### Video Transcript

In this lesson, weβll learn how to
graph and analyze a piecewise-defined function and study itβs different
characteristics.

Sometimes we come across a function
that requires more than one function in order to obtain the given output. We call this a piecewise
function. And these are functions in which
more than one function is used to define the output over different parts of the
domain. Each subfunction is then
individually defined over its own domain.

For instance, letβs take π of π₯,
which is a piecewise-defined function. And itβs given by two π₯ plus one
if π₯ is less than negative one and three π₯ if π₯ is greater than or equal to
negative one. We see that for values of π₯
strictly less than negative one, we use the function π of π₯ equals two π₯ plus
one. For instance, π of negative two
would be evaluated as two times negative two plus one, which is negative three. But then, with values of π₯ greater
than or equal to negative one, we use the function π of π₯ equals three π₯. So, for instance, π of zero would
be three times zero, which is simply zero.

We need to make sure that we can
identify the graphs of these functions as well as sketch them and define the
function when given a graph. So letβs see what that would look
like with an example.

What kind of function is depicted
in the graph? Is it (A) an even function, (B) a
logarithmic function, (C) a piecewise function, or (D) a polynomial function?

Letβs begin by providing a
definition for each of these terms. If a function π of π₯ satisfies
the criteria π of negative π₯ equals π of π₯ for all π₯ within the domain of the
function, then itβs said to be even. We also know that these functions
have reflectional symmetry about the π¦-axis or the line π₯ equals zero. Then we look at logarithmic
functions. These are of the form log base π
of π₯. Theyβre the inverse to exponential
functions. Itβs worth also noting that the
domain of a logarithmic function is the set of positive real numbers, and then the
range is the set of all real numbers.

Then we have piecewise
functions. And these are functions in which
more than one subfunction is used to define the output over different parts of the
domain. Each subfunction is then
individually defined over its own domain. Finally, we have polynomial
functions. Now, these are ones made up of the
sum or difference of constant terms, variables, and positive integer exponents such
as two π₯ cubed plus five π₯. The domain of a polynomial function
is the set of all real numbers. And we know that their graphs are
both continuous and smooth. In other words, there are no gaps
in the graph, which we might call a discontinuity, and there are no sharp corners on
the graph. So letβs look at our graph and
compare these definitions to it.

Firstly, we note that there is no
symmetry about the line π₯ equals zero. And so we see that the function
cannot be an even function. We also see that our graph is
certainly defined for values of π₯ greater than or equal to negative 10 and less
than or equal to eight. It might even be defined outside of
this interval, but we canβt be sure at this stage. What this does tell us is that the
domain is different to that of a logarithmic function, which is simply positive real
numbers. And so our graph cannot be the
graph of a logarithmic function.

And so weβre limited to piecewise
functions and polynomial functions. Now, in fact, we said that the
graph of a polynomial function is smooth; there are no sharp corners. And that means itβs differentiable
at every point. We can quite clearly see that our
graph has two sharp corners. And so itβs not smooth. Our graph cannot therefore be a
polynomial function. And so weβre left with (C) a
piecewise function. In fact, if we look carefully, we
see that there are three parts to this piecewise function. The first part is for values of π₯
less than negative three. We then have values of π₯ between
negative three and zero. And finally, our third subfunction
is values of π₯ greater than zero and certainly, from what we can see, up to
eight.

So weβve seen what the graph of a
piecewise-defined function might look like. Weβre now going to see how we might
determine the domain of a piecewise-defined function given its graph.

Determine the domain of the
function represented by the given graph.

Letβs begin by recalling what we
mean by the word βdomain.β The domain of a function is the set
of possible inputs that will yield real outputs. In other words, itβs the set of
π₯-values that we can substitute into the function. Now, when we look at the graph of a
function, we can establish its domain by considering the spread of values in the
π₯-direction. We will need to be a little bit
careful because if we look at the graph of our function, we notice we have these
empty circles. Sometimes called an open circle, it
tells us that the function canβt be defined by this part of the line at this
point. Since we have a piecewise-defined
function, that is, a function thatβs defined by more than one subfunction, weβll
find the domain of each subfunction first.

