hi guys,
some calculus:
f(x) has domain [0,infinity] and range [-1,1]
g(x) = f^-1(x)
f^2(x) = F(x)
is inversion possible? F^-1(x) = ?

Question

hi guys,
some calculus: 
f(x) has domain [0,infinity] and range [-1,1]
g(x) = f^-1(x)
f^2(x) = F(x)
is inversion possible? F^-1(x) = ?

anonymous · Answer

doesn't seem likely since if the range of f is [-1,1] then the range of 
$$f^2$$ would be [0,1]

anonymous · Answer

but that is just a guess. i would be better to think up an example.   need a one to one function form (0, infinity) to [-1,1] then maybe we could do it by example

anonymous · Answer

suppose for example that 
$$f(a)=-\frac{1}{2}$$ then 
$$f^2(a)=\frac{1}{4}$$ but if the range of f is really [-1,1] then 
$$\frac{1}{4}$$ already has a preimage not equal to a in 
$$(0,\infty)$$ so i would say no, function would not be one to one