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Strange Function

  1. Jul 25, 2008 #1
    Here's a problem

    Let k: R\{-1} → R be given by [tex]k(x) = \frac{2x-1}{x+1}[/tex]

    Prove that k is neither even nor odd.

    That is strange!! But to prove it we go back to the definition;
    A function, [tex]f: (-a,a) \rightarrow R[/tex] is said to be even if for all [tex]x \in (-a,a)[/tex] => [tex]f(x) = f(-x)[/tex]
    And it is odd if [tex]f(x) = -f(-x)[/tex].

    So: if k is even then [tex]k(-x) = k(x)[/tex] for any x
    and [tex]k(x) = -k(-x)[/tex] if it is odd.

    I get the feeling that my proof doesn't seem right and somewhat unimplemented... (1) is there a better way to prove and conclude that k is neither even nor odd?
    And (2) if it is neither even nor odd, then what could it be?
    Last edited: Jul 25, 2008
  2. jcsd
  3. Jul 25, 2008 #2


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    Well, as you have stated it, all you need to do to prove that is to show that k(x) doesn't satisfy either the odd or even criterion. And there's nothing strange about a function being neither odd nor even since a whole lot of functions are neither odd nor even, such as e^x, ln(x) etc.
  4. Jul 25, 2008 #3


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    I'm not sure why you think that function is "strange". It's a simple rational function. If it is because it is neither even nor odd, well most functions are neither even nor odd.

    To be even, a function must satisfy, as you say, f(x)= f(-x) for all x.
    To be odd, a function must satisfy, as you say, f(x)= -f(-x) for all x.

    To prove something is NOT "true for all x", you only need to find a counterexample.

    If f(x)= (2x-1)/(x+1), what is f(2)? What is f(-2)?

    That's all you need.
  5. Jul 28, 2008 #4
    Hall, that's because I thought since even functions are their own reflection in the y-axis. Even functions are symmetric about the y-axis.
    Whereas other functions which are not even (odd functions) aren't like that.
    Yeah, I understood the error I made.

    Yes, I get it. Hi@Defennder

    [tex]\frac{2 \times 2-1}{2+1}[/tex] = [tex]\frac{2 \times -2-1}{-2+1}[/tex]
    1 ≠ -5
    But they don't equal, thus they don't satisfy neither of the conditions [tex]f(x) = f(-x)[/tex] or [tex]f(x) = -f(-x)[/tex].

    ^ is that a valid proof?
    Last edited: Jul 28, 2008
  6. Jul 28, 2008 #5


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    Yes, that is a perfectly valid proof: to show that a general statement is NOT true, it is sufficient to give a 'counter example': one example where it is not true.

    Yes, it is true that an even function is symmetric about the x-axis. It is NOT true that if a function is not symmetric about the x-axis- that is that it is not even- it is necessarily "odd". An odd function is symmetric through the origin: if (x,y) is on the graph then extending a line from (x,y) through (0,0) an equal distance on the other side of (0,0), that is to (-x,-y), you are again on the graph: f(-x)= -f(x).
  7. Jul 30, 2008 #6
    If one maps the excluded point (here -1) to the origin such the (new) domain is R\{0}, things fall back to the old definition of f(x) and f(-x). Which in this case would be [tex]\frac{2x-3}{x}[/tex]. This is pretty simple now :)
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