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Is the function n(x)=2x5+x3n(x) = 2x^5 + x^3 even, odd, or neither?\newlineChoices:\newline(A)even\newline(B)odd\newline(C)neither

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Q. Is the function n(x)=2x5+x3n(x) = 2x^5 + x^3 even, odd, or neither?\newlineChoices:\newline(A)even\newline(B)odd\newline(C)neither
  1. Calculate n(−x)n(-x): Calculate n(−x)n(-x) by substituting −x-x for xx in n(x)=2x5+x3n(x) = 2x^5 + x^3.\newlinen(−x)=2(−x)5+(−x)3n(-x) = 2(-x)^5 + (-x)^3
  2. Simplify function: Simplify the right side of the function. n(−x)=2(−x)5+(−x)3=−2x5−x3n(-x) = 2(-x)^5 + (-x)^3 = -2x^5 - x^3
  3. Compare n(x)n(x) and n(−x)n(-x): Compare n(x)n(x) and n(−x)n(-x). We have n(x)=2x5+x3n(x) = 2x^5 + x^3 and n(−x)=−2x5−x3n(-x) = -2x^5 - x^3. Since n(−x)=−n(x)n(-x) = -n(x), n(x)n(x) is an odd function.

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