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I've come across a question that I'm not too sure how to answer. If someone could explain the answer to this, I would be most grateful.

If $f(x)$ is a function defined for all real numbers, then for all $x$, $f(x^2 + 3x) = f(x^2) + f(3x)$.

Is this statement true or false?

Thanks!

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    What is the question about the function your being asked? – 2017-02-01
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    This doesn't make sense without more information. What is the context here? Where did you encounter this statement? This is certainly not a property of an arbitrary real variable function. – 2017-02-01
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    Oh sorry - its a true or false question. – 2017-02-01
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    @manthanomen So its false then? Could explain why? – 2017-02-01
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    Are you told anything more about the function? – 2017-02-01
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    @theGreatWhatever I think $f(x) = a.x $ will do ,where $a \in \mathbb{R} $ or in $\mathbb{C}$ as you have not specified the codomain of $f$. – 2017-02-01
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    @BAYMAX Any linear function like $f(x) = ax$ will always have the property mentioned in the question. – 2017-02-01
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    @manthanomen yes true . if the question asks for all functions then it would be false and if it asks for if there exists a function with the above property then it will be true as there exists linear function with the above property. – 2017-02-01
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    @theGreatWhatever You haven't told us if $f$ must have any specific properties. Assuming it's just an arbitrary function $f : \mathbb R \to \mathbb R$, define $f$ by $f(1) = 1$ and $f(x) = 0$ for $x \neq 1$. Then $f(1^2 + 3(1)) = f(4) = 0$ but $f(1^2) + f(3(1)) = 1 + 0 = 1$. – 2017-02-01
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    @BAYMAX As stated in the question (the only requirement for $f$ is that the domain is $\mathbb R$), the claim is false. theGreatWhatever asks whether the statement is true or false, not whether there exists such a function. – 2017-02-01
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    @manthanomen ok. – 2017-02-01

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If $f$ is any real-valued function, then the statement is false. For example the constant function $f(x)=1$ makes the equation equivalent to $1=1+1$ for any $x$.

However, if a function's codomain is $\{0\}$ then yes, for all $x$ the equation is satisfied as $0=0+0$, and the statement is true.

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    manthanomen answered first, but yours is comprehensive. Thanks! – 2017-02-01