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March 25, 2017

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Auppose f is even but g is an arbitrary function (possibly neither even nor odd).
(a) Is f o g still even? Show why or why not.
(b) Is g o f even? Show why or why not.

  • Math - ,

    fog is even, since f is even.

    since f(-x) = f(x), f(-g) = f(g)

    gof is unknown. for example,

    f(x) = x^2
    g(x) = x
    g(f) = f = x^2 is even

    g(x) = √x
    g(f) = √f = x is odd

    g(x) = √x + 1
    g(f) = √f + 1 = x+1 neither even nor odd

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