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How to See Symmetry about x = -1?

  1. Jun 1, 2012 #1
    Hi.

    Question: Sketch [tex] f(x) = \sqrt{1 - x} + \sqrt{3 + x} [/tex]

    On this test question from calculus, I got full marks for my answer. But I'm posting in this forum because I'd like to know how to analyze the symmetry of this function (posted below), which I actually didn't notice until I read the solution. After looking at my graph on the test again, I somewhat see it.

    But without using a graph or testing points, how would I
    (I) know that f(x) was symmetric?
    (II) compute that the line of symmetry is x = 1 ?

    My guess is that if there exists a L such that for all d, [tex] f(L + d) = f(L - d)[/tex], then [tex] x = L [/tex] is the line of symmetry. So I calculate

    [tex] f(L + d) = f(L - d)[/tex]

    [tex] \Longrightarrow \sqrt{1 - L - d} + \sqrt{3 + L + d} = \sqrt{1 - c + d} + \sqrt{3 + L - d} [/tex]

    But this doesn't seem useful...

    Of course, if I plug L = 1 then I get an identity. But I want to know how to compute L = 1 !

    Thank you...

    10f2btc.png
     
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2

    tiny-tim

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    hi seniorhs9! :smile:
    (you mean √(1-x) + √(3+x) :wink:)

    put x = z + d

    that gives you (1 - d) - z and (3 + d) + z :smile:
     
  4. Jun 2, 2012 #3
    Thanks tiny-tim... I fixed it.

    But I still don't see how to calculate x = 1 as a line of symmetry?

    I put [tex] x = L + d [/tex]

    and [tex] x = L - d [/tex]

    already into f(x) in my original post? Or am I missing something...
     
  5. Jun 2, 2012 #4
    If you shift f(x) to the right one unit by finding f(x-1), you'll get √(2-x) + √(2+x) and you might be able to tell that the two square roots are mirrors of each other along the line x=0.
     
  6. Jun 6, 2012 #5
    Thank you Buhrok... Appreciate your answer.

    I fully understand and get the algebra needed to do this transformation, but how would you even suspect/guess/know to do to this? You would need to suspect that [tex] f(x) = \sqrt{1 - x} + \sqrt{3 - x} [/tex] COULD have symmetry to start?
     
  7. Jun 6, 2012 #6
    You mean √(3+x) :wink:

    The two parts are of the form √(a ± x) as opposed to c√(a ± bx). The only effects that a and ± have on the graphs of the roots is, respectively, a horizontal shift or a reflection based off the original graph of √x. There is no vertical or horizontal stretching of the graph which we would get if we had numbers for b or c other than what we see in the given function.

    Lastly, √(a + x) and √(b - x) "go" in opposite directions because of the different signs, and so √(1 - x) and √(3 + x) go in opposite directions. Taken altogether, you could decuce that there will be symmetry reflected across a vertical line, but not the y-axis in the case a≠b.
     
  8. Jun 6, 2012 #7

    Ray Vickson

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    One possibility: if x = L is a symmetry line then L is either a max or a min of f, so must be a stationary point. The only stationary point is L = -1, so L = -1 is the symmetry point.

    RGV
     
  9. Jun 7, 2012 #8
    Thank you very much Buhrok and Ray Vickson...

    Buhrok...You're right! I mixed up the positive sign again!

    Ray Vickson...Does that argument always work? Even if a stationary point is a saddle point?
     
  10. Jun 7, 2012 #9

    Ray Vickson

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    No. A saddle point means the function is not symmetric (except, of course, for the constant function). However, you originally asked how one could find symmetry points (if they exist) and *searching among the stationary points is enough*. Given a stationary point, you need to do more work to see if it is a symmetry point or not. In fact, just because a function has, say, a maximum or minimum at some point x = L does not mean L is a symmetry point, because most functions having maxima or minima do not have any symmetry at all.

    RGV
     
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