We see we have one subfunction that
takes values less than negative four. Then we have another subfunction
that takes values greater than negative four. Since the initial or starting point
of each line of our subfunction is represented by that open circle, we see that π₯
equals negative four is not defined within the domain of our function. And so the domain is actually going
to be the set of real numbers not including this number. One way to represent that would be
to use inequality notation and write π₯ can be less than negative four and π₯ can be
greater than negative four.

Alternatively, we can use set
notations, where this funny-looking β represents the set of real numbers and these
squiggly brackets or braces tell us the set contains one single element, and thatβs
negative four. And so the domain of this function
is the set of real numbers not including the set containing the element negative
four.

Now that weβve established that the
spread of values in the π₯-direction tells us the domain of the function, letβs look
at how to find the range of a piecewise-defined function.

Find the range of the function.

Letβs begin by recalling what we
mean by the range of a function. Just as the domain is the set of
possible inputs to our function, the range is the set of possible outputs. In other words, itβs the set of
π¦-values we achieve when the domain of π₯-values have been substituted into the
function. This means that graphically weβre
looking at the spread of values in the π¦-direction to help us calculate the range
of the function.

Looking at the graph, we see that
the values of π¦ begin at negative one. And thatβs when we input π₯-values
less than or equal to four. Then at π₯ equals four, the values
of π¦ steadily increase, and this arrow here tells us that the increase to β. We can therefore say that the
range, the set of possible outputs, is all values of π¦ greater than or equal to
negative one. To use set notation to define the
same interval, we use the left-closed right-open interval from negative one to
β. Note that the round bracket tells
us that β isnβt really a defined number. And so the range of this function,
which is the set of possible π¦-values, is the left-closed right-open interval from
negative one to β.

Up until this stage, weβve
considered what it means for a function to be piecewise-defined and how to determine
its domain and range from its graph. Weβll now look at how to define the
entire piecewise function given the graph of that function.

Give the piecewise definition of
the function β whose graph is shown.

Weβre told that the graph weβre
given is a piecewise-defined function. And we know that a
piecewise-defined function is made up of multiple subfunctions. In fact, by looking at the graph of
this function, we might notice that there are going to be two subfunctions. These are also going to be linear
since the graph of each subfunction is a straight line. And so this means that if we can
calculate the slope π and find one point which each line passes through, we can use
the equation π¦ minus π¦ one equals π times π₯ minus π₯ one to find the equation of
each line.

Letβs begin with the first part of
this subfunction. We notice that this subfunction is
defined up to and including π₯-values of two. So that will give us a hint as to
what its domain is. Then we could use the formula for
slope π equals π¦ two minus π¦ one over π₯ two minus π₯ one to find the slope of
this line. Alternatively, we can use the
triangle method. Choosing a point on the line, in
this case, the π¦-intercept, and then moving exactly one unit to the right, we see
we have to move one unit down to get back to our point on the line. That means the slope of this line
must be negative one. It also passes through the point
zero, three. Remember, this is the π¦-intercept
of the line.

And so substituting everything we
know about this first function into our equation for a straight line, we get π¦
minus three equals negative one times π₯ minus zero. Distributing the parentheses on the
right-hand side, and we get negative π₯. And then weβre going to make π¦ the
subject by adding three to both sides. Remember, π¦ is the output. So itβs going to be β of π₯
essentially. And so the first line is defined by
the equation π¦ equals three minus π₯.

Letβs repeat this process with the
second line. Now we always need to be a little
bit careful using the triangle method for fractional slope values. In this case, when we pick a point
on the line, move one unit to the right, we then have to move a half a unit up to
get back to our point on the line, meaning that the slope of our second line is
one-half. To convince ourselves that this is
true, we could choose the two points given on the line, which have coordinates four,
two and six, three, respectively. Then π¦ two minus π¦ one over π₯
two minus π₯ one, which is change in π¦ over change in π₯, is three minus two over
six minus four, which is one-half as we saw.

