### Video Transcript

Graphs of Inverses of Functions

In this video, we will learn how to
use a graph to find the inverse of a function and analyze the graphs for the inverse
of a function. To begin, letβs consider what is
meant when we say the inverse of a function. A function π can be inverted by
the function π when π of π₯ is equal to π¦ and π of π¦ is equal to π₯. We can think of these two functions
as undoing the actions of each other. This means that if we were to apply
π and then we were to apply π, it would be the same as doing nothing and the
result would be our starting value. We can mathematically represent
this as a function composition as follows. Here we have considered a single
input value π₯ and a single output value π¦. However, if π of π₯ is a
one-to-one function, it can be inverted over its entire domain.

Weβll return to this βone to oneβ
constraint later in this video. But for now, we should understand
that when π of π₯ is a one-to-one function, we can find a uniquely defined inverse
function. Rather than calling this function
π, itβs common to use the following notation. We would read this as π
inverse. We should also note that π is the
inverse function of π inverse. So the relationship goes both
ways. Returning to our function
composition, we can say that applying both π and π inverse in either order will
give us the same value that we started with.

Before moving forward, letβs
highlight one quick point about notation. We should be very careful not to
mistake this minus one for an exponent. As a very quick example, if we had
some function π of π₯ equals two π₯ minus three, π inverse would not be two π₯
minus three all raised to the power of minus one. This video will not go into any
detail on how to algebraically find the inverse of a function. But itβs enough to say this common
misconception should be avoided. Okay, now that weβve recapped the
inverse of a function, letβs see how it relates to graphs. Consider some invertable function
π. When graphing functions, we usually
consider the π₯-axis as the domain of the function, which we can think of as an
input. We consider the π¦-axis as the
range, which we can think of as an output.

Letβs say we were to input some
value π into our function which gave an output of π. More formally, we might denote this
as π of π equals π. If we were to plot the graph of our
function π of π₯, we would say that the point with coordinates π, π lies on the
line or curve. Now letβs consider the function π
inverse. As we saw earlier, π inverse
undoes the action of π. This means that if we were to input
the value of π into π inverse, the resulting output must be π. We can follow a similar line of
reasoning to conclude that the point with coordinates π, π must lie on the graph
of π inverse.

Letβs consider what we have just
found. If we find a coordinate π, π
which lies on the graph of π of π₯, we can reverse the ordered pair to find a
corresponding coordinate π, π which lies on the graph of π inverse. We can also think of this as a swap
in the π₯- and π¦-coordinate. Since we followed a general method
to get here for any given point on the graph of π or indeed of π inverse, this
relationship must be true.

Another interesting point, since
this relationship is true for all values, we might be able to see that the domain of
π is the same as the range as π inverse and the range of π is the same as the
domain of π inverse. Letβs now pick some values and see
what all of this looks like on a graph. Consider the function π of π₯
equals two π₯ minus six. If we wanted to pick some values to
plot in the top-right quadrant for π, we could use the coordinates three, zero;
four, two; and five, four. Now remember, to find points on the
graph of π inverse, we can simply swap our ordered pairs. This means that the points zero,
three; two, four; and four, five all lie on the graph of π inverse.

At this point, we might begin to
notice a pattern or perhaps even a link to function transformations. The change that we have described
with the ordered pairs switches the π₯-coordinate for the π¦-coordinate and vice
versa. This change actually corresponds to
a reflection in the line π¦ equals π₯. Looking at the graph, we can see
that all three of the points weβve plotted display this rule. Another thing we might notice is
that if our function π intersects the line of reflection π¦ equals π₯ at any point,
π inverse will also pass through this point. On our graph, this happens at the
point six, six. Joining up all of our points, we
should be able to see our reflection more clearly.

Again, note that this video will
focus on graphs of inverse functions rather than algebraic manipulation. The important rule that we have
just found relating to this is that inverse functions have graphs which are
reflections of each other in the line π¦ equals π₯ and the coordinates of the points
on these lines or curves are reversed ordered pairs. Let us now take a look at an
example of a question.

The following is the graph of π of
π₯ equals two π₯ minus one. Which is the graph of the inverse
function π inverse of π₯?

And just to be clear for this
question, π of π₯ is the blue line, and our options for π inverse of π₯ are shown
here. When trying to find graphs of
inverse functions, we should remember the following important rule. Inverse functions have graphs which
are reflections in the line π¦ equals π₯ and have corresponding coordinates which
are reversed ordered pairs. One approach to solving this
problem would be to draw the line π¦ equals π₯ onto our diagram. We would then reflect the graph of
π of π₯ using the line π¦ equals π₯ as our line of symmetry. Note here that the question has
tried to throw in some confusing factors. For the graph given by the
question, the π₯-axis goes from negative four to four and the π¦-axis goes from a
negative six to six. This is not the same for the axes
in all of our options.

