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| Student Learning Objectives: |
|---|
|
| | 20 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The diagram shows two graphs, each representing a different relation. Push the button to show a vertical line, then move the line along the graphs.
A function is a relation in which each input is assigned to exactly one output. The set of all possible inputs is called the domain of the function and the set of all possible outputs is called the range. If x represents the inputs and y the outputs of a function, it is often said that y is a function of x
or that y depends on x.
y = f(x)
This way of representing the dependent variable is called function notation. A function can be represented using a table, a mapping diagram, an equation, or a graph.
| Determining Whether a Relation Is a Function | |
|---|---|
| If represented as | Use |
| a set of coordinates or a table of values | a mapping diagram |
| a graph in the coordinate plane | the vertical line test |
Given a relation, a mapping diagram can be used to determine whether the relation is a function. For example, consider the relation given by the following set of coordinates. {(4,3),(0,-5),(-3,3),(8,0),(4,-5)} To figure out if a relation is a function, there are four steps to follow.
If the relation is given as a vertical table, the inputs are the values in the first column.
If the relation is given as a vertical table, the outputs are the values in the second column.
Here, the input 4 has two different outputs, 3 and -5. By definition, for a relation to be a function, every input must be assigned to exactly one output. As such, the given relation is not a function.
Kevin's teacher Maya is organizing a class trip to the Aquatic Wonders World Aquarium. She examines the aquarium's website and observes the number of visitors recorded over the past week.
| Day, x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Number of Visitors, N | 1300 | 1500 | 1800 | 1300 | 1900 | 2500 | 1900 |
The numbers in the first row represent the days of the week, with the number 1 indicating Monday.
Now each element in the first column can be connected to its corresponding element in the second column with an arrow.
Notice that the same number of visitors appears on certain days, but there are no two visitor counts on a single day. In other words, each input is assigned to exactly one output. Therefore, this relation is a function.
The vertical line test is a graphical method to determine whether a given relation is a function. For example, consider the following relations.
| Relation I | Relation II | Relation III |
|---|---|---|
| (y - 2)(y + 1) = x | y=x(x+1)(x-2) | |c|c|c|c|c|c|c|c| x & -2.25 & -1.75 & -1 & -1 & 0 & 1 & 2 y & -2 & 0 & 2 & -1 & 1 & -0.75 & 0.5 |
To determine whether the relations are functions, follow these two steps.
Notice that l_3 cuts the first graph at two different points. The line m_2 also passes through two different points. This means that neither Relation I nor Relation III is a function. However, all of the vertical lines drawn over Relation II only intersect the graph one time at most. Because of this, Relation II is a function.
Keep in mind that before stating whether a relation is a function, the vertical lines drawn have to cover the entire domain to ensure that no vertical line cuts the graph more than once.
Points with the same x-value belong to the same vertical line.
This is why drawing a vertical line and moving it across the graph reveals if the graph is a function or not.
Note that when determining whether a relation is or is not a function, it must be assumed that the graph of a relation continues without any significant change beyond the boundaries of the coordinate plane. If this were not the case, it could never be determined from a graph whether a relation is a function.
Maya and her students are about to start their visit. Their journey begins with a short seminar that gives cool facts about the aquarium. During the presentation, a graph that shows the fish population over the past 8 months was presented.
However, on the screen in front of Kevin, the axes were positioned in reverse. Consider the shape of the graph. It appears quite different.
Which of the two graphs represents a function?
No vertical line ever intersects the graph more than once. This indicates that the first graph represents a function. Now repeat the same procedure for the second graph.
Unlike the first graph, here, there is at least one vertical line that intersects the graph more than once. This graph does not represent a function.
The following applet displays a relation as a set of ordered pairs, a group of coordinate points in the coordinate plane, or a curve in the coordinate plane. Determine whether the relation is a function or not.
Function notation is a special way to write functions that explicitly shows that y is a function of x — in other words, that y depends on x. Function notation is symbolically expressed as y=f(x) and read y equals f of x.
