Question Video: Identifying Whether Two Functions are Inverse From Their Graphs Mathematics

By sketching the graphs of 𝑓(π‘₯) = 2π‘₯Β² and 𝑔(π‘₯) = √(π‘₯/2) for π‘₯ β‰₯ 0, determine whether they are inverse functions.

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Video Transcript

By sketching the graphs of 𝑓 of π‘₯ equals two π‘₯ squared and 𝑔 of π‘₯ is equal to the square root of π‘₯ over two for π‘₯ is greater than or equal to zero, determine whether they are inverse functions.

Remember, the inverse of a function essentially undoes the original function. For instance, suppose we have some function 𝑓 of π‘₯. The inverse 𝑓 of 𝑓 of π‘₯ is equal to π‘₯ for all values of π‘₯ in the domain of the function. But what does that mean when we’re talking about the graphs of our functions? Suppose we have the graph of 𝑦 equals 𝑓 of π‘₯. The graph of its inverse 𝑦 is equal to the inverse 𝑓 of π‘₯ is found by reflecting the original graph across the line 𝑦 equals π‘₯ and, of course, vice versa.

Now, what this means about specific points on the graph is that the π‘₯- and 𝑦-coordinates essentially switch. For example, suppose we have the point π‘Ž, 𝑏 that lies on the line 𝑦 equals 𝑓 of π‘₯. On the graph of 𝑦 equals the inverse 𝑓 of π‘₯, this point is mapped onto the point with coordinates 𝑏, π‘Ž. So let’s sketch the graphs of 𝑓 of π‘₯ and 𝑔 of π‘₯.

We might observe that the domain of our functions has been restricted to nonnegative real numbers. Now this is important because 𝑓 of π‘₯ itself is, in fact, a many-to-one function. This means if we don’t restrict its domain, its inverse would be one to many. Now, of course, a mapping which is one to many is not a function. So we restrict the domain to ensure that two π‘₯ squared is one to one so we can invert it. Specifically, the graph of 𝑦 equals 𝑓 of π‘₯ or 𝑦 equals two π‘₯ squared looks like this. It passes through the origin, the point zero, zero. And we might pick a point that also lies on the line, for example, the point with coordinates two, eight.

But what does the graph of the square root of π‘₯ over two look like? Well, we know the general shape of the graph of 𝑦 equals the square root of π‘₯. So we can perform a simple transformation to map this onto the graph of 𝑦 equals the square root of π‘₯ over two. This is a horizontal stretch by a scale factor of two. So the shape remains roughly the same. In fact, we can plot that on our graph as shown. It certainly does appear as if 𝑓 of π‘₯ and 𝑔 of π‘₯ could be mapped onto one another by reflection in the line 𝑦 equals π‘₯, but we need to check.

One thing we can do to confirm is to see whether the point with coordinates eight, two lies on the line 𝑔 of π‘₯. To do so, we’ll let π‘₯ equal eight. Remember, we’re doing this because we know that the point two, eight lies on the graph of 𝑓 of π‘₯ equals two π‘₯ squared. Then 𝑔 of eight is the square root of eight divided by two. Well, that’s the square root of four, which is simply equal to two. So we can confirm that the point two, eight maps onto the point eight, two by reflection across the line 𝑦 equals π‘₯. We know both lines pass through the point zero, zero. And what’s the other point of intersection?

Well, let’s solve the equations 𝑦 equals π‘₯ and 𝑦 equals two π‘₯ squared simultaneously. This will tell us the point of intersection of the line 𝑦 equals π‘₯ with our first function. And then we can establish whether that also lies on the line 𝑔 of π‘₯ equals the square root of π‘₯ over two. To solve these simultaneously, let’s set them equal to one another. So two π‘₯ squared equals π‘₯. Then we’ll subtract π‘₯ from both sides and then factor the left-hand side. So we get π‘₯ times two π‘₯ minus one equals zero.

One solution to this equation is π‘₯ equals zero. Now we already know that these lines intersect at the origin. The other possible solution is when two π‘₯ minus one equals zero. So if we add one and divide by two, we find π‘₯ is equal to one-half. Then, substituting this into either equation, ideally the equation 𝑦 equals π‘₯, and we see that the line 𝑦 equals π‘₯ and 𝑦 equals two π‘₯ squared intersect at the point one-half, one-half.

Now it doesn’t really matter that our graph is not to scale; this is just a sketch. But what we do need to check is whether the point one-half, one-half lies on the graph 𝑔 of π‘₯ is equal to the square root of π‘₯ over two. We’ll substitute into this function once again. This time π‘₯ equals one-half. 𝑔 of one-half is the square root of one-half divided by two, or the square root of one-quarter. Now to find the square root of a fraction, we can simply root the numerator and the denominator. So we get one-half. This does confirm that the point half, half lies on the function 𝑔 of π‘₯ equals the square root of π‘₯ over two.

So to recap, we sketched each graph and established whether it looked like one graph could be mapped onto the other by a reflection across the line 𝑦 equals π‘₯. We then checked this with a single point. We saw that the point two, eight mapped onto the point eight, two. Because the values of their π‘₯- and 𝑦-coordinates were switched, this represents a reflection across the line 𝑦 equals π‘₯. We also checked that the functions 𝑓 of π‘₯ and 𝑔 of π‘₯ intersected at the same points along the line 𝑦 equals π‘₯. In doing so, we have determined that 𝑓 of π‘₯ and 𝑔 of π‘₯ are inverse functions.

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