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Question Video: Stating the Parity of a Polynomial Function Mathematics • 12th Grade

Is the function 𝑓(π‘₯) = 8π‘₯Β³ + 8π‘₯, βˆ€ π‘₯ ∈ ℝ even, odd, or neither even nor odd?

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Video Transcript

Is the function 𝑓 of π‘₯ equals eight π‘₯ cubed plus eight π‘₯, for all real values of π‘₯, even, odd, or neither even nor odd?

We recall that any real-valued function is even if 𝑓 of negative π‘₯ is equal to 𝑓 of π‘₯. A function is odd, on the other hand, if 𝑓 of negative π‘₯ is equal to the negative of 𝑓 of π‘₯. In order to work out whether this function, eight π‘₯ cubed plus eight π‘₯, is even or odd, we need to find an expression for 𝑓 of negative π‘₯. We do this by replacing all our values of π‘₯ in the initial function with negative π‘₯.

Cubing negative π‘₯ means multiplying negative π‘₯ by negative π‘₯ by negative π‘₯. This is equal to negative π‘₯ cubed. This means that eight multiplied by negative π‘₯ cubed is equal to negative eight π‘₯ cubed. Multiplying eight by negative π‘₯ gives us negative eight π‘₯, as multiplying a positive by a negative gives a negative term. As both these terms are negative, we can factor or factorize out negative one. This can be rewritten as negative eight π‘₯ cubed plus eight π‘₯.

As the part inside the parentheses or bracket, eight π‘₯ cubed plus eight π‘₯, is equal to our initial function 𝑓 of π‘₯, then 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. We can therefore conclude that the function eight π‘₯ cubed plus eight π‘₯, for all real values of π‘₯, is odd.

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