Find all functions, for all real x, that satisfy the following functional equation:

$$ xf(x) + f(1-x) = x^3 - x $$


closed as off-topic by SchrodingersCat, Watson, user223391, Cyclohexanol., user7530 Sep 17 '16 at 21:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Watson, Community, Cyclohexanol., user7530
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I think $x=1-u$ would help. $\endgroup$ – Ahmed S. Attaalla Sep 12 '16 at 4:31
  • $\begingroup$ If you have no idea where to start, a reasonable thing to do is just plug in some values for $x$. $\endgroup$ – grndl Sep 12 '16 at 4:32
  • $\begingroup$ Sorry if this may sound a bit naive, but if I don't know what the function is then how would I sub in values? $\endgroup$ – user193203821309 Sep 12 '16 at 4:34
  • $\begingroup$ When $x = 0$ then $(0)f(0)+f(1)=0^3-0\implies f(1) = 0$ is the solution. Try to build off of that. $\endgroup$ – Mark Sep 12 '16 at 4:35
  • $\begingroup$ For example, let $x=0$. Then, $f(1) = 0$. Let $x=1$, then what? $\endgroup$ – grndl Sep 12 '16 at 4:35

We have \begin{align*} xf(x) + f(1-x) &= x^3-x\\ (1-x)f(1-x) + f(1-(1-x)) &= (1-x)^3 - (1-x) \end{align*} and hence \begin{align*} f(x) + (1-x)f(1-x) = (1-x)^3 - (1-x) \end{align*} Multiplying the first equation by $1-x$ and subtracting from the third equation, we get \begin{align*} (x(1-x) - 1)f(x) &= (x^3-x)(1-x) - (1-x)^3 + (1-x)\\ (-x^2+x-1)f(x) &= x^3-x - x^4 + x^2 - (1-3x+3x^2-x^3) + 1-x \\ &= -x^4 + 2x^3-2x^2 +x \end{align*} Hence \begin{align*} f(x) = \frac{-x^4 + 2x^3-2x^2 +x }{-x^2+x-1} = x^2 - x = x(x-1) \end{align*}

  • $\begingroup$ What do you mean by 'eliminate'? $\endgroup$ – user193203821309 Sep 12 '16 at 4:40
  • $\begingroup$ I have updated the answer. $\endgroup$ – user348749 Sep 12 '16 at 4:47
  • 2
    $\begingroup$ For completeness, it should be added that the result needs be verified to satisfy the given equation (which it does, indeed, in this case) because the entire derivation is based on the implicit assumption that such $f(x)$ exists, but that assumption has not been otherwise proved. $\endgroup$ – dxiv Sep 12 '16 at 4:52
  • $\begingroup$ Thank you very much for your solution; it is indeed very elegant and thought-out. Does this mean that only one such function exists for this problem? $\endgroup$ – user193203821309 Sep 12 '16 at 4:57
  • 2
    $\begingroup$ @dxiv Yes, we need to verify that the obtained $f$ satisfies the given condition. $\endgroup$ – user348749 Sep 12 '16 at 5:00

Plug in $x=1$ and $x=0$ in the given relation to get $f(0)=f(1)=0$ and plug in $x=1/2$ to get $f(1/2)=-1/4$. Apply Lagrange's formula for interpolation:


  • 1
    $\begingroup$ How do we know that $f$ must be a polynomial so that we can apply Lagrange's formula? $\endgroup$ – user348749 Sep 17 '16 at 4:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.