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Derivatives& the Slope of the Graph: Inflection Point

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    For the function
    f(x)=(x^2-3)/(x-2),
    determine the locations of any points of inflection, if any
     
    Last edited: Mar 29, 2010
  2. jcsd
  3. Mar 29, 2010 #2
    Your function is [tex] f(x)=(x^2-3)(x-2) [/tex] right? Maybe if you foil it out first before you take the derivative?
     
    Last edited: Mar 29, 2010
  4. Mar 29, 2010 #3
    The second derivative is right though; I entered it in and it was correct.
     
  5. Mar 29, 2010 #4
    [tex] f(x)=(x^2-3)(x-2) [/tex] is a polynomial and therefore it's domain should be all real number and this should carry on to it's derivatives as well.

    [tex] f(x)=(x^2-3)(x-2) = x^3-2x^2-3x+6 [/tex] right? So there shouldn't be a quotient at all.
     
  6. Mar 29, 2010 #5
    oh the original function is: f(x)=(x^2-3)/(x-2)
     
  7. Mar 29, 2010 #6
    okay, so we have [tex] f(x)=\frac{(x^2-3)}{(x-2)} and f''(x)=\frac{2(x-2)}{(x-2)^4} [/tex] Even though x=2 is a vertical asymptote it is still a critical point on the second derivative
     
  8. Mar 29, 2010 #7

    Mark44

    Staff: Mentor

    Is there a typo in your functions? You have f(x)=(x^2-3)(x-2), but from what you have for f''(x), did you mean f(x) = (x^2 - 3)/(x -2)?
     
  9. Mar 29, 2010 #8
    sorry, yes that was a typo
    f(x)= (x^2−3)/(x-2)
    f '(x)= [(x−3)(x−1)]/(x-2)^2
    f ''(x) = (2(x−2))/(x−2)^4

    the derivatives are all correct
     
  10. Mar 29, 2010 #9
    yes it is a critical point and it's where the function shifts from negative to positive, but it is an inflection point?
    It's not continuous at x=2.
     
  11. Mar 29, 2010 #10
    You're right it shouldn't be an inflection point because the original functions tangent line doesn't at x=2, but possibly it's asking for all critical points on the second derivative? Is this some kind of online homework program?
     
  12. Mar 29, 2010 #11
    yes it is. This is the exact question (copied and pasted):
    determine the locations of any points of inflection, if any.. Give your answer as a list of x values, e.g., {6,7,8,9}
     
  13. Mar 29, 2010 #12

    Mark44

    Staff: Mentor

    No, that's not true. 2(x - 2)/(x - 2)^4 is identically equal to 2/(x - 2)^3. x = 2 is not in the domain of either function, so is not a critical number for the 2nd derivative. It's true that the concavity changes on either side of x = 2 for the original function, but for x = c to be an inflection point, f has to be defined at c. IOW, I agree with you that the function has no inflection point.
     
  14. Mar 29, 2010 #13
    Isn't the definition of a critical point a point on f(x) that occurs at x0 if and only if either f '(x0) is zero or the derivative doesn't exist at that point?
     
  15. Mar 29, 2010 #14
    yeah, I think 2 is a critical point because I used that to find concavity and I got the concavity questions right.
     
  16. Mar 29, 2010 #15
    This is impossible.
     
  17. Mar 29, 2010 #16

    Mark44

    Staff: Mentor

    For x0 in the domain of f, x0 is a critical point if either f'(x0) = 0 or f' is not defined at xp.

    The trouble is that x = 2 is not in the domain of the original function, f(x) = (x2 - 3)/(x - 2), so couldn't possibly be an inflection point.
     
  18. Mar 29, 2010 #17
    then what else can be the inflection point?
     
  19. Mar 29, 2010 #18

    Mark44

    Staff: Mentor

    There isn't one. The only place the concavity changes is at x = 2, but x = 2 isn't in the domain of the original function.
     
  20. Mar 29, 2010 #19
    I e-mailed to my professor about it and we were right, just so you know :)
     
    Last edited: Mar 29, 2010
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