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Graph drawing

  1. Feb 28, 2013 #1
    how can I draw y = arcsec(e^x) where x > 0 and 0 < y < pi/2

    I've found the derivative which I got to be [itex] \frac{1}{\sqrt{e^{2x} -1 }} [/itex] however, the graph doesn't seem to have any minimum or maximum points.

    I don't really know how to test this about, as I don't know what arcsec(e^x) is defined as, so I'm not sure how I can sketch it without being able to test some values of x..
    thanks
     
  2. jcsd
  3. Feb 28, 2013 #2

    tiny-tim

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    hi phospho! :smile:
    so you know it has no bumps!

    ok, then all you need is to plot a few specimen values, and join them smoothly :wink:

    (it might help calculations is you rewrite it as secy = ex, or x = … ?)
     
  4. Feb 28, 2013 #3

    HallsofIvy

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    I'm a bit confused as to what you are saying. arcsec is, of course, the inverse function to secant which is itself the reciprocal of cosine. If y= arcsec(x) then x= sec(y)= 1/cos(y). So cos(y)= 1/x and y= arccos(1/x). arcsec(e^x), then, is arccos(e^{-x}) and any calculator should have those functions.
    For example, if x= 0, arcsec(e^x) is arccos(e^0)= arccos(1)= 0. If x= 1, arcsec(e^x)= arccos(e^{-1})= 0.9331, etc.
     
  5. Feb 28, 2013 #4

    pasmith

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    What you have there is [itex]\frac1{\cos(y)} = e^x[/itex], which can be re-arranged to give [itex]\cos(y) = e^{-x}[/itex] so that [itex]x = -\ln(\cos(y))[/itex] which might be easier to sketch.
     
  6. Feb 28, 2013 #5
    got it, thank you
     
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