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When is a function bounded using differentiation

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data
    how do i determine whether a function is bounded using differentiation

    eg: f(x)=x/(2^x)
    2. Relevant equations



    3. The attempt at a solution

    i know it has something to do with maximums and minimums but i cant figure out how to do it.

    any help would be appreciated. thank you
     
  2. jcsd
  3. Jan 16, 2008 #2

    Dick

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    You could look at the limits of the function as it approaches plus and minus infinity. If both exist and are finite, and if the function is defined and continuous for all x, then it is bounded.
     
  4. Jan 16, 2008 #3

    EnumaElish

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    You can use differentiation to investigate the behavior of f. Say, the function is f(x) = x/2^x on x > 0. Then f'(x) = 2^-x (1 - x Log[2]), which has roots 1/Log[2] and +infinity. At x = 1/Log[2], f''(x) = 2^-x Log[2] (-2 + x Log[2]) is < 0, so you have the maximum. Note that f(x) > 0 for x > 0 and f(0) = 0. As x --> +infinity, f(x) --> 0 from above; but f(0) = 0 so x = 0 is the minimum. Since you can "account for" both the maximum and the minimum, f is bounded on x > 0.
     
    Last edited: Jan 16, 2008
  5. Jan 16, 2008 #4
    thank you very much.
    what if we have:
    f(x)=(-2x^2)/(4x^2-1)
    i know that it's not bounded but i dont know why
     
  6. Jan 16, 2008 #5

    Tom Mattson

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    The graph of that function has two vertical asymptotes. Functions don't get much more "unbounded" than that!
     
  7. Jan 16, 2008 #6
    actually i think it is bounded because there's no value of x that would make that function greater than 1
    or is there?
     
  8. Jan 16, 2008 #7

    Tom Mattson

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    Sure there is. As I said, the graph of that function has 2 vertical asymptotes. You can find values of x for which the function blows up to infinity, and down to negative infinity.

    Do you know what I mean when I say "vertical asymptote"?
     
  9. Jan 16, 2008 #8
    yes i do know what vertical assymptotes are.
    umm but i still didnt understand what u meant. you can find values of x for which the function blows down to -ve infinity but not up.
    ?
     
  10. Jan 16, 2008 #9

    Tom Mattson

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    The function certainly does blow up to positive infinity, as you approach -1/2 from the right and as you approach +1/2 from the left.
     
  11. Jan 16, 2008 #10
    oh thank u very much
    that helps.
    just one last question:
    same question as before but with function:
    sqrt(x)/1000

    is it not bounded since n continues to increase to infinity?
     
  12. Jan 17, 2008 #11

    Tom Mattson

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    You're right that it's not bounded (on [itex][0,\infty)[/itex] that is--we really should be specifying an interval when making these statements).

    But what's "n"?
     
  13. Jan 17, 2008 #12

    HallsofIvy

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    Now, I'm confused as to what function you are talking about. The original function was f(x)= x/(2^x) which is definitely bounded on [itex][0, \infty)[/itex]. It is bounded "above" but not bounded "below" so is not bounded. I don't see any asymptotes when I graph it.
     
  14. Jan 17, 2008 #13

    Tom Mattson

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    Posts 1 through 3 pertain to f(x)=x/(2^x).
    Posts 4 through 9 pertain to f(x)=(-2x^2)/(4x^2-1).
    Posts 10 and 11 pertain to f(x)=sqrt(x)/1000.
     
  15. Jan 17, 2008 #14

    EnumaElish

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    Post 3 pertains to f(x)=x/(2^x) for x > 0.
     
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