Question Video: Stating the Parity of a Function Mathematics

Is the function 𝑓(π‘₯) = π‘₯⁡ tan⁴ (6π‘₯) even, odd, or neither even nor odd?

03:09

Video Transcript

Is the function 𝑓 of π‘₯ equals π‘₯ to the fifth power times tan of six π‘₯ to the fourth power even, odd, or neither even nor odd?

Let’s recall how we check the parity of a function. The first thing we do is check the domain of the function. We need that to be centered at π‘₯ equals zero. Then, if the answer is no, we can say that the function is neither even nor odd without performing any further tests. If the answer is yes, though, we say that it will be even if it satisfies 𝑓 of negative π‘₯ equals 𝑓 of π‘₯. And it will be odd if it satisfies 𝑓 of negative π‘₯ equals negative 𝑓 of π‘₯. Then, of course, if it satisfies neither of these, it will be neither even nor odd.

So let’s think about the domain of our function. Our function is the product of two functions. It’s the product of π‘₯ to the fifth power and tan of six π‘₯ to the fourth power. And so the domain of 𝑓 of π‘₯ will be the intersection of the domains of the respective parts of the function. Well, π‘₯ to the fifth power is a polynomial, so its domain is the set of real numbers or the open interval from negative ∞ to ∞. But what about the domain of the trigonometric part? Well, it’s all real numbers, except those that make cos of six π‘₯ equal to zero. But since the values of π‘₯ that make cos of six π‘₯ equal to zero are symmetrical about the 𝑦-axis, then we can say that the domain of tan of six π‘₯ to the fourth power must be centered at π‘₯ equals zero.

Since both domains are centered at π‘₯ equals zero, then we can answer yes to this first question, and we’re able to move on. We now see that it’s even if 𝑓 of negative π‘₯ is equal to 𝑓 of π‘₯ and odd if it’s equal to negative 𝑓 of π‘₯. And so let’s evaluate 𝑓 of negative π‘₯. To do so, we replace each instance of π‘₯ in our original function with negative π‘₯. And we get 𝑓 of negative π‘₯ is negative π‘₯ to the fifth power times tan of negative six π‘₯ to the fourth power. We’ll evaluate each part in turn. Let’s begin with negative π‘₯ to the fifth power. Since the exponent is odd, when we multiply this out, we’re going to get a negative result. Negative π‘₯ to the fifth power is as shown.

But what about the tan function? Well, we can actually quote the result that tan of π‘₯ is odd, meaning that tan of negative π‘₯ is equal to negative tan of π‘₯ and, in turn, the tan of negative six π‘₯ is equal to negative tan of six π‘₯. But of course, we’re raising this to the fourth power. We’re raising it to an even exponent. And we know when we raise a negative number to an even exponent, the result is positive. And so tan of negative six π‘₯ to the fourth power is just tan of six π‘₯ to the fourth power.

And so 𝑓 of negative π‘₯ is therefore equal to negative π‘₯ to the fifth power times tan of six π‘₯ to the fourth power. So does this satisfy either of our criteria, is it even or odd? Well, yes. If we look at it carefully, we see it’s the same as negative 𝑓 of π‘₯. 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. And so the function must be odd.

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