In spite of this, we can still
begin to eliminate some options. We can first observe that the graph
of π inverse of π₯ has a positive gradient. Looking at options (b) and (c), we
can see both of these have a negative gradient. They can therefore be eliminated as
they cannot be the graph of π inverse. Next, we can observe that the graph
of π inverse appears to have a positive π¦-intercept. Of our remaining two options,
option (a) does have a positive π¦-intercept, whereas option (d) does not. This means that we can eliminate
option (d). This means that the answer to our
question must be option (a).

We might also want to consider that
there are other ways that we could have solved this problem. In particular, we can use the fact
that corresponding coordinates are reversed ordered pairs. Imagine we had not reflected π of
π₯ in the line π¦ equals π₯. Instead, we could eliminate the
incorrect options by cleverly picking a point on the graph of π of π₯. Letβs pick the π¦-intercept which
is the point negative one, zero. Reversing this ordered pair gives
us the coordinates of a corresponding point which we know must lie on the line of π
inverse of π₯. That coordinate is zero, negative
one. We can then use this information by
looking at all of the options for the graph of π inverse of π₯ and seeing which of
these graphs appears to pass through the point zero, negative one.

Of course, we already know the
answer. The graph of option (a) does pass
through the point zero, negative one. The graphs of option (b), (c), and
(d) do not. Again, we have confirmed that the
graph of π inverse of π₯ is option (a).

Letβs now take a look at another
example.

Shown is the graph of π of π₯
equals five π₯ cubed plus six. Find the intersection of the
inverse function π inverse of π₯ with the π₯-axis.

To begin, we might remember the
following rule. Inverse functions have graphs which
are reflections of each other in the line π¦ equals π₯. One approach to solving this
problem might be to draw the line π¦ equals π₯. We could then reflect the graph of
π of π₯ using π¦ equals π₯ as our line of symmetry and then find the intersection
of π inverse of π₯ with the π₯-axis. This alternate approach relies on
the fact that coordinates of points which lie on the graphs of inverse functions are
reversed ordered pairs. In other words, if we swap the π₯-
and π¦-coordinates for a point on the graph of π, we get the coordinates of a
corresponding point which we know is on the graph of π inverse. Also note that this relationship
works both ways.

Okay, so weβre looking for the
intersection of the graph of π inverse with the π₯-axis. This intersection point will have a
π¦-coordinate of zero. Letβs say this intersection point
on the graph of π inverse has the coordinates π, zero. Remember that we can swap this
ordered pair to get the corresponding point on the graph of π. The corresponding point on the
graph of π will therefore have the coordinates of zero, π. In other words, it will be the
π¦-axis intercept on the graph of π. Okay, so our method will then be to
find the π¦-axis intercept on the graph of π and to swap the ordered pair to get
the π₯-axis intercept on the graph of π inverse.

Looking at the graph of π given in
the question, we can clearly see that the π¦-axis intercept has coordinates zero,
six. Swapping this ordered pair, we
conclude that the corresponding point on the graph of π inverse will have the
coordinates six, zero. With this, we have answered our
question. The graph of π inverse of π₯
intersects the π₯-axis at the point six, zero.

Okay, earlier in this video, we
briefly mentioned one-to-one functions and how they related to inverse
functions. Letβs explore this in more detail
now. We said that if π is a one-to-one
function, it has an inverse function. It is also true that if π is not a
one-to-one function, it does not have an inverse. Let us explore why this is the case
using the following graph. This function π of π₯ is not a
one-to-one function. One of the ways that we can test if
something is a one-to-one function is using the horizontal line test. This says that if we can draw a
horizontal line that intersects the graph at more than one point, the graph does not
represent a one-to-one function, and we say it fails the test.

For our graph of π of π₯, we can
clearly see that a horizontal line can be drawn such that it intersects the graph at
more than one point. This means that π of π₯ fails the
test, and we confirm it is not a one-to-one function. Okay, we know that reflecting a
graph in the line π¦ equals π₯ gives us the graph of its inverse. Let us see what happens in this
case. Now, we could choose to do this on
the same set of axes, but letβs do this on a different set for clarity. We will tentatively call this new
graph π inverse of π₯. But as weβll see soon, this is not
the case.