Equations that are functions can be written using function notation.
ccc
Equation & & Function Notation [1ex]
y=-5x+4 & & f(x) = -5x+4
Notice that y has been replaced by f(x). In function notation, x represents an element of the domain and f(x) represents the element of the range that corresponds to x. When written in function notation, the expression that describes how to convert an input into an output — the right-hand side expression — is called the function rule.
Besides f, other letters such as g or h can be used to name the function. Similarly, letters other than x can name the independent variable.
The domain of a function is the set of all x-values, called inputs, for which the function is defined. As an example, consider the following functions. f(x) & = 3x [0.2cm] g(x) & = sqrt(x) [0.3em] h(x) & = 1/x Their domains can be written by analyzing the definition of each function.
| Function | Analysis | Domain |
|---|---|---|
| f(x) = 3x | Multiplying by 3 is defined for all real numbers. | All real numbers |
| g(x) = sqrt(x) | Square roots are not defined for negative numbers. | All non-negative numbers — that is, x≥ 0 |
| h(x) = 1/x | Dividing by zero is undefined. | All real numbers except 0 — that is, x≠ 0 |
The domain of a function can be determined through a variety of methods depending on how the function is represented.
The range of a function is the set of all y-values, called outputs, of the function. The range depends on both the domain and the function itself. For example, consider the following functions and their defined domains.
| Function | Domain |
|---|---|
| f(x)=2x | All integers |
| g(x)=x^2 | All real numbers |
| h(x)=4 | All real numbers |
The ranges of each function can be determined by analyzing the definition of each function along with the given domains.
| Function | Domain | Analysis | Range |
|---|---|---|---|
| f(x) = 2x | All integers | The function takes any integer input and produces an output that is an even number, as each input is multiplied by 2. | All even numbers |
| g(x) = x^2 | All real numbers | The function takes any real number input and produces an output that is a non-negative number, as each input is squared. | All non-negative numbers. That is, y≥ 0 |
| h(x) = 4 | All real numbers | The function takes any real number input and sends it to 4. | Only the number 4. That is, the range is {4} |
The method used to determine the range of a function can vary depending on how that function is represented.
Kevin, feeling thirsty after the seminar, buys water from a nearby vending machine. There are six types of products sold in the vending machine: orange juice, water, chocolate, donuts, sandwiches, and pizza.
Notice that the buttons on the vending machine are labeled with the first letters of the food items. When a button is pressed, the machine will give the corresponding food.
Is it a function? Yes
Range: {Chocolate, Donut, Orange Juice, Pizza, Sandwich, Water}
Now each element in the first column can be connected with an arrow to its corresponding element in the second column.
Notice that each element of the input set is paired with exactly one element of the output set, so this is one-to-one mapping. Therefore, this relation is a function. This conclusion makes sense because it is not logical that pressing a button would result in two different snacks being dispensed.
When sorted in alphabetical order, the domain and range of the function can be written as follows. Domain: &{C,D,O,P,S,W } [0.8em] Range: & { l Chocolate,Donut,Orange Juice, Pizza,Sandwich,Water }
Maya and her students start their tour of the aquarium with a visit to the famous dolphins. The students excitedly gather around the dolphin exhibit and watch the playful animals swimming in the water.
The guide shares interesting facts about these intelligent creatures, capturing the students' attention and sparking their curiosity about the animals in the aquarium. Kevin notes this information as a set of ordered pairs. K = {(1.9,120),(2.3,138),(2.5,140),(2,138) } Here, every pair of numbers represents the length of a dolphin in meters and its weight in kilograms. For example, the pair (2,138) indicates that a dolphin is 2 meters long and weighs 138 kilograms.
{(1.9,120),(2.3,138),(2.5,140),(2,138) } It is stated that the first coordinates indicate the lengths of dolphins, while the second coordinates represent their weights. This means that the x-axis corresponds to length and the y-axis corresponds to weight. The points can now be plotted.
Now the vertical line test can be used to determine whether the relation represents a function. Draw a vertical line and move it horizontally across the graph.
As the vertical line moves across the graph, no two points appear on the line at the same time. This indicates that the relation K is a function.
