10
$\begingroup$

Determine whether there exists a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f\big(x^3+x\big)\le x\le\big(f(x)\big)^3+f(x)$$ for all $x\in\mathbb{R}$.


Source: Math Excalibur Volume 22 No. 4 Page 3 Problem 536 rephrased

$\endgroup$
9
$\begingroup$

Let $g(x) := x^3+x$. Then the problem asks for $f$ such that $f(g(x)) \leq x \leq g(f(x))$.

Since $g$ is bijective, choose $f = g^{-1}$, and the inequality becomes $x \leq x \leq x$, which is satisfied for all $x$.

EDIT: Since it was mentioned that there is

only one solution, here's a quick proof of that too:

Let $x$ be arbitrary, let $y := g^{-1}(x)$. We have $g(f(x)) \geq x$, but we also have $f(g(y)) \leq y$, so $g(f(x)) = g(f(g(y)) \leq g(y) = x$ by monotony of $g$. Hence, $g(f(x)) = x$, so $f(x) = g^{-1}(x)$.

$\endgroup$
1
  • 1
    $\begingroup$ Correct! And the only possible solution is the inverse of $x^3+x$! $\endgroup$ May 27 '20 at 4:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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