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Math Help - check if the function is odd or even..

  1. #1
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    check if the function is odd or even..

    i know that if f(x)=f(-x) and if f(-x)=-f(x) then its odd.

    i tried to do this check on this cases
    but its not working

    http://img126.imageshack.us/img126/3942/74448798fv8.gif
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by transgalactic View Post
    i know that if f(x)=f(-x) and if f(-x)=-f(x) then its odd.

    i tried to do this check on this cases
    but its not working

    http://img126.imageshack.us/img126/3942/74448798fv8.gif
    for 2A, what is your -f(x) = -\log \dfrac{1-x}{1+x}?
    2B is an example of a function which is neither odd nor even..
    2C.. you haven't done it.. i'll do it for you..

    if you replace x with -x, then..

    f(-x) = \left\{\begin{array}{cl} -x + 1,& \mbox{for } -x < -1 \\ 0, &\mbox{for }-1\leq -x \leq 1 \\ -x - 1, &\mbox{for } -x > 1  \end{array}\right.
    or
    f(-x) = \left\{\begin{array}{cl} -x + 1,& \mbox{for } x > 1 \\ 0, &\mbox{for }1\geq x \geq -1 \\ -x - 1, &\mbox{for } x < -1 \end{array}\right.

    but this is precisely -f(x).. check it..
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  3. #3
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    f(x) = \log\left(\frac{1-x}{1+x}\right)

    f(-x) = \log\left(\frac{1+x}{1-x}\right) = \log\left(\frac{1-x}{1+x}\right)^{-1} = -\log\left(\frac{1-x}{1+x}\right)

    odd function


    f(x) = \sin\left(x + \sqrt{1+x^2}\right)

    f(-x) = \sin\left(-x + \sqrt{1+(-x)^2}\right) = \sin\left(-x + \sqrt{1+x^2}\right)

    this function and the last one are neither even or odd
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by skeeter View Post
    f(x) = \log\left(\frac{1-x}{1+x}\right)

    f(-x) = \log\left(\frac{1+x}{1-x}\right) = \log\left(\frac{1-x}{1+x}\right)^{-1} = -\log\left(\frac{1-x}{1+x}\right)

    odd function


    f(x) = \sin\left(x + \sqrt{1+x^2}\right)

    f(-x) = \sin\left(-x + \sqrt{1+(-x)^2}\right) = \sin\left(-x + \sqrt{1+x^2}\right)

    this function and the last one are neither even or odd
    i think, no i believe, the last function is an odd function..
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  5. #5
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    how its odd?
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  6. #6
    MHF Contributor kalagota's Avatar
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    well, negation of a piecewise function is negation of the components ONLY of the piecewise function and the negation process does not include the domain of each component..

    why don't you negate each component of your function, and you will see that it is precisely the f(-x) i did in the previous post..
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  7. #7
    MHF Contributor kalagota's Avatar
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    well, here it is...

    -f(x) = \left\{\begin{array}{cl} -(x + 1), &\mbox{for } x < -1  \\ 0, &\mbox{for }-1\leq x \leq 1 \\ -(x - 1),& \mbox{for } x > 1 \end{array}\right.

    f(-x) = \left\{\begin{array}{cl} -x + 1,& \mbox{for } x > 1 \\ 0, &\mbox{for }1\geq x \geq -1 \\ -x - 1, &\mbox{for } x < -1 \end{array}\right.
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  8. #8
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    i ment the sinus function
    how can you conclude that its odd?
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  9. #9
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    graph of the sine function has no symmetry with respect to the y-axis or the origin. it is neither even or odd.

    the piece-wise function is odd ... I mistook the last "piece" as x = 1 instead of x-1.
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