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Math Help - f(t) = t^2 defined for 0 < t <= 2 pi. Even or not?

  1. #1
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    f(t) = t^2 defined for 0 < t <= 2 pi. Even or not?

    f(t) = t^2 defined for 0 &lt; t &lt;= 2 pi. Even or not?-sketch.gif
    If I substitute -t for t I get back t^2, so this function would appear to be even. However if I sketch the function (attached) it is not symetrical in the y-axis.
    So is the sketch correct and if so, is this function even or not?

    Thanks for the help
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    f:A\subset \mathbb{R}\rightarrow \mathbb{R} is an even function iff:

    (i) t\in A \Leftrightarrow -t\in A

    (ii) f(-t)=f(t)\;\;\forall t\in A

    Your function does not satisfy (i).


    Fernando Revilla
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    Thanks for the quick response!
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