### Video Transcript

In this lesson, weβre going to
learn how to decide whether a function is even, odd, or neither, both by considering
the graph of that function and from the equation itself. Weβre going to begin, though, by
looking at the definition of the parity of a function. Parity describes whether a function
is even or odd. But what does it actually mean for
that function to be even or odd? Well, to decide whether a function
is odd or even or indeed neither, the first thing we do is we check the domain of
the function. We need the domain to be centered
at the origin. If we answer no to this question,
then the function can be neither odd nor even.

For instance, letβs imagine the
domain of our function is the open interval from negative eight to eight. π₯ equals zero is exactly halfway
between these, and so the domain is indeed centered at π₯ equals zero. But what about the domain the
left-closed right-open interval from negative four to two? Well, no. π₯ equals zero isnβt exactly
halfway through here. And so a function with this domain
would be neither odd nor even. However, if weβre able to answer
yes to that earlier question, then we could say that the function is even if π of
negative π₯ is equal to π of π₯. But itβs odd if π of negative π₯
is equal to negative π of π₯. And if the function π of π₯
satisfies neither of these criteria, then it is neither odd nor even.

And so one technique we have to
check whether a function is odd or even or indeed neither is to substitute negative
π₯ into the function and see what we get. However, there is also a geometric
approach. Letβs see where that approach comes
from.

Determine whether the function
represented by the following figure is even, odd, or neither even nor odd.

And then we have a graph of the
function shown. So letβs recall how to check the
parity of a function, how to check whether itβs even or odd. Well, the first thing we do is ask
ourselves, is the domain of this function centered at π₯ equals zero? We might recall that the domain of
a function is the set of possible inputs, the set of values of π₯, that we can
substitute into the function. And we can read that domain from
the graph.

Now we do need to be a little bit
careful because the graph doesnβt actually appear to be defined at π₯ equals
zero. In fact, the domain is the union of
the left-closed right-open interval from negative eight to zero and the left-open
right-closed interval from zero to eight. This is centered at π₯ equals
zero. Zero is exactly halfway through
this domain. And so we can now say yes to this
question. And we are able to move on to the
next part.

We can say that if π of negative
π₯ is equal to π of π₯, the function is even, and itβs odd if π of negative π₯ is
equal to negative π of π₯. Well, one way we can establish
whether either of these is true is to choose a value of π₯. For instance, letβs use the point
π₯ equals five. When π₯ is equal to five, the value
of our function, the π¦-value, is negative one. So π of five is negative one. And then this means that negative
π₯ must be negative five. And so we need to read the π¦-value
when π₯ is negative five. π of negative five is also
negative one. So it does look like this might be
an even function. But letβs check with another
value.

Letβs choose π₯ equals one. π of one is roughly equal to
negative 4.1. Then negative π₯ will be equal to
negative one. And once again, π of negative one
is roughly equal to negative 4.1. And so for the two values weβve
tried, π of negative π₯ is equal to π of π₯. But actually, if we look carefully,
we see that that function itself has reflectional symmetry about the π¦-axis. And so indeed, every value of π of
π₯ must be equal to every value of π of negative π₯. And so we can say that the function
itself must be even.

In fact, we can generalize
this. We can say that an even function
has reflectional symmetry about the π¦-axis. Take, for instance, the graph of π
of π₯ equals cos of π₯. This is completely symmetrical
about the π¦-axis, and so cos of π₯ must be an even function. This is a general result that we
can quote. The same cannot be said for an odd
function, though. An odd function does have symmetry,
but it has rotational symmetry about the origin order two. In other words, itβs unchanged when
itβs rotated by 180 degrees about the origin. Take, for instance, the function π
of π₯ equals sin π₯. If we rotate the graph of this
function 180 degrees about the origin, it will end up looking exactly the same as it
did originally.

The same can be said for the
tangent function. Despite its asymptotes, if we
rotate this 180 degrees about the origin, about zero, zero, it will look exactly the
same, as long as, of course, its domain is also symmetrical. Its domain must be centered at π₯
equals zero. Now that we have these graphical
representations, letβs have a look at another example.

Is the function represented by the
figure even, odd, or neither even nor odd?

