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Math Help - Symmetry about a vertical line (like x = c, c not zero).

  1. #1
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    Symmetry about a vertical line (like x = c, c not zero).

    Hi all ----

    I actually get part (ii) of this question - but I'm just curious - how can I formally prove the symmetry? The question doesn't ask this but I'm just curious. The green box is the solution.



    Part (ii) --- I can see that \int_0^1 2^{-(x - c)^2} reaches maximum when c = 1/2.

    But it also looks like 2^{-\left(x - \frac{1}{2})^2\right} is symmetric about x = 1/2. How would I prove this?

    I know how to do this for f(x) = 2^{-x^2} ---

    f(x) is symmetric about x = 0 because f(-x) = 2^{-(-x)^2} = f(x) = 2^{-x^2} so f(-x) = f(x).

    Thanks a lot ---
    Last edited by mathminor827; September 18th 2011 at 06:18 AM.
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  2. #2
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    Re: Symmetry about a vertical line (like x = c, c not zero).

    y = 2^{-\left(x - \frac{1}{2})^2\right} is symmetric about x=1/2, but not x=1.
    try to translate y = 2^{-\left(x - \frac{1}{2})^2\right} 1/2 unit left
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  3. #3
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    Re: Symmetry about a vertical line (like x = c, c not zero).

    Quote Originally Posted by piscoau View Post
    y = 2^{-\left(x - \frac{1}{2})^2\right} is symmetric about x=1/2, but not x=1.
    try to translate y = 2^{-\left(x - \frac{1}{2})^2\right} 1/2 unit left
    Hi piscoau ---

    Thanks - I fixed it now.

    But I know how to show symmetry for f(x) = 2^{-x^2}. I did this already in my original post. I want to prove symmetry for 2^{-\left(x - \frac{1}{2})^2\right} about x = 1/2.
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  4. #4
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    Re: Symmetry about a vertical line (like x = c, c not zero).

    do you know how to translate the graph of a function horizontally? If the question want you to prove f(x) is symmetry about x=c, then you need to translate the graph c unit left/right.
    For example, you want to prove function (x-1)^4 is symmetry about x=1. you need to translate the function 1 unit left, the function will become (x-1+1)^4 = x^4. Obviously it's symmetry about x=0, i.e. x=0 is its symmetry axis of x^4. If you translate x^4 back to (x-1)^4, the whole graph will move 1 unit right, i.e. the symmetry axis will move 1 unit right too. Therefore, (x-1)^4 is symmetry about x=1.
    If a figure is translated either horizontally or vertically, only its position changed, its shape, size, pointing direction DOESN'T change. So I think you know how to do it know.
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  5. #5
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    Re: Symmetry about a vertical line (like x = c, c not zero).

    Quote Originally Posted by mathminor827 View Post
    Hi piscoau ---

    Thanks - I fixed it now.

    But I know how to show symmetry for f(x) = 2^{-x^2}. I did this already in my original post. I want to prove symmetry for 2^{-\left(x - \frac{1}{2})^2\right} about x = 1/2.
    1. Take 2 x-values equidistant from x = \tfrac12 . for instance x = \tfrac12 - d~\text{or}~x=\tfrac12 + d~,~d \in \mathbb{R}

    2. If f\left(\tfrac12 - d  \right) = f\left(\tfrac12 + d  \right)

    then the graph of f is symmetric about x = \tfrac12
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