By listing the numbers from least to greatest and including each element only once, the domain and range of K are as follows. Domain ofK &= {1.9,2,2.3,2.5} Range ofK &= {120,138,140}
One way to understand functions is to think of functions as machines. Inputs are like raw materials that go through the processing stage of the function. Outputs are the final product. In this applet, four preset inputs are available. The machine specifically processes numbers between -100 and 100 as materials. Try plugging in a few values and see the outcomes!
In the context of functions, the input is often referred to as the independent variable because it can be chosen arbitrarily from the domain. Conversely, the output is called the dependent variable because its value depends on the value of the independent variable. For instance, if the price of oranges is $ 2.50 per pound, the total cost is determined by the product of the unit price and the weight in pounds. ccccc Cost & & Unit Price & & Weight [0.4em] y & = & 2.50 & * & x
As shown, the total cost of oranges depends on how many pounds of fruit are purchased. Therefore, the cost of oranges y is the dependent variable and the number of pounds purchased x is the independent variable.Evaluating a function involves determining the value of the function when its independent variable is set to a specific value. This is done by substituting the given input value for the variable and evaluating the function rule. As an example, consider the value of the following function when x=4. f(x)=3x+4 To evaluate a function for a particular input, there are two steps to follow.
As shown, when the input is 4, the output of the function is 16.
The dolphin trainer claims that the dolphin in front of him eats 15 kilograms of fish per day.
Graph:
Independent Variable: & d Dependent Variable: & k Therefore, the total amount of fish eaten by the dolphin k can be written as a function of d. It will be equal to the product of 15 and the number of days d. Equation k = 15 d This equation can also be expressed using function notation. In this case, it would be appropriate to use the notation k(d) as it shows that k is a function of d. Function Notation k(d) = 15d Next, make a table of values to help graph the function. For example, evaluate the function when d is 1, 3, 5, and 7.
| d | 15d | k |
|---|---|---|
| 1 | k=15( 1) | 15 |
| 3 | k=15( 3) | 45 |
| 5 | k=15( 5) | 75 |
| 7 | k=15( 7) | 105 |
Now plot the ordered pairs ( d, k) as points in a coordinate plane.
Finally, connect the points with a line. Note that negative numbers have no meaning in the context of the problem, so the graph will only be in the first quadrant.
k(d)=15d Substitute d = 15 into the function and evaluate the right-hand side.
As shown, when the input is 15, the output is 225. In the context of the dolphin, this means that the dolphin eats 225 kilograms of fish in 15 days.
Given a function, it is possible to find the input that produces a certain output. This is done by substituting the given output value for the dependent variable and then solving for the independent variable. For the following function, try finding the x-value for which f(x)=21. f(x)=4x-3 To find the input that produces a certain output, there are two steps to follow.
The aquarium has an exhibit showing a planned expansion. The exhibit shows a model of a new pool. The walls of the pool will be 1 meter thick on each side. The exterior of the pool will be x+2 meters long, 10 meters wide, and 6 meters tall.
The volume of a prism is the product of its dimensions, so the volume of the pool V can be written as a function of x in this case. V(x) = x * 8 * 5 ⇔ V(x) = 40x Use the fact that the pool is designed to hold 520 cubic meters of water can be used to find the value of x.
The amount of concrete needed to construct the pool can also be described as a function of x. The amount of cement needed is the difference between the volumes of the outer and inner dimensions of the pool.
Simplify the right-hand side of the function and substitute x = 13.
The construction of the pool requires 380 cubic meters of concrete.
R(d) = 520 - 2d To determine how many days it will take for the amount of water in the pool to drop to 490 cubic meters, substitute R(d) = 490 into the equation and solve for d.
This means that after 15 days, there will be 490 cubic meters of water left in the pool.
Consider the follow graph of a relation. If a vertical line is drawn between x_1=-4 and x_2=-2, it will intersect the graph multiple times, which means that the relation fails the vertical line test and therefore cannot be classified as a function. However, it is possible to transform the relation into a function by redefining certain characteristics, such as its domain.
