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Is the function w(x)=−2x5+7x2w(x) = -2x^5 + 7x^2 even, odd, or neither?\newlineChoices:\newline(A)even\newline(B)odd\newline(C)neither

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Q. Is the function w(x)=−2x5+7x2w(x) = -2x^5 + 7x^2 even, odd, or neither?\newlineChoices:\newline(A)even\newline(B)odd\newline(C)neither
  1. Calculate w(−x)w(-x): Calculate w(−x)w(-x) by substituting −x-x for xx in the function w(x)w(x).w(−x)=−2(−x)5+7(−x)2w(-x) = -2(-x)^5 + 7(-x)^2
  2. Simplify w(−x)w(-x): Simplify the function w(−x)w(-x).\newlinew(−x)=−2(−x)5+7(−x)2w(-x) = -2(-x)^5 + 7(-x)^2 becomes w(−x)=−2(−1)5x5+7(−1)2x2w(-x) = -2(-1)^5x^5 + 7(-1)^2x^2\newlinew(−x)=−2(−1)x5+7x2w(-x) = -2(-1)x^5 + 7x^2\newlinew(−x)=2x5+7x2w(-x) = 2x^5 + 7x^2
  3. Compare w(x)w(x) and w(−x)w(-x): Compare w(x)w(x) and w(−x)w(-x). We have w(x)=−2x5+7x2w(x) = -2x^5 + 7x^2 and w(−x)=2x5+7x2w(-x) = 2x^5 + 7x^2. Since w(−x)≠w(x)w(-x) \neq w(x) and w(−x)≠−w(x)w(-x) \neq -w(x), w(x)w(x) is neither even nor odd.

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