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Show there exists a function $f: \Bbb R \to \Bbb R$ such that $ f(x)^5+f(x)^4+f(x)^3+f(x)^2+6f(x)=x, $ for all $x \in \Bbb R$.

Using linear algebra, if the 'system' is invertible then such function exists.
I can't recall this theorem.
I'm not sure it is the right approach either.

3 Answers 3

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Hint, ok an extended hint :-): Use the fact that the polynomial $p(y)=y^5+y^4+y^3+y^2+6y$ is continuous and satisfies the limit constraints $\lim_{y\to\infty}p(y)=\infty,\qquad\lim_{y\to-\infty}p(y)=-\infty,$ so it must be surjective by the intermediate value theorem. Hence for all reals $x$ there exists a $y$ such that $p(y)=x$.

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    Correct, @Robert. But I don't think that the OP asked for a continuous function $f$.2012-12-26
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How about this:

For all $x$, there exists $y$ such that $y^5+y^4+y^3+y^2+6y=x$ (because the polynomial has an odd degree).

Let $f(x)=\max\{y\in R|y^5+y^4+y^3+y^2+6y=x\}$

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There must be a real root to the stated equation in $x$ because the polynomial in $f(x)$ is of odd degree, and $f$ is assumed to be a real-valued function.