Need help with this!

Question

lolyeha · Answer

What is the question?

samigupta8 · Answer

Let f be a continuous real function such that f(11)=10 and for all x,
f(x)f(f(x))=1 then evaluate f(9)

kainui · Answer

well here's a bit of playing around:

$$f(f(x))=\frac{1}{f(x)}$$
now we can write this in two ways:

$$f(f(f(x))) = f(\tfrac{1}{f(x)}) = \frac{1}{f(f(x))}$$
However, the second part there is f(x) so:

$$f(x) = f(\tfrac{1}{f(x)})$$

kainui · Answer

Start here, and successively plug in x=f(x) 
$$f(x)f^2(x)=1$$$$f^2(x)f^3(x)=1$$$$f^3(x)f^4(x)=1$$$$f^4(x)f^5(x)=1$$$$\cdots$$
Now if you flip the equations, multiply them together, we have:
$$f(x)f^2(x) = f^2(x)f^3(x)$$$$f(x)=f^3(x)$$
We can repeat this process another step to get:
$$f(x)f^4(x)=1$$
So we have sorta an infinite family of ridiculous and probably useless equations:

$$f(x)f^{2n}(x)=1$$and$$f(x)=f^{2n+1}(x)$$

kainui · Answer

wait a sec...

samigupta8 · Answer

From where did you get this?
I didn't get this latter post of yours.

kainui · Answer

From my mind just now? I tried to put all my reasoning there.

kainui · Answer

Hahahahaha! Check this out.

samigupta8 · Answer

Lol!
Sorry but i won't be able to stay up at the moment.

kainui · Answer

$$f(x)f(f(x))=1$$
let $f(x)=9$

$$9*f(9)=1$$
$$f(9)=\frac{1}{9}$$

samigupta8 · Answer

Uhhh! That easy ...

parthkohli · Answer

That'd also mean$$f(10) = 1/10$$Which is not true.
Basically your function is not surjective.

anonymous · Answer

SOMEONE IS ALREADY ANSWERING THIS, PLEASE GO HELP OTHER STUDENTS.

kainui · Answer

Uhh I was wrong @BrownieCat, chill

anonymous · Answer

Okay, that just means one person can stay and watch for errors/correct the ones you've made. Everyone else can go home.

samigupta8 · Answer

Did you mean to say that f(11)=1/11 if we go by his method which is not true since it's value is different?

parthkohli · Answer

Yes.

kainui · Answer

OK I think I have it:

"Let f be a continuous real function such that f(11)=10 and for all x,
f(x)f(f(x))=1 then evaluate f(9)"

since for all x, f(x)f(f(x))=1 then certainly for x=11 we have: f(11)f(f(11))=1 We can exactly evaluate f(11)=10 so we do to get:

$$f(10)=\frac{1}{10}$$
Since we know $$f(11)=10$$ and this is a continuous, real function on R, we know by continuity that for the range 10 to 11,
$$10 < x < 11$$ that $$\frac{1}{10} < f(x) < 11$$ so this implies that there exists an f(x)=9 by continuity. I think they call this like mean value theorem or something maybe idk, whatever.

Because of that we are allowed to now plug in f(x)=9
$$f(9)=\frac{1}{9}$$

samigupta8 · Answer

Thanks kainui.