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Family of functions etc

  1. Apr 23, 2007 #1
    The problem statement, all variables and given/known data
    [​IMG]

    The attempt at a solution

    [​IMG]
    I want to find the sign (i.e. max or min) of the function, however I'm not quite sure if my last step (deriv>0 for x>a/2) and hence deriv<0 for for x<a/2 is correct. I think this is a better idea than finding the double derivative because it starts to get messy.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2007 #2

    mjsd

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    NB: x=0 is also a critical point
    look at the 2nd derivative to check the nature of turning point.. and it doesn't get that messy... product rule is a bit like binomial theorem
     
  4. Apr 23, 2007 #3
    Oh yeh, you get the x=0 by dividing 0 by that exponential correct! :approve:

    Using double derivative method I get...

    [​IMG]

    But for f''(a/2) the answer will be in terms of a. Hope you can read what I've written, but how would I determine if f''(x) is positive? (I know a>0, but I'm still confused)
     
  5. Apr 23, 2007 #4

    Dick

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    Also the nonzero critical point is not at a/2. It's at 2/a. You might find it's a lot easier to recognize the sign of f'' if you put in the right number for the critical point.
     
  6. Apr 23, 2007 #5
    silly me....:yuck:
     
  7. Apr 23, 2007 #6
    I figure out f"(2/a) = -2e^(-2)
    which is a negative number hence at x=2/a a maximum occurs

    *thumbsup*

    For part c) I'm assuming I just solve for f" = 0?
     
  8. Apr 23, 2007 #7

    HallsofIvy

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    By the way, you very first statement, that "the limit, as |x| goes to infinity is 0, because e-x is 0 is wrong. x2e-x goes to 0 as x goes to + infinity, but increases without bound as x goes to negative infinity.
     
  9. Apr 24, 2007 #8
    Thanks for pointing that out. I'm still stuck on c). Is it simply letting f"(x) = 0?
     
  10. Apr 24, 2007 #9

    HallsofIvy

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    "Points of inflection" are places where the derivative changes sign. Those can only occur, of course, at places where the second derivative is 0 but you should then check that the first derivative actually does change sign.
     
  11. Apr 24, 2007 #10
    Yeh, for the first derivate at x=0 a minimum occurs and at x=2/a a maximum occurs.

    Im having problems solving for f"(x) = 0 though. I take out e^-ax as a common factor but I'm then left with (-2xa) + (x^2a^2) + 2 - 2x = 0

    :confused: :confused:
     
  12. Apr 24, 2007 #11

    Dick

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    I don't quite agree with your result for f''(x) with the exponential factored out. But once you get it right, then this looks like a job for the quadratic formula, doesn't it?
     
  13. Apr 24, 2007 #12
    I just double checked and it still seems right to me. Can you suggest what you think is wrong?
     
  14. Apr 24, 2007 #13

    Dick

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    I get a -4*a*x and no 2*x term.
     
  15. Apr 25, 2007 #14
    yeah, you are right, the an a went missing in the -2*x term which would have made it -4xa by grouping with the other -2xa.

    so after all that I get...
    [​IMG]
    somebody hinted to use quadratic formula, so a= a^2, b= -4x, c =2 etc etc

    figure all that out to get...
    [​IMG]

    correct? (crosses fingers)
     
  16. Apr 25, 2007 #15

    Dick

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    You're getting there. Do you mean (sqrt(2))*a or sqrt(2*a). The former, I hope.
     
  17. Apr 25, 2007 #16
    Yeh, just extended the roof over a in my working...
    silly error :yuck:

    Now for part d). I'm really puzzled as to how you would prove something like this mathematically especially given the unknown a.
     
  18. Apr 25, 2007 #17

    Dick

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    You have tons of information about the function. Sketch a graph and label significant points in terms of a. Also examine the limiting behavior of the function as x->+/- infinity.
     
  19. Apr 25, 2007 #18
    I can't figure out the limiting behaviour as x-> - infinity.

    x^2 will become infinity, but at the same time e^-(ax) will become a very small number. In graphical terms, what does that mean?
     
  20. Apr 25, 2007 #19

    Dick

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    What kind of a number is -a*x if a>0 and x<0? I don't think e^(-ax) gets small at all. If you mean x->+infinity do you know L'Hopital's rule (judging from the problem you are working on you might not).
     
  21. Apr 25, 2007 #20
    oh, my bad! I don't know that rule, can you please explain?
     
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