Then letβs pick this point. We know our line passes through the
point with coordinates two, one. And so the equation of our line is
π¦ minus one equals a half times π₯ minus two. Then when we distribute the
parentheses on the right-hand side, we get that one-half times π₯ minus two is the
same as one-half π₯ or π₯ over two minus one. We can then finally add one to both
sides, eliminating that negative one. And so the second line has the
equation π¦ equals π₯ over two. Now that we have the equations that
represent our subfunctions, weβre going to pop this back together using piecewise
definition.

β is given by three minus π₯ for
values of π₯ less than two. And π₯ over two of π₯ is greater
than or equal to two, which of course is the same as writing two is less than or
equal to π₯. Note, of course, that the function
could have been defined at the point π₯ equals two by either subfunction. Itβs generally convention that we
choose the second function to define that point, although it would have been just as
correct to write three minus π₯ if π₯ is less than or equal to two and π₯ over two
if π₯ is greater than two. The piecewise definition of β is
three minus π₯ if π₯ is less than two and π₯ over two if two is less than or equal
to π₯.

In our final example, weβll look at
how to define a piecewise function given a graph that also includes a
discontinuity.

Give the piecewise definition of
the function π whose graph is shown.

Weβre told that this graph
represents the graph of a piecewise-defined function. And this makes a lot of sense. We see that itβs made up of three
different parts. We have a linear function over here
given by a single straight line and a other linear function here given by another
straight line. But then we have something really
strange here. We have a single dot at this
point. And weβll consider what that means
for our piecewise definition in a moment.

For now, weβre going to begin by
finding the equation of our two straight lines. We use the formula π¦ minus π¦ one
equals π times π₯ minus π₯ one, where π is the slope of the graph and π₯ one, π¦
one is a single point it passes through. Then the slope is given by change
in π¦ divided by change in π₯, which is π¦ two minus π¦ one over π₯ two minus π₯
one. And so letβs begin by finding the
slope of our first line, we can choose any two points on this line. Letβs choose the points with
coordinates negative three, six and one, two. Then change in π¦ divided by change
in π₯ is six minus two over negative three minus one. Of course, we could write two minus
six over one minus negative three and get the same result.

This gives us four divided by
negative four, which is negative one. Then substituting everything we
know about our first straight line into the formula for a straight line, and we get
π¦ minus six equals negative one times π₯ minus negative three. Distributing the parentheses on the
right-hand side, and this simplifies to negative π₯ minus three. Finally, we add six to both sides,
and we find π¦ is equal to negative π₯ plus three or three minus π₯. And so for values of π₯ strictly
less than two, we can use the equation π¦ equals three minus π₯ to draw its
graph.

Next, we choose two points on our
second line. Letβs choose the points with
coordinates four, four and six, five. Change in π¦ divided by change in
π₯ here is five minus four over six minus four, which is equal to one-half. So the slope of our second line is
one-half. Substituting then π equals
one-half and π₯ one, π¦ one equals four, four into our formula for a straight line,
and we get π¦ minus four equals a half times π₯ minus four. And that right-hand side simplifies
to π₯ over two minus two. Then we add four to both sides. And we see that our second line has
the equation π¦ equals π₯ over two plus two. Now, this time, thatβs for values
of π₯ strictly greater than two. So we now have the equations of our
two straight lines. These are three minus π₯ if π₯ is
less than two and π₯ over two plus two if π₯ is greater than two.

Weβre not finished though; there
was a third subfunction that weβre interested in. And this subfunction is represented
graphically by a single point. This point has coordinates two,
two. In other words, if π₯ is exactly
equal to two, the function yields an output of two. And so that is our third
subfunction. Noting that we can alternatively
write the domain on our third line as two is less than π₯, we now have the piecewise
definition of our function. Itβs π of π₯ is equal to three
minus π₯ if π₯ is less than two, two of π₯ is equal to two, and π₯ over two plus two
if two is less than π₯.

Weβll now recap some of the key
points from our lesson. In this lesson, we learned that a
piecewise-defined function is a function defined by multiple subfunctions. We then saw how each one of those
subfunctions is defined over a given interval of the main functions domain. We can call that a subdomain. And we saw how by carefully
considering their definition, we can identify the domain and range from a function
and from its graph.