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Inverse function

  1. Sep 4, 2011 #1
    f(x) = (3 - e^(2x))^(1/2)

    y^2 = 3 - e^(2x)

    -(y^2 - 3) = e^(2x)

    ln(-(y^2 - 3) ) / 2 = x

    What am i doing wrong?
    In the back of the book it says....
    ln(-(x^2 - 3) ) / 2 = y
  2. jcsd
  3. Sep 4, 2011 #2


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    [or y= (3 - e^(2x))^(1/2)]
    Here you should switch the x and y, and then solve for y. The result will give you the inverse. So instead of
    y^2 = 3 - e^(2x)
    you should have
    x^2 = 3 - e^(2y)
    ... and so on.
  4. Sep 4, 2011 #3


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    As eumyang said, it is just a matter of notation. We typically write a function as "y= f(x)", not "x= f(y)". But, of course, as long as f is the same function "y= f(x)", "x= f(y)", or "b= f(a)" all refer to the same function.
  5. Sep 4, 2011 #4
    Knowing that log functions are the inverse of exponential functions, I thought all I had to do was find the log of the expo? This is obviously incorrect for me to do?
  6. Sep 5, 2011 #5


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    [tex]x\rightarrow f \rightarrow y[/tex]
    [tex]y\rightarrow f^{-1} \rightarrow x[/tex]

    which is why your notation is the opposite, in the answer of your book they redefined the inverse function to be a function of x
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