9
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ $f(-1)<f(1)$ 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
  • 1
    $\begingroup$ 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
13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $f(3)$.

Try:

$f(x)=1+x^{2}$ 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 $f(\frac{31}{18})=3$.

$\endgroup$

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.