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Proving even and odd functions

  1. Jan 15, 2009 #1
    Can someone prove even and odd fucntions for me not through examples but by actually proving them?

    Thanks
     
  2. jcsd
  3. Jan 15, 2009 #2
    Well how do you define even and how do you define odd functions?
     
  4. Jan 16, 2009 #3
    DEFINITION. A function f is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f.

    A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f

    I need an algebraic proof using angles and algebra... its for my trig class.
     
  5. Jan 16, 2009 #4
    Angles and algebra? Why don't you do it by plugging -x in? Like you said if f(-x) = f(x) then f is even if f(-x) = -f(x) then f is odd. Where are you getting lost?
     
  6. Jan 16, 2009 #5
    Is that a proof though?
     
  7. Jan 16, 2009 #6
    Yes... you are taking a definition and using it. What's a proof in your opinion?
     
  8. Jan 16, 2009 #7
    You are defining even and odd functions to have those properties. There is no need for a proof.
     
  9. Jan 16, 2009 #8
    IS there a way to prove an even and odd function?
     
  10. Jan 16, 2009 #9
    How can you prove in general what an even or odd function is? First you have to define what you mean by that, but once you do that, there is no need for a proof.
     
  11. Jan 16, 2009 #10

    symbolipoint

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    Start with a specific function and test it for identity of evenness and oddness according to the definition for even and odd functions.
     
  12. Jan 16, 2009 #11
    I'm not sure you understand what a proof is. You have a bunch of axioms, and you use those to arrive from part A to part B. Your axiom is what an even and what an odd function are. Using them, you show that some functions are even, some are odd, some are neither.
     
  13. Jan 16, 2009 #12
    so could an axiom be f(x)=|x|?
     
  14. Jan 16, 2009 #13
    What axiom would that be? If you are trying to DEFINE an absolute value function then you can say, f(x) = |x| means f(x) = x for x >= 0 and -x for x <= 0
     
  15. Jan 16, 2009 #14

    symbolipoint

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    USE the Definitions of EVEN functions and ODD functions. Does one statement or the other become an identity?

    Check:
    |-x|=|x|

    Check:
    |-x|=-|x|

    It one of those or both of those or neither of those true? What is the meaning?
     
  16. Jan 16, 2009 #15

    Mentallic

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    What about proving that if an even function by definition being [tex]f(x)=f(-x)[/tex] is a reflection in the y-axis and an odd function defined as [tex]f(-x)=-f(x)[/tex] has symmetry about the origin?
     
  17. Jan 16, 2009 #16

    HallsofIvy

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    Then you have to start with definitions of "reflection in the y-axis" and "symmetry about the origin"! It should be obvious that you can't prove anything about "X" if you don't know what "X" means and, further, math definitions are "working definitions"- you use the precise words of the definitions themselves in working with the concepts.

    The reflection of (x,y) in the y-axis is the point (-x, y). If f is an even function and y= f(x), what are (x, y) and (-x, y) in terms of the graph of f? Are they both on the graph?

    The point "symmetric" to (x, y) in the origin is (-x, -y). If f is an odd function and y= f(x), what are (x, y) and (-x, -y) in terms of the graph of f? Are they both on the graph?
     
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