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Lesson Explainer: Understanding Functions Mathematics

In this explainer, we will learn how to define and evaluate functions and find the roots and the maximum or minimum of quadratic functions.

function is a mathematical relationship that takes inputs and maps each input onto exactly one output. Functions can be defined over many different objects; however, in this explainer, we will only focus on functions over sets of numbers.

It is important to note that when we input a value into a function, we always get the same output; we can never get a different output value. We can represent functions using function notation. If π‘₯ is the input value to a function 𝑓, then we can represent the output of this function at π‘₯ as 𝑓(π‘₯). There are many ways of representing functions, including an expression, a diagram, or a rule.

Definition: Function

A function is mathematical relationship that takes inputs and maps each input onto exactly one output.

For example, the function might double a value. Then, when we input 5, the function will double this value to output 10. We can describe this function using a rule and express the inputs and corresponding outputs using function notation. If we say that the input is π‘₯, then the output would be 2π‘₯. We can write this function as 𝑓(π‘₯)=2π‘₯, where 𝑓 is the name of the function, π‘₯ is the input, and 2π‘₯ is the output.

We can also represent functions using diagrams. For instance, let’s say that we have a function that can take inputs of 1, 2, and 3 and the function adds 2 to each value. We can calculate the outputs of the function as 3, 4, and 5. We show this by connecting the inputs to the outputs using an arrow.

Of course, we could extend this function to include more input values; however, we are told that the only valid inputs are 1,2, and 3. In general, we say that the domain is the set of all valid input values.

For example, consider the function 𝑓(π‘₯)=1π‘₯. We can evaluate the function by substituting values of π‘₯ into the function. If we let π‘₯=2, then we see that 𝑓(2)=12.

Thus, 2 is a possible input for the function. However, we cannot input every number. If we tried to input π‘₯=0 using the same method, we would divide by zero. This is not allowed, so we can conclude that 0 is not a valid input for this function. We call the set of all input values the domain of the function. So, the domain of this function is β„βˆ’{0}.

In our example of 𝑓(π‘₯)=2π‘₯, we can input any real value of π‘₯, so we can say that its domain is ℝ.

Let’s now consider the outputs of the function 𝑓(π‘₯)=2π‘₯. Is any output possible? Let π‘Ž be any real number. We want to find an input value of π‘₯ so that the function outputs π‘Ž. Inputting π‘₯=π‘Ž2 into the function gives π‘“ο€»π‘Ž2=2ο€»π‘Ž2=π‘Ž.

Thus, any real value π‘Ž is a possible output of the function. The set of all output values of a function is called its range. So, the range of the function 𝑓(π‘₯)=2π‘₯ is ℝ. We can summarize these definitions as follows.

Definition: Domain and Range of a Function

The set of all possible input values of a function is called its domain.

The set of all possible output values of a function is called its range.

Let’s now see an example of finding the sum of two functions evaluated at different values.

Example 1: Finding the Value of the Addition of Two Functions at Given Values

Given that 𝑓(π‘₯)=π‘₯+3 and 𝑔(π‘₯)=βˆ’2π‘₯+3, find the value of 𝑓(1)+𝑔(2).

Answer

We evaluate functions by substituting the input values into the functions. This means that we can evaluate 𝑓(1) by substituting π‘₯=1 into the given function 𝑓(π‘₯)=π‘₯+3: 𝑓(1)=1+3=4.

Similarly, we can evaluate 𝑔(2) by substituting π‘₯=2 into the given function 𝑔(π‘₯)=3βˆ’2π‘₯: 𝑔(2)=3βˆ’2(2)=3βˆ’4=βˆ’1.

We can now evaluate 𝑓(1)+𝑔(2) by adding these values together. We get 𝑓(1)+𝑔(2)=4+(βˆ’1)=3.

Before we move on to our next example, there is another useful definition we can consider for functions.

Definition: Root of a Function

We call any input value π‘₯ for which the output of the function is 0 a root of the function.

This means if π‘₯ is a root of the function 𝑓, then we have 𝑓(π‘₯)=0. Thus, the roots of a function 𝑓 are all the solutions of the equation 𝑓(π‘₯)=0. It follows that we can find roots of polynomial functions by factoring the polynomial; the roots of the function are then given by the π‘₯-values that make one of the factors 0.

Let’s see an example of using factoring to determine the roots of a quadratic function.

Example 2: Finding the Roots of a Quadratic Function by Solving It

Find all of the roots of the function 𝑓(π‘₯)=π‘₯+5π‘₯+4.