As can be seen by its graph, the relation y=±sqrt(x) is not a function. However, it can be transformed into a function by setting restrictions or modifying some of its characteristics. Move x_1 and x_2 to see if restricting the domain will make the relation a function.
| Side Length | 2 | 5 | 8 | |
|---|---|---|---|---|
| Perimeter | 32 | 36 |
Write the numbers in ascending order.
| Side Length | 1 | 5 | 6 | |
|---|---|---|---|---|
| Area | 16 | 36 |
Write the numbers in ascending order.
We will write a function for the perimeter of a square. Recall that the perimeter of a square is the sum of its side lengths. If we let s be the side length of the square, then the perimeter of the square will be 4 times s. P(s) = 4s In this function, the independent variable is the side length s and the dependent variable is the perimeter P(s).
Let's use the function P(s) = 4s to complete the table.
| Side Length | 2 | 5 | 8 | |
|---|---|---|---|---|
| Perimeter | 32 | 36 |
To find the perimeter of a square with a side length of 2 units, we substitute 2 for s into the perimeter function.
We will follow the same procedure to determine P(5).
Finally, we need to find the side length of a square whose perimeter is 36. Let's substitute 36 for P(s) and solve for s.
Now we can fill in the table!
| Side Length | 2 | 5 | 8 | 9 |
|---|---|---|---|---|
| Perimeter | 8 | 20 | 28 | 36 |
We need to submit these numbers in ascending order. Let's list them in order. 8,9,20
We will write a function for the area of a square. Recall that the area of a square is its side length squared. If the side length of a square is s, then the area of the square will be s^2.
A(s) = s^2
In this function, the independent variable is the side length s and the dependent variable is the area A(s).
Just like we did in Part B, let's complete the table for the function A(s)=s^2.
| Side Length | 1 | 5 | 6 | |
|---|---|---|---|---|
| Area | 16 | 36 |
To find the area of a square with a side length of 1 unit, we substitute 1 for s into the area function.
Now let's substitute 5 for s to find the area of a square with a side length of 5 units.
Finally, we need to find the side length of a square whose area is 16 square units. Let's substitute 36 for A(s) and solve for s.
We can ignore the negative solution here because length cannot be negative. We now have all the missing numbers!
| Side Length | 1 | 4 | 5 | 6 |
|---|---|---|---|---|
| Area | 1 | 16 | 25 | 36 |
We need to submit these numbers in ascending order, so let's order our values. 1,4,25
Now we can use the points from the tables in the previous parts to draw the graphs of the functions.
We can see that both graphs start from the origin, so Statement IV is true. IV. Both graphs pass through the origin. ✓ Notice also that the graph of the perimeter is a line, but the graph of the area is a curve. This indicates that the perimeter is a linear function of the square's side length, while the area is a non-linear function of the side length. Statement III is also true! rl III. & The graph of the second function is & non-linear. ✓ Since it is a non-linear function, the area function does not have a constant rate of change. It increase more and more rapidly the as the value of s increases.
The rate of change varies across the graph. We know from this that statement II is false. cl II.& The second function has a constant & rate of change. * Lastly, let's find the slope of the first function.
We can see that the function has a constant rate of change between any pair of its points and that this rate of change is equal to 4. Statement I is also correct. rl I.& The graph of the first function is a & straight line with a slope of 4. ✓ As a result, Statements I, III, and IV are true.
Let's draw the graphs of the functions.
| First Function | Second Function |
|---|---|
| P(s) = 4s | A(s) = s^2 |
The domains of both functions consist of non-negative real numbers because lengths cannot be negative. Similarly, their ranges are also all non-negative real numbers.
| P(s) = 4s | A(s) = s^2 | |
|---|---|---|
| Domain | s ≥ 0 | |
| Range | P(s) ≥ 0 | A(s)≥ 0 |
Let's keep these values in mind as we draw the functions, starting with P(s)=4s. We are given four points in the table from Part B. (2,8), (5,20), (8,28), and(9,36) In our coordinate plane, the x-axis will represent the independent variables, while the y-axis will represent the dependent variables. Let's plot the points and connect them.
Now we will graph A(s)=s^2 by plotting the points found in Part C. (1,1), (4,16), (5,25), and(6,36) We can plot the points and draw a curve through the points.
Let's see these graphs on the same plane.