Question

Show that there exists a unique bijection from the set of positive real numbers with values in the set that verifies f'(x)=g(x), where g(x) is the functional inverse of f(x).

Answer 1

f'(x)=g(x), where g(x) is the functional inverse of f(x).
The only function that suits the above is the function of y=x. As both y=x is the inverse and non-inverse.

so looking at the line y=x


You can conclude that for every value of x there is a unique value of y, gving one-to-one correspondence, that is necessary for a bijection function

Answer 2

What? Did you miss f'(x)? The derivative of f(x)?

Answer 3

omgomgomg. yesh.

Answer 4

The proof is fugly, lol.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=70&t=37916
There it is if you're interested.

Answer 5

thanks!