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Posers and Puzzles

Posers and Puzzles

  1. 15 Apr '08 22:48
    f(x)= 2 for all real numbers x. What is f(x+2)?
  2. 15 Apr '08 22:53
    2
  3. Subscriber AThousandYoung
    It's about respect
    16 Apr '08 00:05
    Originally posted by GregM
    2
    Unless x is not a real number...
  4. Subscriber joe shmo On Vacation
    Strange Egg
    16 Apr '08 00:21 / 1 edit
    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
  5. 16 Apr '08 09:33
    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.
  6. Standard member forkedknight
    Defend the Universe
    21 Apr '08 16:23 / 1 edit
    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.
  7. 22 Apr '08 02:18
    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.
  8. 23 Apr '08 23:17
    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
  9. 23 Apr '08 23:37
    Missing an end parentheses on that equality.

    It would be useful to know where it is before solving begins.
  10. 24 Apr '08 00:03
    Very sorry:
    Solve this f:R->R

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

    for all reals x,y
  11. 24 Apr '08 02:42
    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.
  12. 24 Apr '08 09:53
    There is another solution, which includes f(x)=1
    f(x)=1 for all x>=-2/3 and |f(x)|=1 for all x