Is the function w(x) = -2x^5 + 7x^2 even, odd, or neither? Choices: (A)even (B)odd (C)neither

Calculate w(-x): Calculate w(-x) by substituting -x for x in the function w(x). w(-x) = -2(-x)^5 + 7(-x)^2Simplify w(-x): Simplify the function w(-x). w(-x) = -2(-x)^5 + 7(-x)^2 becomes w(-x) = -2(-1)^5x^5 + 7(-1)^2x^2 w(-x) = -2(-1)x^5 + 7x^2 w(-x) = 2x^5 + 7x^2Compare w(x) and w(-x): Compare w(x) and w(-x). We have w(x) = -2x^5 + 7x^2 and w(-x) = 2x^5 + 7x^2. Since w(-x) ≠ w(x) and w(-x) ≠ -w(x), w(x) is neither even nor odd.

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Is the function $w(x) = -2x^5 + 7x^2$ even, odd, or neither? $\newline$ Choices: $\newline$ (A)even $\newline$ (B)odd $\newline$ (C)neither

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Algebra 2

Even and odd functions

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Q. Is the function

w(x) = -2x^5 + 7x^2

even, odd, or neither?

\newline

Choices:

\newline

(A)even

\newline

(B)odd

\newline

(C)neither

Calculate $w(-x)$ : Calculate $w(-x)$ by substituting $-x$ for $x$ in the function $w(x)$ . $w(-x) = -2(-x)^5 + 7(-x)^2$
Simplify $w(-x)$ : Simplify the function $w(-x)$ . $\newline$ $w(-x) = -2(-x)^5 + 7(-x)^2$ becomes $w(-x) = -2(-1)^5x^5 + 7(-1)^2x^2$ $\newline$ $w(-x) = -2(-1)x^5 + 7x^2$ $\newline$ $w(-x) = 2x^5 + 7x^2$
Compare $w(x)$ and $w(-x)$ : Compare $w(x)$ and $w(-x)$ . We have $w(x) = -2x^5 + 7x^2$ and $w(-x) = 2x^5 + 7x^2$ . Since $w(-x) \neq w(x)$ and $w(-x) \neq -w(x)$ , $w(x)$ is neither even nor odd.