Answer

We first recall that the roots of a function are the input values for which the output of the function is 0. This means that to find the roots of 𝑓(π‘₯), we want to solve the equation 𝑓(π‘₯)=0. We can substitute in the quadratic we are given to get the equation π‘₯+5π‘₯+4=0.

This is a quadratic equation, and we recall that we can solve these by factoring. In particular, we want to find two numbers that multiply to give 4 and add to give 5. Since 4Γ—1=4 and 4+1=5, we can factor the quadratic as (π‘₯+4)(π‘₯+1)=0.

We now have that the product of two factors is equal to 0. This can only happen if one of the factors is 0. Thus, we either have π‘₯+4=0 or π‘₯+1=0. We can solve each equation separately to get π‘₯=βˆ’4 or π‘₯=βˆ’1.

Hence, the roots of the function 𝑓(π‘₯) are π‘₯=βˆ’4 and π‘₯=βˆ’1.

It is worth noting that we can check that the answers we found are indeed roots by substituting them back into the function. For example, we can verify π‘₯=βˆ’4 is a root of 𝑓(π‘₯)=π‘₯+5π‘₯+4 by evaluating 𝑓(βˆ’4): 𝑓(βˆ’4)=(βˆ’4)+5(βˆ’4)+4=16βˆ’20+4=0.

We have verified that 𝑓(βˆ’4)=0, so π‘₯=βˆ’4 is a root of 𝑓(π‘₯).

In our next example, we will determine the roots of a polynomial function by rewriting it as a quadratic function.

Example 3: Finding the Roots of a Disguised Quadratic

Find all the real roots of the function 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’12οŠͺ.

Answer

We first recall that the roots of a function are the input values for which the output of the function is 0. This means that to find the roots of 𝑓(π‘₯), we want to solve the equation 𝑓(π‘₯)=0. We can substitute in the polynomial we are given to get the equation π‘₯βˆ’π‘₯βˆ’12=0.οŠͺ

There are many ways we could solve this polynomial equation. The easiest way is to note that this is a quadratic equation in π‘₯ since π‘₯=ο€Ήπ‘₯οŠͺ. This allows us to rewrite the equation as ο€Ήπ‘₯ο…βˆ’π‘₯βˆ’12=0.

If we now substitute π‘Ž=π‘₯ into the equation, we get π‘Žβˆ’π‘Žβˆ’12=0.

This is now a quadratic equation in π‘Ž. We can solve this by factoring. We note that (βˆ’4)Γ—3=βˆ’12 and βˆ’4+3=βˆ’1. This allows us to factor the quadratic as follows: (π‘Žβˆ’4)(π‘Ž+3)=0.

We now have that the product of two factors is equal to 0. This can only happen if one of the factors is 0. Thus, we either have π‘Žβˆ’4=0 or π‘Ž+3=0. We can solve each equation for π‘Ž to get π‘Ž=4 or π‘Ž=βˆ’3.

However, we are not looking for values of π‘Ž; we need to find values of π‘₯. So, we substitute π‘Ž=π‘₯ into each equation. We get π‘₯=4 or π‘₯=βˆ’3. Since π‘₯β‰₯0, we know there are no real solutions for π‘₯=βˆ’3. Thus, we only need to solve π‘₯=4. Taking square roots of both sides of the equation yields π‘₯=βˆ’2 or π‘₯=2.

Hence, the roots of the function 𝑓(π‘₯) are π‘₯=βˆ’2 and π‘₯=2.

In our next example, we will determine all of the input values for which two given functions are equal.

Example 4: Finding the π‘₯-Values for Which Two Quadratic Functions Are Equal

If 𝑓(π‘₯)=3π‘₯+4π‘₯ and 𝑔(π‘₯)=π‘₯βˆ’4π‘₯βˆ’6, find all of the real values of π‘₯ such that 𝑓(π‘₯)=𝑔(π‘₯).

Answer

To determine the values of π‘₯ such that 𝑓(π‘₯)=𝑔(π‘₯), we can substitute the given expressions for 𝑓(π‘₯) and 𝑔(π‘₯) into the equation to get 3π‘₯+4π‘₯=π‘₯βˆ’4π‘₯βˆ’6.

This is a quadratic equation; we can simplify the equation by rearranging and combining like terms. Let’s start by subtracting π‘₯ from both sides of the equation and adding 4π‘₯+6 to both sides of the equation: 3π‘₯+4π‘₯βˆ’π‘₯+4π‘₯+6=π‘₯βˆ’4π‘₯βˆ’6βˆ’π‘₯+4π‘₯+6.

