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Video: Determining the Parity of a Graphed Rational Function

Alex Cutbill

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

03:58

Video Transcript

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

This is a question about the parity of the function in the graph, and there are three options here. The function is either even, or it’s odd, or it’s neither. Let’s remind ourselves what the terms even and odd mean in the context of functions.

A function 𝑓 of π‘₯ is even if 𝑓 of minus π‘₯ equals 𝑓 of π‘₯. That must hold for all values of π‘₯. So 𝑓 of negative one must be 𝑓 of one, 𝑓 of negative seven must be 𝑓 of seven, 𝑓 of negative πœ‹ must be 𝑓 of πœ‹, and so on.

The definition of an odd function is very similar: the function 𝑓 of π‘₯ is odd if 𝑓 of minus π‘₯ equals minus 𝑓 of π‘₯. So for an odd function, 𝑓 of negative one is minus 𝑓 of one, 𝑓 of negative seven is minus 𝑓 of seven, and 𝑓 of negative πœ‹ is minus 𝑓 of πœ‹.

And unlike for whole numbers, our function 𝑓 of π‘₯ doesn’t have to be either even or odd; it can be neither. Let’s first check if the function is even. In other words, is 𝑓 of minus π‘₯ equal to 𝑓 of π‘₯ for all values of π‘₯? Well, let’s try a value of π‘₯, say one, and find 𝑓 of negative one.

So we look at negative one on the π‘₯-axis, we go up until we hit the graph, and we see that 𝑓 of negative one is one. So for 𝑓 of π‘₯ to be even, 𝑓 of one has to be equal to one as well. We go through the same process to find 𝑓 of one, going down this time until we hit the graph. And reading off from the 𝑦-axis, we conclude that 𝑓 of one is equal to negative one. So clearly, 𝑓 of negative one is not equal to 𝑓 of one, and hence 𝑓 of π‘₯ is not even.

Remember this doesn’t mean that 𝑓 of π‘₯ is odd; 𝑓 of π‘₯ could be neither even nor odd. We now check if 𝑓 of π‘₯ is odd, so is 𝑓 of minus π‘₯ equal to minus 𝑓 of π‘₯ for all values of π‘₯? We can use the values of the function that we found already β€” it’s a shame to waste them β€” and we can see that 𝑓 of negative one is equal to minus 𝑓 of one.

On the left-hand side, we have one, and on the right-hand side, we have minus negative one, which is also equal to one. And so it’s true that 𝑓 of negative π‘₯ is equal to minus 𝑓 of π‘₯ when π‘₯ is one at least, but it’s not enough for 𝑓 of minus π‘₯ to be equal to minus 𝑓 of π‘₯ for one value of π‘₯; it has to be true for all values of π‘₯.

Checking another value of π‘₯, let’s say seven. While it’s hard to read off the values of 𝑓 of negative seven and 𝑓 of seven, it’s very plausible that their opposites 𝑓 of negative seven is just a bit greater than zero and 𝑓 of seven is just a bit less than zero. And we could continue trying values of π‘₯ until we’ve convinced ourselves that 𝑓 of π‘₯ is indeed an odd function.

Here we’ve used the definition of an odd function to show that our function is indeed odd, but there’s another way to do it; we can use a property of odd functions. A function is odd if its graph has a 180-degree rotational symmetry about the origin.

We can see that this is true for our graph: if we rotate our graph by 180 degrees about the origin, we get the same graph back. And there’s a corresponding fact for an even function: a function is even if its graph is symmetric in the 𝑦-axis. So if you reflect the graph of an even function in the 𝑦-axis, you get the same graph back.

Looking at our graph, we can see that our graph isn’t symmetric in the 𝑦-axis, and so our function isn’t even, but as stated before, because our graph has a 180-degree rotational symmetry about the origin, our function is odd.