0
$\begingroup$

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

3

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 $p(y)=x$.

  • 0
    Also add that it is invertible!2012-12-26
  • 2
    Is it? May be? Do we need that?2012-12-26
  • 0
    But from the question is seems the PO want a function s.t there is an equality for all $x$...I don't understand how this answers the question2012-12-26
  • 1
    Define $f$ by mapping $x$ to $y$, where $y$ is any solution of $p(y)=x$. Pick one if there are more! No need for $p$ to be invertible.2012-12-26
  • 1
    @Belgi: I only wanted to give a hint.2012-12-26
  • 0
    @JyrkiLahtonen You are right2012-12-26
  • 0
    But you do want it to be invertible if you want a continuous function $f$.2012-12-26
  • 0
    Thanks! I didn't understand that this gives a way to define $f$ (+1)2012-12-26
  • 0
    Correct, @Robert. But I don't think that the OP asked for a continuous function $f$.2012-12-26
2

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

2

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.