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Concavity, inflection point, maxima, minima

  1. Nov 20, 2011 #1
    1.Find the vertical and horizontal asymptotes (if any), intervals of increase and decrease, local maximum and minimum values, intervals of concavity and inflection points for f(x) = 1 / (L - x^1/2)




    3. I found the vertical asymptote to be "x=0" , I found the horizontal asymptote to be "y=0"
    So to find intervals of increase and decrease you have to find the derivative, which I found to be 2*x/(L^2-2*x^2*L+x^4) from here I plugged in values.

    To find local maximum and minimum values you have to find the second derivitive, which I found to be (2*L+6*x^2)/(L^3-3*x^2*L^2+3*x^4*L-x^6)

    Then I came to the conclusion that there is no inflection point because there are no local maximum's meaning no change in concavity

    I'm not sure if I did these right, would really appreciate some help, thanks.
     
  2. jcsd
  3. Nov 20, 2011 #2
    Look over your methods of finding critical points and points of inflection.
    What does the first derivative mean?
     
  4. Nov 20, 2011 #3
    Yes, I think I stated it wrong in my first post I used the first derivative to find the intervals of increasing and decreasing values and also for minimum and maximum values. I used the second derivative to find the concavity.
     
  5. Nov 20, 2011 #4

    Dick

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    Do you really mean f(x)=1/(L-sqrt(x))? Why do you think there is a vertical asymptote at x=0? And I don't think any of your derivatives are correct either. Or did you mean something else altogether?
     
  6. Nov 20, 2011 #5
    No I meant L - x^2
     
  7. Nov 20, 2011 #6

    Dick

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    Ok, so f(x)=1/(L-x^2), right? Still doesn't have a vertical asymptote at x=0. But at least your derivatives are making more sense. When you are thinking about this, I wouldn't expand powers of (L-x^2). Just leave them as they are.
     
  8. Nov 20, 2011 #7
    When taking the derivative, remember L is not a variable, just a number.
     
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