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Homework Help: How to find the max,min,sup,inf of these cases

  1. Dec 20, 2008 #1
    i made a limit on both infinity and minus infinity for them

    and i tried to find but its not working

    http://img201.imageshack.us/img201/5458/23597303em5.gif [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 20, 2008 #2


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    Well, I haven't reviewed all of your work, but I can give you a hint for finding what value x, f(x) = 1/(1+(lnx)^2) has a max at. Assuming you restrict the value of ln(x) to real values, (ln(x))^2 is always greater than zero, therefore your denominator must be greater than or equal to one. If this is true, what must the denominator equal in order to maximize f(x) and what value must x be for this to be true.
  4. Dec 20, 2008 #3


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    Alright, since no one else is replying, I'll give hints for the rest of them.

    1) I already gave you an idea of how to find the max and sup of the function. Now, assuming ln(x) is restricted to real values only, we consider the domain 0 < x < infinity.
    Evaluate what happens as x tends to infinity and you find the function asymptotically approaches 0. What about when x approaches zero? What does that suggest about the min. and inf?

    2) Based on what you've showed with limits, it should be fairly simple to deduce something about the max. min. sup and inf.

    3) As sin(x) is less than or equal to one for all values x, the greatest value the function could possibly assume is 1 (for the values it is defined); however, the smallest positive value x for which sin(x) is one is pi/2. Let's now consider the degenerate case when x = 0. Finding the limit of the function at that point yields 1. What does that suggest about the sup of the function?
  5. Dec 21, 2008 #4
    regarding 3:

    when the highest value is 1
    is it max or sup
  6. Dec 21, 2008 #5


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    Do you understand the difference between "max" and "sup"? If a set of numbers has a maximum, then max= sup.
  7. Dec 21, 2008 #6
    1) max=sup=1
    inf=0, min doesn't exist
    How did I find it?

    Use the fact that Vf=Df-1

    Vf is the set of values, that y can have.

    Df-1 is the values, that x, from the inverse function of f(x) can have.

    2) Vf = [-∞, -2] U [2, +∞)

    3)sup=1, min=inf=0, there isn't maximum.

  8. Dec 21, 2008 #7


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    Actually, regarding 3. The minimum and infimum of the function are definately not zero as sin(x)/x can be negative.
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