Remember, we can check the parity
of a function by considering its graph. Before we do, though, we do need to
decide whether the domain is centered at π₯ equals zero. If the answer is yes, then we move
on to the next stage. But if itβs not, the function
cannot be even nor odd. So letβs find the domain of our
function. We remember that the domain is the
set of possible inputs to the function, the set of values of π₯ that go into the
function. The smallest value of π₯ that is in
our function is π₯ equals two, and the largest possible value of π₯ is π₯ equals
six. And so we see that the domain is
the closed interval from two to six. This domain is actually centered at
π₯ equals four. The halfway point is π₯ equals
four. And so actually, weβre answering no
to this first question. And so the function is neither even
nor odd.

But letβs have a look at a common
misconception here. When we think about the graphical
representation of functions, we know that these functions are even if they have
reflectional symmetry about the π¦-axis. And theyβre odd if they have
rotational symmetry order two about the origin. Now, our graph does appear to have
some rotational symmetry. If we take the center to be four,
one, then it does indeed have rotational symmetry order two. If I rotate that graph 180 degrees,
it will look exactly the same. But this is not about the
origin. Its center is four, one. And so that confirms to us that
this graph is neither even nor odd.

In our next example, weβll look at
how to decide whether a function is even or odd given the equation.

Is the function π of π₯ equals π₯
to the fifth power times tan of six π₯ to the fourth power even, odd, or neither
even nor odd?

Letβs recall how we check the
parity of a function. The first thing we do is check the
domain of the function. We need that to be centered at π₯
equals zero. Then, if the answer is no, we can
say that the function is neither even nor odd without performing any further
tests. If the answer is yes, though, we
say that it will be even if it satisfies π of negative π₯ equals π of π₯. And it will be odd if it satisfies
π of negative π₯ equals negative π of π₯. Then, of course, if it satisfies
neither of these, it will be neither even nor odd.

So letβs think about the domain of
our function. Our function is the product of two
functions. Itβs the product of π₯ to the fifth
power and tan of six π₯ to the fourth power. And so the domain of π of π₯ will
be the intersection of the domains of the respective parts of the function. Well, π₯ to the fifth power is a
polynomial, so its domain is the set of real numbers or the open interval from
negative β to β. But what about the domain of the
trigonometric part? Well, itβs all real numbers, except
those that make cos of six π₯ equal to zero. But since the values of π₯ that
make cos of six π₯ equal to zero are symmetrical about the π¦-axis, then we can say
that the domain of tan of six π₯ to the fourth power must be centered at π₯ equals
zero.

Since both domains are centered at
π₯ equals zero, then we can answer yes to this first question, and weβre able to
move on. We now see that itβs even if π of
negative π₯ is equal to π of π₯ and odd if itβs equal to negative π of π₯. And so letβs evaluate π of
negative π₯. To do so, we replace each instance
of π₯ in our original function with negative π₯. And we get π of negative π₯ is
negative π₯ to the fifth power times tan of negative six π₯ to the fourth power. Weβll evaluate each part in
turn. Letβs begin with negative π₯ to the
fifth power. Since the exponent is odd, when we
multiply this out, weβre going to get a negative result. Negative π₯ to the fifth power is
as shown.

But what about the tan
function? Well, we can actually quote the
result that tan of π₯ is odd, meaning that tan of negative π₯ is equal to negative
tan of π₯ and, in turn, the tan of negative six π₯ is equal to negative tan of six
π₯. But of course, weβre raising this
to the fourth power. Weβre raising it to an even
exponent. And we know when we raise a
negative number to an even exponent, the result is positive. And so tan of negative six π₯ to
the fourth power is just tan of six π₯ to the fourth power.

And so π of negative π₯ is
therefore equal to negative π₯ to the fifth power times tan of six π₯ to the fourth
power. So does this satisfy either of our
criteria, is it even or odd? Well, yes. If we look at it carefully, we see
itβs the same as negative π of π₯. π of negative π₯ is equal to
negative π of π₯. And so the function must be
odd.

Letβs now consider how this process
might work, given a piecewise function.

Determine whether the function π
is even, odd, or neither, given that π of π₯ is equal to negative nine π₯ minus
eight if π₯ is less than zero and nine π₯ minus eight if π₯ is greater than or equal
to zero.

