### Video Transcript

In this video, we will learn how to
identify the domain and range of functions from their equations.

We begin by recalling that a
function π assigns to members of one set, for example, π₯, values in a second set
π. We think of π₯ as the source of
inputs to the function and π as the target of its outputs. For every input π₯ that exists in
set π, we have an output π of π₯ that exists in set π. This leads us to our definitions
for the domain and range. The domain of π is the set of all
inputs π₯ that exist in set π for which π of π₯ is defined. And the range of π is the set of
all outputs π of π₯ as π₯ varies over the domain.

In our first example, we will
identify the ordered pairs of a function given the domain, range, and the equation
of the function.

π and π are two sets of numbers
where set π contains the values 10, one, two, and eight and set π contains the
values 12, seven, 60, six, 48, and four. The function π of π₯ is equal to
six π₯, where π maps elements of π onto elements of π. Find the ordered pairs that satisfy
the function and its range.

One way of answering this question
is to consider a mapping diagram that represents the function π. There are four elements in set
π. They are the numbers 10, one, two,
and eight. Set π contains six elements, the
values 12, seven, 60, six, 48, and four. We are told that the function π of
π₯ is equal to six π₯. We recall that the domain of π is
the set of inputs for which π of π₯ is defined. And the range of π₯ is the set of
corresponding outputs.

Our first value in set π is
10. And since π of π₯ is equal to six
π₯, π of 10 is equal to six multiplied by 10. This is equal to 60 and means that
an input of 10 gives an output of 60. The first ordered pair that
satisfies the function is 10, 60. π of one is equal to six
multiplied by one, which equals six. This gives us a second ordered pair
one, six. Repeating this for the third value
in set π, we have π of two is equal to 12, giving us a third ordered pair two,
12. Finally, π of eight is equal to
48. And we have a fourth ordered pair
eight, 48. The four ordered pairs that satisfy
the function are 10, 60; one, six; two, 12; and eight, 48.

We are also asked to give the range
of the function. As already mentioned, this is the
set of all outputs of π of π₯. The range of the function is
therefore the set of the four values 60, six, 12, and 48. And we have now answered the two
parts of this question.

In our next question, we will
identify the range of a function represented by a Cartesian diagram.

The figure below shows the graph of
a function π. What is the range of the
function?

We can see from the figure that the
function π contains four ordered pairs. They are one, negative two; two,
negative three; three, zero; and four, negative three. We are asked to find the range of
this function.

We begin by recalling that the
domain of a function is the set of all input or π₯-values for which the function π
of π₯ is defined. These are the π₯-values of our
ordered pairs. The numbers one, two, three, and
four. The range of a function is the set
of all output values. These are the π¦-coordinates in our
ordered pairs. We have the values negative two,
negative three, zero, and negative three once again.

Despite the fact that this value
appears twice, we only need to include it once when writing the range. The range of the function shown in
the figure is the set of values negative two, negative three, and zero. And writing these in ascending
order, we have negative three, negative two, zero.

We could also find these values
directly from the figure by drawing horizontal lines through all the points on the
graph. These three lines intersect the
π¦-axis at negative three, negative two, and zero.

In our next two examples, we will
identify the domain and range of a function given its graph.

Determine the domain of the
function represented in the graph below.

We begin by recalling that the
domain of a function π is the set of all π₯-values or inputs for which π of π₯ is
defined. The solid dot at the far left of
our curve lies at the point with coordinates four, one. This means that when π₯ equals
four, π¦ is equal to one. And using function notation, π of
four equals one.

The arrow at the other end of our
curve tells us that the function is defined for all real numbers to the right of
this point. And we can therefore conclude that
the function is defined for all values of π₯ greater than or equal to four. Had the dot at the point four, one
been hollow, then π of π₯ wouldβve been defined on the strict inequality π₯ is
greater than four. This inequality π₯ is greater than
or equal to four is the domain of the function. And we can also write this using
interval notation. The domain of the function
represented in the graph is the left-closed, right-open interval from four to β.

Whilst we could also find the range
of the function from the graph, we will do this in the next example.

Determine the range of the function
represented by the graph below.

We begin this question by recalling
that the range of a function π is the set of all its outputs or π¦-values. We need to find all the π¦-values
that are represented by the curve. Our graph is in the shape of a
parabola. And it appears that the function is
quadratic. It has a maximum at the point with
coordinates negative seven, one. This means that π of negative
seven is equal to one, and the maximum value of the function is one. There is no π₯-value such that π
of π₯ is greater than one.

Looking at the graph, it does
appear that the parabola takes on all values less than or equal to one. And we can therefore conclude that
the range of the function is the set of real π¦-values such that π¦ is less than or
equal to one. This can also be written using
interval notation as the left-open, right-closed interval from negative β to
one.

In our final two examples, we will
identify the domain and range of a function given its equation.

Determine the domain of the
function π of π₯ which is equal to the square root of the absolute value of π₯
minus 33.

We begin by recalling that the
domain of a function is the set of inputs for which π of π₯ is defined. We know that the square root is
only valid for nonnegative numbers. This means that the absolute value
of π₯ minus 33 must be greater than or equal to zero. Adding 33 to both sides of our
inequality, we have the absolute value of π₯ is greater than or equal to 33. Recalling the graph of the equation
π¦ is equal to the absolute value of π₯ and then drawing the horizontal line π¦ is
equal to 33, we see that the absolute value of π₯ is greater than or equal to 33
when π₯ is less than or equal to negative 33 or π₯ is greater than or equal to
33. These inequalities can be rewritten
as the union of two intervals: the left-open, right-closed interval from negative β
to negative 33 and the left-closed, right-open interval from 33 to β.

Whilst this is a perfectly valid
answer for the domain of the function, this is not an interval. We can, however, write it as the
complement of one. The domain of the function is the
set of real values minus the set on the open interval from negative 33 to 33.

We could once again also find the
range of this function. However, weβll do this in one final
example.

If π maps elements on the closed
interval from two to 21 to the set of real numbers, where π of π₯ is equal to three
π₯ minus 10, find the range of π.

In this question, we are given a
linear function π of π₯, which is equal to three π₯ minus 10. We know that the graph of π¦ is
equal to three π₯ minus 10 is a straight line as shown. We know that the domain of a
function is the set of input or π₯-values. And in this question, weβre told
that these lie on the closed interval from two to 21. The function π of π₯ is therefore
defined for all values between the two points shown.

We can calculate the corresponding
π of π₯ values by substituting π₯ equals two and π₯ equals 21 into the
function. When π₯ is equal to two, π of π₯
is equal to three multiplied by two minus 10. This is equal to negative four. Likewise, when π₯ is equal to 21,
π of π₯ is equal to three multiplied by 21 minus 10. This is equal to 53. Since the range of a function π is
the set of outputs or π¦-values, we can conclude that π of π₯ or π¦ is greater than
or equal to negative four and less than or equal to 53. Using interval notation, the range
of the function π on the given domain is the closed interval from negative four to
53.

We will now summarize the key
points from this video. In this video, we introduced the
definitions for the domain and range of a function. The domain of a function is the set
of inputs or π₯-values for which the function is defined. And the range is the set of
corresponding outputs or π¦-values. We considered examples where we
identified the domain and range of a function given a mapping, graph, or the
equation of the function.