Posers and Puzzles

Posers and Puzzles

  1. Joined
    19 Sep '06
    Moves
    3530
    15 Apr '08 22:48
    f(x)= 2 for all real numbers x. What is f(x+2)?
  2. Joined
    13 Dec '06
    Moves
    792
    15 Apr '08 22:53
    2
  3. SubscriberAThousandYoung
    iViva la Hispanidad!
    tinyurl.com/y2c6j2t3
    Joined
    23 Aug '04
    Moves
    24791
    16 Apr '08 00:05
    Originally posted by GregM
    2
    Unless x is not a real number...
  4. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    7733
    16 Apr '08 00:211 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. Joined
    06 Apr '08
    Moves
    1871
    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 memberforkedknight
    Defend the Universe
    127.0.0.1
    Joined
    18 Dec '03
    Moves
    16172
    21 Apr '08 16:231 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. Joined
    15 Feb '07
    Moves
    667
    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. Joined
    12 Sep '07
    Moves
    2668
    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. Joined
    15 Feb '07
    Moves
    667
    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. Joined
    12 Sep '07
    Moves
    2668
    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. Joined
    15 Feb '07
    Moves
    667
    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. Joined
    12 Sep '07
    Moves
    2668
    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
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