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Math Help - Proving if a function is odd or even

  1. #1
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    Proving if a function is odd or even

    A function f is even if: f(-x) = f(x)
    A function f is off if: f(-x) = -f(x)

    a) Suppose that f and g are both odd. Prove that y = f(x)g(x) is even.
    b) suppose that f is even and g is odd. Prove that y = f(x)g(x) is odd.
    c) Suppose that f and g are botheven. Prove that y = f(x)g(x) is odd.

    My attempt for a), but I'm not sure if I did it correctly.
    f(-x)g(-x) = -f(x) (-g(x))
    Then I brought the (-)'s out on the left side
    -f(x) (-g(x))= -f(x) (-g(x))

    How would I do b and c) if it's not correct?
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  2. #2
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    a
    f(-x)g(-x)=-f(x)*-g(x)=f(x)g(x)

    b
    f(-x)g(-x)=f(x)*-g(x)=-f(x)g(x) lets call f(x)g(x)=h(x) -h(x)=h(-x)

    c
    f(-x)g(-x)=f(x)g(x)
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  3. #3
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    a) f(-x)g(-x) = [-f(x)][-g(x)]=(-1)(-1)f(x)g(x) = f(x)g(x) (you essentially did this correct, but never multiplied the two -1's together at the end)

    looks like you have the idea. Try b and c yourself, and I'll tell you if you got it right.
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  4. #4
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    Quote Originally Posted by DrSteve View Post
    a) f(-x)g(-x) = [-f(x)][-g(x)]=(-1)(-1)f(x)g(x) = f(x)g(x) (you essentially did this correct, but never multiplied the two -1's together at the end)

    looks like you have the idea. Try b and c yourself, and I'll tell you if you got it right.
    Okay.
    So I always start the left side with f(-x) or g(-x)?

    b) f(-x) * g(-x) = f(x) * (-g(x))
    Then I take out the -1's like you said, for left side:
    (-1) (-1) f(x) g(x) = - f(x)g(x)
    f(x) g(x) = -f(x)g(x)

    ...? lol.

    c) f(-x) g(-x) = f(x) g(x)
    (-1)(-1) f(x) g(x) = f(x) g(x)
    f(x) g(x) = f(x) g(x)
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  5. #5
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    Quote Originally Posted by TN17 View Post

    b) f(-x) * g(-x) = f(x) * (-g(x))
    Then I take out the -1's like you said, for left side:
    (-1) (-1) f(x) g(x) = - f(x)g(x)
    f(x) g(x) = -f(x)g(x)

    ...? lol.
    No, no, no....there's no "taking out -1's on the left." This doesn't make sense. You simply apply the def of even and odd.

    f(-x)g(-x) = f(x)[-g(x)]=f(x)(-1)g(x)=(-1)f(x)g(x) = -f(x)g(x).

    I'm simply working from left to right here.
    Across the first equal sign I'm applying the definitions of even and odd.
    Across the second equal sign I simply revealed the "hidden" 1.
    Across the third equal sign I just used the fact that multiplication is commutative to move the -1 to the front.
    And finally I hid the 1 again.
    (In general hiding and unhiding the 1 is unnecessary. I just did it here because I think that this may be what is confusing you.)


    To summarize, to check algebraically if a function is even or odd you replace x by -x, and see what happens,
    The function we are checking here is f(x)g(x).
    So I am replacing x in this function by -x. Thus I'm "checking" what f(-x)g(-x) is equal to.

    Note: I am abusing notation a bit here to make things less confusing. This particular abuse of notation is pretty standard however, so I won't confuse you further by writing the technical details.
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