Lesson Video: Graphs of Inverses of Functions Mathematics

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.

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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, let’s consider what is meant when we say the inverse of a function. A function 𝑓 can be inverted by the function 𝑔 when 𝑓 of π‘₯ is equal to 𝑦 and 𝑔 of 𝑦 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 𝑓 and then we were to apply 𝑔, 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 𝑦. However, if 𝑓 of π‘₯ is a one-to-one function, it can be inverted over its entire domain.

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

Before moving forward, let’s 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 𝑓 of π‘₯ equals two π‘₯ minus three, 𝑓 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 it’s enough to say this common misconception should be avoided. Okay, now that we’ve recapped the inverse of a function, let’s see how it relates to graphs. Consider some invertable function 𝑓. 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 𝑦-axis as the range, which we can think of as an output.

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

Let’s consider what we have just found. If we find a coordinate π‘Ž, 𝑏 which lies on the graph of 𝑓 of π‘₯, we can reverse the ordered pair to find a corresponding coordinate 𝑏, π‘Ž which lies on the graph of 𝑓 inverse. We can also think of this as a swap in the π‘₯- and 𝑦-coordinate. Since we followed a general method to get here for any given point on the graph of 𝑓 or indeed of 𝑓 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 𝑓 is the same as the range as 𝑓 inverse and the range of 𝑓 is the same as the domain of 𝑓 inverse. Let’s now pick some values and see what all of this looks like on a graph. Consider the function 𝑓 of π‘₯ equals two π‘₯ minus six. If we wanted to pick some values to plot in the top-right quadrant for 𝑓, we could use the coordinates three, zero; four, two; and five, four. Now remember, to find points on the graph of 𝑓 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 𝑓 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 𝑦-coordinate and vice versa. This change actually corresponds to a reflection in the line 𝑦 equals π‘₯. Looking at the graph, we can see that all three of the points we’ve plotted display this rule. Another thing we might notice is that if our function 𝑓 intersects the line of reflection 𝑦 equals π‘₯ at any point, 𝑓 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 𝑦 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 𝑓 of π‘₯ equals two π‘₯ minus one. Which is the graph of the inverse function 𝑓 inverse of π‘₯?

And just to be clear for this question, 𝑓 of π‘₯ is the blue line, and our options for 𝑓 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 𝑦 equals π‘₯ and have corresponding coordinates which are reversed ordered pairs. One approach to solving this problem would be to draw the line 𝑦 equals π‘₯ onto our diagram. We would then reflect the graph of 𝑓 of π‘₯ using the line 𝑦 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 𝑦-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 𝑓 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 𝑓 inverse. Next, we can observe that the graph of 𝑓 inverse appears to have a positive 𝑦-intercept. Of our remaining two options, option (a) does have a positive 𝑦-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 𝑓 of π‘₯ in the line 𝑦 equals π‘₯. Instead, we could eliminate the incorrect options by cleverly picking a point on the graph of 𝑓 of π‘₯. Let’s pick the 𝑦-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 𝑓 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 𝑓 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 𝑓 inverse of π‘₯ is option (a).

Let’s now take a look at another example.

Shown is the graph of 𝑓 of π‘₯ equals five π‘₯ cubed plus six. Find the intersection of the inverse function 𝑓 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 𝑦 equals π‘₯. One approach to solving this problem might be to draw the line 𝑦 equals π‘₯. We could then reflect the graph of 𝑓 of π‘₯ using 𝑦 equals π‘₯ as our line of symmetry and then find the intersection of 𝑓 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 𝑦-coordinates for a point on the graph of 𝑓, we get the coordinates of a corresponding point which we know is on the graph of 𝑓 inverse. Also note that this relationship works both ways.

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

Looking at the graph of 𝑓 given in the question, we can clearly see that the 𝑦-axis intercept has coordinates zero, six. Swapping this ordered pair, we conclude that the corresponding point on the graph of 𝑓 inverse will have the coordinates six, zero. With this, we have answered our question. The graph of 𝑓 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. Let’s explore this in more detail now. We said that if 𝑓 is a one-to-one function, it has an inverse function. It is also true that if 𝑓 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 𝑓 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 𝑓 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 𝑓 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 𝑦 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 let’s do this on a different set for clarity. We will tentatively call this new graph 𝑓 inverse of π‘₯. But as we’ll 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. We’re 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 𝑓 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 𝑓 inverse of π‘₯ is not a function at all. Since this is not a function, we say that our original function 𝑓 of π‘₯ does not have an inverse. Another way to think about this is as follows. Consider our original function 𝑓. There are two values in its domain, π‘Ž one and π‘Ž two, which correspond to the same value in its range, 𝑏 one. This means that for the relation which we supposed was the inverse of 𝑓, which we now know is not a function, the value of 𝑏 one, which is now in its domain, must correspond to two values, π‘Ž one and π‘Ž two, in its range.

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

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

To answer this question, we’re 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 𝑔 of π‘₯. We have just found that 𝑔 of π‘₯ fails the horizontal line test. This means that it’s 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). 𝑔 of π‘₯ does not have an inverse function, but the other three options do since they passed the horizontal line test.

Let’s 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 𝑦 equals π‘₯. This means that a graph which is the inverse of itself has symmetry about the line 𝑦 equals π‘₯. What this means is that if a function is the inverse of itself when reflected over the line 𝑦 equals π‘₯, the resulting function will be the same as the original function. You may already be able to spot the answer, but let’s reflect all of our functions in the line 𝑦 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. It’s 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, let’s go through some key points. If 𝑓 is a one-to-one function, it has an inverse function which can be represented using the following notation. We would say this as 𝑓 inverse. If 𝑓 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 𝑦 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 𝑦 equals π‘₯.

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