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Posers and Puzzles

Joined 19 Sep '06 Moves 3530 f(x)= 2 for all real numbers x. What is f(x+2)?

Joined 13 Dec '06 Moves 792

tinyurl.com/y2c6j2t3

Joined 23 Aug '04 Moves 24791 Originally posted by GregM
2 Unless x is not a real number...

podunk, PA

Joined 10 Dec '06 Moves 7733 Originally posted by golfer1
f(x)= 2 for all real numbers x. What is f(x+2)? I case you dont know f(x)= 2 is a horizantal line. the (+2) technically shifts the graph of f(x) =2 to the left 2 units, but on a horizantal line nothing happens, so in general f(x + a ) = 2 for all real numbers a....

or at least I hope so..lol

Joined 06 Apr '08 Moves 1871 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.

127.0.0.1

Joined 18 Dec '03 Moves 16172 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]generally when someone says f(x) = ..., for all real numbers x, I assume R to be the domain.

Joined 15 Feb '07 Moves 667 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.

Joined 12 Sep '07 Moves 2668 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

Joined 15 Feb '07 Moves 667 Missing an end parentheses on that equality.
It would be useful to know where it is before solving begins.

Joined 12 Sep '07 Moves 2668 Very sorry:
Solve this f:R->R
f(x^2+y^2+2f(xy))=f(x+y)^2
for all reals x,y

Joined 15 Feb '07 Moves 667 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.

Joined 12 Sep '07 Moves 2668 There is another solution, which includes f(x)=1
f(x)=1 for all x>=-2/3 and |f(x)|=1 for all x