help me with this Q:
a) determine whether f with f(x)=f[(x)^1/2].f[(x)^1/2] where x=0 is a function or not? help me with this Q:
a) determine whether f with f(x)=f[(x)^1/2].f[(x)^1/2] where x=0 is a function or not? @Mathematics

Question

help me with this Q:
a) determine whether f with f(x)=f[(x)^1/2].f[(x)^1/2] where x>=0 is a function or not? help me with this Q:
a) determine whether f with f(x)=f[(x)^1/2].f[(x)^1/2] where x>=0 is a function or not? @Mathematics

anonymous · Answer

$$f(x)=f(\sqrt{x})f \sqrt{x})$$

anonymous · Answer

That is utterly maddening. The function is recursive, so I don't see a way to get even a single value for f(x).

asnaseer · Answer

@Dalvoron is right - this is an infinitely recursive function. So, in a sense, it IS a function, but one that will never finish the calculation it has been asked to perform. It depends on what you mean by "is it a function"?

jhonyy9 · Answer

so this exponent 1/2 sign square-root and sqrt(x) can being always one function with x>=0

anonymous · Answer

i dont know..its written like that..:(