### Video Transcript

By sketching the graphs of the following functions, which is the inverse of itself? Is it (A) one over π₯, (B) π₯ squared, (C) π₯ cubed, or (D) one over π₯ squared?

So, weβve been given four functions and asked to identify which is the inverse of itself. Before we sketch the graphs, letβs just remind ourselves what it means to find the inverse of a function. Suppose we have some function π of π₯. The inverse π with a superscript negative one essentially undoes the original function. In other words, if we apply the inverse to the original function, we get π₯ for any values of π₯ in the domain of the original function. But what does this mean if we think about the graphs of the functions?

Well, we can map a function onto its inverse and vice versa by performing a single reflection across the line π¦ equals π₯. So, letβs sketch the graphs of the functions and see which of these appear to remain unchanged when reflected across that line. We begin with the graph of π¦ equals one over π₯. We might recall the general shape of the reciprocal function. It has asymptotes given by the π₯- and π¦-axes. In fact, it looks a lot like this.

Letβs also add the line π¦ equals π₯ onto this sketch and see what will happen if we reflect this graph across that line. Reflecting the original graph across this line will actually create another graph that looks identical. And so, it does appear that the answer is (A). One over π₯ is the inverse of itself.

But letβs check by looking at the other three graphs. In fact, letβs look at one over π₯ squared first, since this looks a lot like one over π₯. The difference is that the part of the graph that lies in the third quadrant for the graph π¦ equals one over π₯ is reflected into the second quadrant for the graph of π¦ equals one over π₯ squared. This makes a lot of sense of course. When we square the number, we expect a positive result. And one divided by a positive number will always be positive. So, we expect all of our graph to lie above the π₯-axis. Now, of course, if we then reflect this part of the graph across the line π¦ equals π₯, weβll get something like this. So, we can confirm that one over π₯ squared is not the inverse of itself.

Letβs now consider the graph of π¦ equals π₯ squared. The graph of π¦ equals π₯ squared passes through the origin as shown. We have this symmetric parabola shape. Now, we do have a problem when it comes to performing the inverse for the graph π¦ equals π₯ squared. If we reflect π¦ equals π₯ squared across the line π¦ equals π₯, we can first see that the graph cannot be the inverse of itself. Secondly, though, we have a rather unusual-looking graph. In fact, if we perform the vertical line test, we see this is an example of a one-to-many mapping. So, this cannot be a function. This also means we cannot invert π¦ equals π₯ squared without performing some restriction on its domain. Specifically, we restrict the domain of π¦ equals π₯ squared to nonnegative real numbers. Then, it is invertible, but π¦ equals π₯ squared is certainly not the inverse of itself.

Thereβs one more graph to think about, and thatβs π₯ cubed. We can plot π¦ equals π₯ cubed and its reflection across the line π¦ equals π₯ on our coordinate plane. Once again, we see that this graph is not mapped onto itself. So, π₯ cubed is not the inverse of itself. So, we have confirmed that the function which is an inverse of itself is one over π₯. Itβs option (A).