Is it possible to solve this without guessing the function?

[Drao92]

We know f(f(x))=2^x-1;
We must calculate f(1)+f(0)
Its posible to solve this without guessing the function? If i derivate i get to f(0)=f(0) or f(1)=f(1).

 

[vela]

We know f(f(x))=2^x-1;
We must calculate f(1)+f(0)
Its possible to solve this without guessing the function? If i derivate i get to f(0)=f(0) or f(1)=f(1).

What does derivate mean?

Yes, it's possible to solve without knowing what f(x) is.

Hint: Consider f(f(f(x))) to show that $f(2^x-1) = 2^{f(x)}-1$.

 

[Drao92]

Isnt f(2^x-1)=2^f(f(x))-1? And in this case it becamse very easy.
The hint leads me to f(0)=f(0)^f(0)-1 , so f(0)=0 or 1.
Same result for f(1).
I guess the function is injective so the sum is 1?
Actually i solved the function but with diferential equation and obviously i don't know the constant :D.
Is this correct?
f(f(f(x)))=2^(f(f(x)))-1
f(f(f(x)))=f(2^x-1)=2^(f(f(x)))-1=f(2^x-1)=2^(2^x-1)-1
x=0=>f(2^0-1)=2^(2^0-1) -1<=>f(0)=0;
x=1=>f(2^1-1)=2^(2^1-1)-1<=>f(1)=1;
f(1)+f(0)=1;

 

[vela]

How'd you get that?

 

[Drao92]

No, sorry you are right. I messed up the argument of the function :D
Thanks, i understood .