Video: Identifying the Parity of Functions

Determine whether the function 𝑓(π‘₯) = 9π‘₯Β³ is even, odd, or neither even nor odd, given that 𝑓: (βˆ’7, 7] ⟢ ℝ.

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Video Transcript

Determine whether the function 𝑓 of π‘₯ equals nine π‘₯ cubed is even, odd, or neither even or odd, given that 𝑓 is defined for values of π‘₯ greater than negative seven and less than seven and π‘₯ is a real number.

Remember, a function is defined to be even if 𝑓 of negative π‘₯ is equal to 𝑓 of π‘₯ for all values of π‘₯ in the domain of 𝑓 of π‘₯. And it’s said to be odd if 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. Of course, a function is neither odd nor even if neither of these definitions hold. In our question, we see that 𝑓 of π‘₯ is equal to nine π‘₯ cubed. So what’s 𝑓 of negative π‘₯? Well, if we replace π‘₯ with negative π‘₯, we see 𝑓 of negative π‘₯ is nine times negative π‘₯ cubed. Well, negative π‘₯ cubed is negative π‘₯ cubed. So 𝑓 of negative π‘₯ is negative nine π‘₯ cubed.

And so far, so good. 𝑓 of negative π‘₯ does indeed look like it’s equal to negative 𝑓 of π‘₯, making this an odd function. However, we’re going to need to be a little bit careful with the domain of our function. We see that π‘₯ takes values greater than negative seven and less than or equal to seven. So we evaluate 𝑓 of seven. It’s nine times seven cubed, which is 3087. But negative 𝑓 of negative seven is not equal to 𝑓 of seven. And this is because 𝑓 of negative seven is not defined on the domain of 𝑓. π‘₯ equals negative seven is outside the domain of our function.

So we’ve just shown that 𝑓 of negative π‘₯ is not equal to negative 𝑓 of π‘₯ for all values of π‘₯ in the domain of our function and nor is 𝑓 of negative π‘₯ equal to 𝑓 of π‘₯ . This means the function is neither odd nor even.

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