If f(x)=1-x, f(f(x))=x. Can you provide another such f(x), other than the obvious f(x)=x or f(x)=-x?

Can you find a real function g(x) such that g(g(x))=-x? Note that if we allowed complex numbers, g(x)=ix would do the job.

Can you find another real function h(x) so for x≠0, h(h(x))=1/x?

In all cases, if you cannot find a solution that works for all x, a function valid for some ranges is better than nothing!

how about

f(x)=(k- x^(m/n) ) ^ (n/m)

where m and n are basically any numbers.

The solutions that cover all real numbers, for maximum domain and range, however, are those with:

1)m/n positive. A negative exponent causes a problem with divide-by-zero

2)m and n odd. Even numbers cause a problem with roots of negative numbers.

eg m=1 n=1 f(x)=k-x
    m=3 n=1 f(x)=³√(k-x³)
    m=1 n=3 f(x)=(k-³√x)³
    m=3 n=5 f(x)=(k- x^(3/5) ) ^ (5/3)

Specific Solution: f(x)=³√(1-x³)

Posted by Dimmeh on 2005-08-22 10:35:09