The right-hand side of the equation simplifies to give 0, and we can combine like terms on the left-hand side of the equation to get (3βˆ’1)π‘₯+(4+4)π‘₯+6=02π‘₯+8π‘₯+6=0.

We can now divide the equation through by 2 to get π‘₯+4π‘₯+3=0.

This is now a quadratic equation in standard form; we can factor this equation by finding two numbers whose product is 3 and whose sum is 4. We note that 3Γ—1=3 and 3+1=4. This allows us to factor the quadratic to get (π‘₯+3)(π‘₯+1)=0.

We now have that the product of two factors is equal to 0. This can only happen if one of the factors is 0. Thus, we either have π‘₯+3=0 or π‘₯+1=0. We can solve each equation separately to get π‘₯=βˆ’3 or π‘₯=βˆ’1.

Hence, the functions are equal if π‘₯=βˆ’3 or π‘₯=βˆ’1.

There is one final concept we can consider for quadratic functions. We recall that we can rewrite a quadratic function by completing the square. In general, we can rewrite 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ in the form 𝑓(π‘₯)=π‘Ž(π‘₯+𝑝)+π‘žοŠ¨ for some constants 𝑝 and π‘ž.

We know that squaring a number gives a nonnegative number, so for any real value of π‘₯, we must have (π‘₯+𝑝)β‰₯0. We can use this to construct an inequality for 𝑓(π‘₯). We want to multiply the inequality through by π‘Ž, but we do not know the sign of π‘Ž.

If π‘Ž is positive, then π‘Ž(π‘₯+𝑝)β‰₯0.

Adding π‘ž to both sides of the inequality then gives π‘Ž(π‘₯+𝑝)+π‘žβ‰₯π‘ž.

In other words, 𝑓(π‘₯)β‰₯π‘ž for any input value of π‘₯. We can see that the function achieves this minimum output when π‘₯=βˆ’π‘.

Similarly, if π‘Ž is negative, then we need to reverse the inequality: π‘Ž(π‘₯+𝑝)≀0.

Adding π‘ž to both sides of the inequality then gives π‘Ž(π‘₯+𝑝)+π‘žβ‰€π‘ž.

In other words, 𝑓(π‘₯)β‰€π‘ž for any input value of π‘₯. We can see that the function achieves this maximum output when π‘₯=βˆ’π‘.

In our final example, we will find the minimum output of a quadratic function by completing the square.

Example 5: Finding the Minimum Value of a Quadratic Function by Converting It into Vertex Form

By writing the function 𝑓(π‘₯)=π‘₯+6π‘₯+5 in the form (π‘₯+𝑝)+π‘žοŠ¨, find the minimum output of the function.

Answer

To write a quadratic in the form (π‘₯+𝑝)+π‘žοŠ¨, we need to complete the square. First, we note that the coefficient of π‘₯ is 1; this means we need to find half of the π‘₯-coefficient. We have 12Γ—6=3. This gives us (π‘₯+3)+π‘žοŠ¨. If we expand this expression, we get (π‘₯+3)+π‘ž=π‘₯+6π‘₯+9+π‘žοŠ¨οŠ¨. Therefore, for this expression to be equal to π‘₯+6π‘₯+5, we need 9+π‘ž=5. Thus, π‘ž=βˆ’4 and we have 𝑓(π‘₯)=(π‘₯+3)βˆ’4.

We know that for any input value, (π‘₯+3)β‰₯0. This allows us to see that the smallest output of 𝑓(π‘₯) will occur when π‘₯+3=0. Therefore, the minimum output occurs when π‘₯=βˆ’3 and 𝑓(βˆ’3)=βˆ’4.

Hence, the minimum output of the function 𝑓(π‘₯) is βˆ’4.

In the previous example, we found that the minimum output of a quadratic function of the form (π‘₯+𝑝)+π‘žοŠ¨ occurs when π‘₯=βˆ’π‘ and its value is 𝑓(βˆ’π‘)=π‘ž. In the graph of 𝑦=𝑓(π‘₯), this will be the point (βˆ’π‘,π‘ž) and it is called the turning point, or vertex, of the quadratic graph.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • A function is a mathematical relationship that takes inputs and maps each input onto exactly one output.
  • If π‘₯ is the input, then the corresponding output of function 𝑓 is denoted as 𝑓(π‘₯).
  • The set of all possible input values of a function is called its domain.
  • The set of all possible output values of a function is called its range.
  • We call any input value π‘₯ for which 𝑓(π‘₯)=0 a root of the function 𝑓.
  • We can find the maximum or minimum output of a quadratic function by completing the square.

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