Question Video: Determining If a Graphed Function Is Even, Odd, or Neither Mathematics • 12th Grade

Determine whether the function represented by the figure is even, odd, or neither even nor odd.

02:22

Video Transcript

Determine whether the function represented by the following figure is even, odd, or neither even nor odd.

And then we have a graph of the function shown. So let’s recall how to check the parity of a function, how to check whether it’s even or odd. Well, the first thing we do is ask ourselves, is the domain of this function centered at π‘₯ equals zero? We might recall that the domain of a function is the set of possible inputs, the set of values of π‘₯, that we can substitute into the function. And we can read that domain from the graph.

Now we do need to be a little bit careful because the graph doesn’t actually appear to be defined at π‘₯ equals zero. In fact, the domain is the union of the left-closed right-open interval from negative eight to zero and the left-open right-closed interval from zero to eight. This is centered at π‘₯ equals zero. Zero is exactly halfway through this domain. And so we can now say yes to this question. And we are able to move on to the next part.

We can say that if 𝑓 of negative π‘₯ is equal to 𝑓 of π‘₯, the function is even, and it’s odd if 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. Well, one way we can establish whether either of these is true is to choose a value of π‘₯. For instance, let’s use the point π‘₯ equals five. When π‘₯ is equal to five, the value of our function, the 𝑦-value, is negative one. So 𝑓 of five is negative one. And then this means that negative π‘₯ must be negative five. And so we need to read the 𝑦-value when π‘₯ is negative five. 𝑓 of negative five is also negative one. So it does look like this might be an even function. But let’s check with another value.

Let’s choose π‘₯ equals one. 𝑓 of one is roughly equal to negative 4.1. Then negative π‘₯ will be equal to negative one. And once again, 𝑓 of negative one is roughly equal to negative 4.1. And so for the two values we’ve tried, 𝑓 of negative π‘₯ is equal to 𝑓 of π‘₯. But actually, if we look carefully, we see that that function itself has reflectional symmetry about the 𝑦-axis. And so indeed, every value of 𝑓 of π‘₯ must be equal to every value of 𝑓 of negative π‘₯. And so we can say that the function itself must be even.

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