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Is the function h(x)=x3+2xh(x) = x^3 + 2x even, odd, or neither?\newlineChoices:\newline(A)even\newline(B)odd\newline(C)neither

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Full solution

Q. Is the function h(x)=x3+2xh(x) = x^3 + 2x even, odd, or neither?\newlineChoices:\newline(A)even\newline(B)odd\newline(C)neither
  1. Check Even Function: Check if h(x)h(x) is even by substituting −x-x for xx and comparing h(−x)h(-x) to h(x)h(x).
    h(−x)=(−x)3+2(−x)h(-x) = (-x)^3 + 2(-x)
    h(−x)=−x3−2xh(-x) = -x^3 - 2x
  2. Compare h(−x)h(-x) with h(x)h(x): Compare h(−x)h(-x) with h(x)h(x).h(x)=x3+2xh(x) = x^3 + 2xh(−x)=−x3−2xh(-x) = -x^3 - 2xSince h(−x)≠h(x)h(-x) \neq h(x), h(x)h(x) is not even.
  3. Check Odd Function: Check if h(x)h(x) is odd by seeing if h(−x)=−h(x)h(-x) = -h(x).
    h(−x)=−x3−2xh(-x) = -x^3 - 2x
    −h(x)=−(x3+2x)-h(x) = -(x^3 + 2x)
    −h(x)=−x3−2x-h(x) = -x^3 - 2x
    Since h(−x)=−h(x)h(-x) = -h(x), h(x)h(x) is odd.

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