function

f(x)= 2 for all real numbers x. What is f(x+2)?

2

Originally posted by GregM
2

Originally posted by golfer1
f(x)= 2 for all real numbers x. What is f(x+2)?

yeah, but consider a slight modification of the Dirichlet's function

f(x) = 2 if x is real, and f(x) = 0 if x is complex ... now you have two answers...

or f(x) = 2 if x is real, and f(x) = Re(x) if x is complex ... now the range of the function is the entire R.

in other words, the question as posed is incomplete, without stating the domain of the original function.

Originally posted by 3v1l5w1n
yeah, but consider a slight modification of the Dirichlet's function

f(x) = 2 if x is real, and f(x) = 0 if x is complex ... now you have two answers...

or f(x) = 2 if x is real, and f(x) = Re(x) if x is complex ... now the range of the function is the entire R.

in other words, the question as posed is [b]incomplete, without stating the domain of the original function.[/b]

Real numbers fall within the realm of complex numbers, because any real number R could be written as R + 0i.

If x is real, then x+2 is also real, and hence f(x+2) = 2.

F(x) + 2 would equal 4, however.

Solve this f:R->R

f(x^2+y^2+2f(xy)=f(x+y)^2

for all reals x,y

have fun! Three solutions

Missing an end parentheses on that equality.

It would be useful to know where it is before solving begins.

Very sorry:
Solve this f:R->R

f(x^2+y^2+2f(xy))=f(x+y)^2

for all reals x,y

I can find 3 solutions.

f(x) = 0
f(x) = 1
f(x) = x

I do not know how to prove those are the only 3 solutions, or indeed if others may exist.

There is another solution, which includes f(x)=1
f(x)=1 for all x>=-2/3 and |f(x)|=1 for all x