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If $f$ is a function that $f(x)+2f(-x)=x^3$, then $f(x)$

A- Even

B- Odd

C- Neither even nor odd

D- The given is not sufficient to determine the type of f

This is a question from an old test I found.

Well, I tried to assume it's odd once and even once. Well when I assumed it is even, I got a contradiction as $f(x)=\frac{x^3}{3}$, which clearly is not even. And when I assumed it is odd, I got $f(x)=-x^3$. But, is there a function that is neither odd nor even that will suffice for this relation?

Mina Basilious
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1 Answers1

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If $f: \Bbb R \to \Bbb R$ (or $f:I \to \Bbb R$ for some interval $I$ which is symmetric to the origin) satisfies $$ f(x) + 2f(-x) = x^3 $$ for all $x$ then we can substitute $x \mapsto -x$ to get a second equation $$ f(-x) + 2f(x) = -x^3 \, . $$ Adding these two equations gives $3f(x) + 3f(-x) = 0$, so that $f$ is necessarily an odd function.

It follows that $f(x) = -x^3$ is the only solution.

Martin R
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