Bdmo 2014 regionals(a tweaked version of question):

If $f$ is a strictly increasing function over the reals with $f(f(x))=x^2+2$, then $f(3)=?$

Obviously,$f(3)=f(1)^2+2$ but I can't see where we are going to use the 'strictly increasing' fact.I don't think there is a way to reverse-engineer such a function without heavy machinery.I have plugged in loads of values but they have yielded nothing.Some help will be appreciated.

EDIT: As others have noted, such a function is not possible with our current domain, but:

If the function is defined over the positive reals, does $f(3)$ have a definite value?

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    $\begingroup$ $f(-1)<f(1)$ because it is strictly increasing, so we should have $f(f(-1))<f(f(1))$. However your formula contradicts this. It says $f(f(-1))=f(f(1))$ $\endgroup$ – gebruiker Jun 13 '14 at 12:00
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    $\begingroup$ However if $f$ is increasing (not strictly), we can use this argument to show $f(x)= 2$ $\endgroup$ – gebruiker Jun 13 '14 at 12:04
  • $\begingroup$ @gebruiker,indeed.But what happens if we define the function over the positive reals? $\endgroup$ – rah4927 Jun 13 '14 at 12:08
  • $\begingroup$ Then you should say so in the question... $\endgroup$ – gebruiker Jun 13 '14 at 12:09
  • $\begingroup$ @gebruiker,I have added the follow-up question. $\endgroup$ – rah4927 Jun 13 '14 at 12:12

If $f:\mathbb{R}\rightarrow \mathbb{R}$ be a strictly increasing function. Then $f(f(x))$ cannot be even.

Proof: $$x>0\Rightarrow f(-x)<f(x)\Rightarrow f(f(-x))<f(f(x))$$ Contradiction.

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  • $\begingroup$ And hence $f(3)$ is not defined :) $\endgroup$ – Servaes Jun 13 '14 at 12:01
  • $\begingroup$ Ah,indeed,+1.What if the function is defined over the positive reals? $\endgroup$ – rah4927 Jun 13 '14 at 12:03

Any function $f_{base}$ satisfying:

(1) Is defined on $[0,r)$ for some $0<r<2$.

(2) Is continuous and strictly increasing.

(3) $f_{base}(0)=r$ and $\lim\limits_{x\rightarrow r^{-}}f_{base}(x)=2$.

can be extended to a function $f$ on nonnegative real satisfying the above functional equation and is strictly increasing.

This can be proved by induction. We extend to $[r,2)$, $[2,r^{2}+2)$, $[r^{2}+2,6)$ and so on repeatedly using the equation $f(x)=(f^{-1}(x))^{2}+2$. We won't prove it here, as it's not the main point. The main point is that we simply need to specify the function in a small range and trust that it can be extended to the whole nonnegative real. Using this we can show that there are no definite value for $f(3)$.


$f(x)=1+x^{2}$ for $x\in[0,1)$; $x+1$ for $x\in[1,2]$. Then $f(3)=6=2^{2}+2$ since $f(2)=3$.

Now try:

$f(x)=\frac{3}{2}+\frac{2}{9}x^{2}$ for $x\in[0,\frac{3}{2})$; $\frac{9}{2}(x-\frac{3}{2})+2$ for $x\in[\frac{3}{2},2]$. Then $f(3)=(\frac{31}{18})^{2}+2\not=6$ since $f(\frac{31}{18})=3$.

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