Looking at this new graph, we might
notice an important detail. Remember that in order for some
relation to be called a function, each input must correspond to only one output. One way to test for this is using
the vertical line test. Weβre gonna edit this statement
below to now represent the vertical line test. This test says that if we can draw
a vertical line that intersects the graph at more than one point, the graph does not
represent a function. Earlier in this video, you may
recall that we said the domain and range of some function can be said to swap for
its inverse function. If we could draw a horizontal line
that intersected our original graph of π of π₯ at more than one point, it might now
be obvious that we can draw a vertical line that intersects our new graph at more
than one point. We can also clearly see this on our
diagram.

This means that what we have
tentatively called π inverse of π₯ is not a function at all. Since this is not a function, we
say that our original function π of π₯ does not have an inverse. Another way to think about this is
as follows. Consider our original function
π. There are two values in its domain,
π one and π two, which correspond to the same value in its range, π one. This means that for the relation
which we supposed was the inverse of π, which we now know is not a function, the
value of π one, which is now in its domain, must correspond to two values, π one
and π two, in its range.

Consider what would happen if we
try to use this relation to undo the action of our original function π for an
output of π one. Our supposed inverse cannot tell us
for certain whether our original input was π one or π two. For this reason, we say that if
something is not a one-to-one function, it does not have an inverse. Letβs take a look at an example to
illustrate this.

Determine which of the following
functions does not have an inverse.

To answer this question, weβre
going to be using the fact that if a function is not a one-to-one function, it does
not have an inverse. The way that we can test for
one-to-one functions is using the horizontal line test. This tells us that if we can draw a
horizontal line that intersects the graph at more than one point, it does not
represent a one-to-one function. And we say the graph fails the
test.

After carefully examining our
options, it should become clear that there is only one graph that we could draw a
horizontal line on so that it intersects the graph at more than one point. This is the function π of π₯. We have just found that π of π₯
fails the horizontal line test. This means that itβs therefore not
a one-to-one function, and so it does not have an inverse function. The answer to our question is
therefore option (b). π of π₯ does not have an inverse
function, but the other three options do since they passed the horizontal line
test.

Letβs now take a look at one final
example.

By sketching graphs of the
following functions, which is the inverse of itself?

This question has directed us to
sketch these four graphs, and so this should be our first step. At this point, you should be
familiar with sketching graphs. And in order to avoid going into
unnecessary detail, this video will simply provide them. You may, however, wish to verify
these graphs yourself by drawing up a small table of values or by using graphing
software. First, we have the graph of one
over π₯. Next, we have the graph of π₯
squared, forming the familiar shape of a parabola. We then have the graph of π₯ cubed
and, finally, the graph of one over π₯ squared.

In order to move forward with this
question, we recall the following rule. Inverse functions have graphs which
are reflections of each other in the line π¦ equals π₯. This means that a graph which is
the inverse of itself has symmetry about the line π¦ equals π₯. What this means is that if a
function is the inverse of itself when reflected over the line π¦ equals π₯, the
resulting function will be the same as the original function. You may already be able to spot the
answer, but letβs reflect all of our functions in the line π¦ equals π₯ now. We first reflect one over π₯, then
π₯ squared, followed by π₯ cubed, and finally one over π₯ squared.

After doing so, it should be clear
to us that the reflection of the graph one over π₯ is the same. Itβs also the graph of one over
π₯. The reflections of all three of the
other options are different. This means that the inverse
functions are different from the original function. Hence, π₯ squared, π₯ cubed, and
one over π₯ squared are not inverses of themselves. One small side note, for options
(b) and (d), the reflections are not actually functions at all. These graphs would fail the
vertical line test. Since the graphs are not functions
at all, we say that π₯ squared and one over π₯ squared do not have inverses. Okay, back to our question, it
should be clear that option (a) is the only graph for which the original and the
reflection are the same. This means that from our options
one over π₯ is the only function which is the inverse of itself.

To finish off this video, letβs go
through some key points. If π is a one-to-one function, it
has an inverse function which can be represented using the following notation. We would say this as π
inverse. If π is not a one-to-one function,
it does not have an inverse function. We can think of inverse functions
as undoing the action of each other. This can be represented as a
function composition like so. Inverse functions have graphs which
are reflections of each other in the line π¦ equals π₯. The coordinates of points which lie
on the graphs of inverse functions can be thought of as reversed ordered pairs. We saw that this can be interpreted
as the domain and the range of a function switching for its inverse. Finally, some functions are their
own inverse. The graphs of these functions have
reflectional symmetry in the line π¦ equals π₯.