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F(x) = (e^x)/x. at what value of x does f(x) attain its minimum

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data
    f(x) = (e^x)/x. at what value of x does f(x) attain its minimum


    2. Relevant equations



    3. The attempt at a solution
    f'(x) = (x)(e^x) / (e^x)
    then do i set both of those to 0 and solve?
    how o i do that?
    im losttt
     
  2. jcsd
  3. Mar 8, 2009 #2

    jgens

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    Gold Member

    Well, that is something you could do, but it would only work if your derivative is correct and I don't think that it is.
     
  4. Mar 8, 2009 #3
    That's not the derivative. To get the derivative you need to use the product (or quotient) rule:

    (fg)' = fg' + gf'

    This equals e^x/x - e^x/x^2 = (e^x/x)(1-1/x)

    You need to set this equal to 0 and solve for x to find the critical points of f. To find all local minima, find all points for which the 2nd derivative is positive. To find the minimum, you need to compare all the values of the local minima and take the smallest one. You also need to look at boundary points, which in our case would be +infinity and -infinity.

    This problem's solution is easy to see since there is only one critical point.
     
    Last edited: Mar 8, 2009
  5. Mar 8, 2009 #4
    at x=1 we would have the minimum
    because x=1 is the only critical point after we equate first derivative =0
    and second derivative is>0 at x=1 so minima occurs
     
  6. Mar 8, 2009 #5
    "at x=1 we would have the minimum
    because x=1 is the only critical point after we equate first derivative =0
    and second derivative is>0 at x=1 so minima occurs "

    I believe this answer is incorrect, if one answers the specific question the OP is asking. I can throw out a value of x where the f(x) < f(1).

    Your answer may be right, if one means to talk about local extrema.
     
    Last edited: Mar 8, 2009
  7. Mar 8, 2009 #6
    csprof is right. It's a stupid question, because there is no minimum (by definition - either the authors didn't understand this or they did this as a trick question). The infimum is 0, which occurs at x=-infinity. There is no minimum. The only local minimum is at x=1.
     
  8. Mar 8, 2009 #7
    What about x = -0.000000000001? exp(x) ~ 1 and 1/x = -1,000,000,000,000. Looks like f(-0.000000000001) < f(1) < lim f(x) as x -> -inf.
     
  9. Mar 8, 2009 #8

    HallsofIvy

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    Staff Emeritus
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    phreak said local minimum. x= -0.00000000001 is not vry "local" to 1!
     
  10. Mar 8, 2009 #9
    I wasn't taking issue with his statement regarding the local minimum... just pointing out that, as should be clear to everyone, the function goes to -infinity as x approaches zero from below.
     
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