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Homework Help: Could you check this please

  1. Sep 27, 2005 #1

    equation for the axis of symmetry???
    I got x=pi

    f(x)=ln(x^2 - 20)

    describe symmetry??

    I got symmetric arround y axis.


    I got y=sqr of ((x/ln) +20)



    I got y= sqr of (x/ln)
  2. jcsd
  3. Sep 27, 2005 #2


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    Homework Helper

    Since cos x is a repeating function, it has an infinite number of axes of symmetry. They occur at n pi. Something similar needs to be done for exp(cos x).

    For the others, you seem to be using "ln" as a variable of some sort. Remember that ln x is a function, with inverse exp(x). You might want to look at those again.
  4. Sep 27, 2005 #3
    So for others I just change ln to exp?
  5. Sep 27, 2005 #4
    So I get f(x)^-1=e^(X^2-20)


  6. Sep 28, 2005 #5


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    Let me take one of them. You should be able to do the rest.

    y = ln(x^2 - 20)

    exp(y) = x^2 - 20

    x^2 = exp(y) + 20

    x = sqrt(exp(y) + 20)

    So, f^-1(x) = sqrt(exp(x) + 20)

    Note that there are restrictions on the domain and range - these would correspond to the restrictions on the range and domain of the original function.

    Does that help?
  7. Sep 28, 2005 #6
    It does.

    Thank you so much. :smile:
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