Letβs recall the steps that allow
us to check the parity of a function, in other words, whether itβs even or odd. Firstly, we establish whether the
domain is centered at π₯ equals zero. If the answer to this is yes, then
we can say that it is even if π of negative π₯ equals π of π₯ and odd if π of
negative π₯ equals negative π of π₯. Well, we see by looking at the
values here that the domain of our function is simply the set of real numbers. And of course, that set is very
obviously centered at π₯ equals zero. And so we answer yes to the first
question. And so now weβre going to consider
the second criteria.

Since this is a piecewise function,
we need to be really careful. When π₯ is less than zero, weβre
working with π of π₯ equals negative nine π₯ minus eight. And when itβs greater than or equal
to zero, weβre working with nine π₯ minus eight. And so letβs consider both parts of
our function individually. Letβs take negative nine π₯ minus
eight. And weβre now going to find π of
negative π₯. It will be negative nine times
negative π₯ minus eight. But of course, a negative
multiplied by a negative is a positive. So π of negative π₯ is equal to
nine π₯ minus eight.

Notice that that is equal to the
second part of our function. And itβs absolutely fine that itβs
equal to the other part, since weβve changed the sign of the value of π₯. And so weβre moving on to the other
side of the piecewise function. And so for this part, yes, π of
negative π₯ is equal to π of π₯. So that part certainly appears to
be even. But letβs check the second
part. Letβs take π of π₯ equals nine π₯
minus eight and then find π of negative π₯. Itβs nine times negative π₯ minus
eight, which is negative nine π₯ minus eight. Weβve changed the sign of the value
of π₯, and weβve ended up with π of π₯ when π₯ is less than zero. And so this part of the function is
also even. We can say that π of negative π₯
is equal to π of π₯. And we can therefore say that our
piecewise function is even.

In our final example, weβll look at
how the domain can affect the parity of a function.

Determine whether the function π
of π₯ equals nine π₯ cubed is even, odd, or neither even nor odd, given that π maps
numbers from the left-open right-closed interval from negative seven to seven onto
the set of real numbers.

Remember, to check the parity of a
function, we first check whether its domain is centered at π₯ equals zero. If not, then the function will be
neither even nor odd. But if it is, we can say itβs even
if it satisfies π of negative π₯ equals π of π₯. And itβs said to be odd if it
satisfies π of negative π₯ equals negative π of π₯. Now the domain of our function is
the set of possible inputs. Itβs the left-open right-closed
interval from negative seven to seven. And at first glance, it does look
like our domain is centered at the origin. But notice that we have a curly or
round bracket here and a square bracket here. This means that our domain doesnβt
include the value of π₯ equals negative seven, but it does include the value of π₯
equals seven. And so itβs kind of lopsided. Its center will be slightly to the
right of zero. And so we can say that this
function is in fact neither even nor odd.

Notice that had we failed to check
this, we might have concluded that the function is odd. And this is because π of negative
π₯ here is nine times negative π₯ cubed. But negative π₯ cubed is negative
π₯ cubed. And so π of negative π₯ is the
same as negative nine π₯ cubed, which is the same as negative π of π₯. And so this shows us how important
it is that we do check the domain. You might also like to think about
this graphically. A function is odd if it has
rotational symmetry order two about the origin. In other words, if we rotate it 180
degrees about zero, zero, it will look exactly the same. This cannot be the case here
because when we rotate it, we take the point that weβre including and map it onto
the point that weβre not including and of course vice versa. And so it wonβt look exactly the
same. And so we determine that our
function under these restraints is neither even nor odd.

Weβll now clarify the key points
from this lesson. Parity describes whether a function
is even or odd. And we recall that, to check the
parity, we first check whether the domain of the function is centered at π₯ equals
zero. If the answer to this is no, then
the function can be neither even nor odd. But if the answer is yes, we say
that the function is even if it satisfies π of negative π₯ equals π of π₯ and itβs
odd if it satisfies π of negative π₯ equals negative π of π₯. Once again, if it satisfies neither
of these, it can be neither even nor odd.

We saw that we can also use the
graph of the functions to determine whether they are even or odd. An even function will have a
reflectional symmetry about the π¦-axis, whereas odd functions have rotational
symmetry order two about the origin. Theyβll be unchanged when theyβre
rotated by 180 degrees about zero, zero.