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I have $f(x^2-x)=x$ and I would like to find $f(x)$. Is there a systematic way to do it, which also works for similar composite functions?

Bran
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    The function $f$ is not well-defined. For example: $f(2^2-2)=2$; also $f((-1)^2-(-1))=-1$; but $2^2-2=2=(-1)^2-(-1)$. You need to be more specific and give more conditions about $f$. – Earnur Apr 28 '22 at 11:10

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Let $x \in\mathbb R$ and $y = x^2 -x = (x-\frac{1}{2})^2 - \frac{1}{4}$.

Then $x$ is a solution of a second order polynomial equation, so we know that $y\geq -1/4$ and that $$x=\frac{1}{2}(1\pm\sqrt{1+4y})$$

Therefore, $f$ is such that : $$\forall y \geq -\frac 14, f(y) = \frac12(1\pm\sqrt{1+4y})$$

If you assume that $f$ is continuous, then the sign $\pm$ is fixed for all $y$. If you have no condition on $f$, then the sign can depend on $y$.

Also $f$ is arbitrary on $(-\infty, -1/4)$

SolubleFish
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  • Hi @SolubleFish, this is unrelated to your answer, hence I would be deleting this comment soonest. Please could you help out with this question https://math.stackexchange.com/q/4435979/585488 – linker Apr 29 '22 at 19:17
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There's no such function, not without some constraints on domain/codomain: $$x=f(x^2-x)=f((1-x)^2-(1-x))=1-x,$$ so $x$ and $1-x$ can't both belong to the codomain of $f$, if $x\neq1-x$.

You can have a solution for $x$ restricted to $[\frac12,\infty)$, naturally.

wasn't me
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