, the general form becomes a particular case

Detailed explanation

Start practice

Test yourself on domain of a function!

Given the following function:

$\frac{5-x}{2-x}$

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

Practice more now

If you are interested in this article, you may also be interested in the following articles:

Graphical Representation of a Function

Algebraic Representation of a Function

Function Notation

Domain of a Function

Indefinite Integral

Variation of a Function

Increasing Function

Decreasing Function

Constant Function

Functions for Seventh Grade

Intervals of Increase and Decrease of a Function

Table of Values

On the Tutorela website, you will find a variety of articles with interesting explanations about mathematics

Examples and exercises with solutions for assigning numerical value in a function

Exercise #1

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

$\frac{9x}{4}$

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:

$\frac{5}{x}$

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, $x\ne0$

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, $x\ne0$

Exercise #3

Given the following function:

$\frac{65}{(2x-2)^2}$

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

Video Solution

Step-by-Step Solution

The denominator of the function cannot be equal to 0.

Therefore, we will set the denominator equal to 0 and solve for the domain:

$(2x-2)^2\ne0$

$2x\ne2$

$x\ne1$

In other words, the domain of the function is all numbers except 1.

Answer

Yes, $x\ne1$

Exercise #4

Look at the following function:

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

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 values of $x$ . Therefore, there is no domain limits.

Answer

No, the entire domain

Exercise #5

Find the area of domain (no need to solve)

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

Video Solution

Step-by-Step Solution

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

there is multiplication between fractions whose denominators contain algebraic expressions that include the variable of the equation (which we are looking for when solving the equation),

Of course, these fractions are defined as long as the expressions in their denominators are not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, meaning:

For the fraction in parentheses in the expression on the left side we get:

$\boxed{ x\neq0}$

For the fraction in parentheses in the expression on the right side we get:

$x+1\neq0 \\$ Let's solve the second inequality above (in the same way as solving an equation):

$x+1\neq0 \\ \boxed{x\neq-1}$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a trend inequality (meaning it negates equality: $(\neq)$ and does not require a trend: $(<,>,\leq,\geq)$ ) which is solved exactly like solving an equation, this is unlike solving a trend inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which has only first-degree and lower algebraic expressions), is solved almost identically to solving an equation, however any division or multiplication of both sides by a negative number requires reversing the trend.